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February 20

gas central heating

how much is a new natural gas central heating system cost for a 1700 square foot home --Tomjohnson357 (talk) 01:36, 20 February 2011 (UTC)

That depends wildly on how the house is constructed and isolated and which climate the heating system will be dimensioned for, not to mention the labor costs in whichever economy you're located in. Probably easiest simply to ask some local heating contractors for estimates. –Henning Makholm (talk) 01:47, 20 February 2011 (UTC)

2 story house, cold climate, i just need a approximate cost --Tomjohnson357 (talk) 01:53, 20 February 2011 (UTC)

What kind of heat? Hot water? Steam? Air? If it's hot water then it's around $2000 to $4000 for the boiler. Hot air is cheaper, probably around $1000 to $2000. I'm assuming a high efficiency model (about 93% efficient), if you go with the 80% models (if legal in your area) prices are much lower. That's for parts, installation varies a lot but is probably in the $1000 to $3000 range. Ariel. (talk) 02:55, 20 February 2011 (UTC)

it is natural gas --Tomjohnson357 (talk) 03:42, 20 February 2011 (UTC)

I know; you said that. But how is the heat transmitted through the house? Hot water (i.e. radiators) or Air (i.e. air ducts)? Or something else? Steam? Hydronic AKA underfloor? Do you have lots of zones? A few? Just one? Anyway, regardless of the answers to my questions, I hope my estimate was useful. If you really want, then take a high quality picture of your current heater and I'll see if I can figure it out. Take one "overview" shot, and then closer shots of any parts or pipes that look interesting. Ariel. (talk) 09:39, 20 February 2011 (UTC)

I had to pay $4300 for mine in November for a 95% model. Googlemeister (talk) 19:30, 22 February 2011 (UTC)

how old was yours

Lab rats

Would it be reasonable to say that lab rats are bred to be prone to getting cancer? I read once that they were so sensitive to getting cancer, that changing the amount of food they eat can statistically significantly influence their rate of cancer. 75.138.198.62 (talk) 01:51, 20 February 2011 (UTC)

Actually, most lab rats used in modeling cancer are actually synthetically given cancer genes. If you're a doctor studying cancer and using rats to model it, you give your rats cancer, specifically the cancer you want to study. Otherwise, generic lab rats are no less likely to get cancer than you'd expect. --Shaggorama (talk) 03:00, 20 February 2011 (UTC)

(ec)No, that would not be reasonable to say. However there probably is a breed of rat that is very prone to cancer, but it's specific to that breed, not lab rats in general. Plus it has nothing to do with the amount of food they eat - they just get cancer randomly. Ariel. (talk) 03:01, 20 February 2011 (UTC)

I think you're confusing the Oncomouse with lab rats in general. --Mr.98 (talk) 03:25, 20 February 2011 (UTC)

No, the OP is actually correct, lab rats do develop tumors pretty frequently without any special provocation. I don't know of any evidence that they are more prone to it than wild-types, though, but I haven't ever looked into it. Looie496 (talk) 07:49, 20 February 2011 (UTC)

If they are, it's probably due to being highly inbred. Most lab mice & rats aren't truly "wild-type", but are inbred strains, which are chosen specifically so that experiments are affected as little as possible by genetic variation. The principle of heterosis would say that such strains are, in general, more susceptible to most diseases/conditions/ailments than wild-types. So I'd say that, aside from specific strains like oncomouse, it's not "bred to be prone to getting cancer", as the selective pressure was for genetically sameness, not cancer susceptibility. -- 174.21.250.120 (talk) 18:00, 20 February 2011 (UTC)

And note that changing the amount of food people get can also affect their chances of getting certain cancers: [1]. So, it's no surprise that this would happen in mice, too.

Also, if you are headed toward the conclusion that "lab rats get cancer from everything, so you can't draw any conclusions if they get cancer from substance X", then that's plain wrong, since they only look at the increase in cancer cases from an exposure to a given substance, versus the number of cases in the control group. StuRat (talk) 20:59, 20 February 2011 (UTC)

Measuring orbital and rotation periods

If there was nothing in the Universe except the Earth and Sun (in particular, no fixed stars to refer to), and the Earth's orbit around the Sun was perfectly circular, then how would we measure the length of one Earth orbit? If there was nothing in the Universe except the Earth, how would we measure the length of one Earth rotation, and how would we locate the Poles (ignoring the fact that it would be too cold and dark for life to exist)? 86.179.112.215 (talk) 03:04, 20 February 2011 (UTC)

People have argued that it might be impossible (Mach's principle), but I'll ignore that; you can imagine you're simply inside a dust cloud that blocks your view of the stars. In that case, you can measure the (sidereal) day length and find the poles with a Foucault pendulum (though it would be difficult to get the necessary precision), and you can calculate the year length from the difference in length between the sidereal and solar day. -- BenRG (talk) 04:16, 20 February 2011 (UTC)

Linear motion is relative - you can only detect it by comparing to something else, but circular motion is not. It's absolute, and you can measure it without reference to anything else. (Although you may run into measurement problems if the motion is very slow, it's just a matter of the necessary accuracy, not a problem in principle.) Ariel. (talk) 04:38, 20 February 2011 (UTC)

If this is a technologically advanced society, then they could also launch their own spaceships, and use those as a reference. Or, if they can measure the distance to the Sun, the length of a year could be determined by that, I believe, as each orbit has a distinct orbital speed. (However, without any other sources of info, they may not know about this relationship.) StuRat (talk) 20:51, 20 February 2011 (UTC)

Donating blood, good for health?

Can donating blood be good for your health? The article about that suggest that it can reduce the amount of iron in your blood, which can be good in the case of people with too much of it. It also suggest that it can improve the heart conditions on men. But, why is the latter plausible? And, wouldn't the reduction of iron in all men be beneficial? 212.169.189.114 (talk) 04:09, 20 February 2011 (UTC)

Haven't heard about the iron idea, but I reckon that a free mini-medical checkup every three months is good for me. HiLo48 (talk) 06:03, 20 February 2011 (UTC)

I remember reading some years ago a study that suggested that long time blood donors replenished their red blood cells faster after a blood loss. Don't ask me to find it, though.Sjö (talk) 08:46, 20 February 2011 (UTC)

Iron overload#treatment - the treatment is bloodletting. I don't know if the blood would be suitable for transfusion, though. I thnk I've heard something similar to Sjö regarding blood donors recovering blood cells faster, but I've been unable to Google anything up, so I'd be skeptical. Vimescarrot (talk) 11:49, 20 February 2011 (UTC)

Bloodletting is, from a medical perspective, equivalent to donating blood. However, it has come into disuse as a treatment along the centuries. 81.47.150.216 (talk) 13:00, 20 February 2011 (UTC)

...except for in cases like iron overload, as explained in the link that Vimescarrot gave in the very comment you are replying to. I've certainly read of cases where people regularly donated blood, and only later found that they suffered from iron overload, implying that the donating of blood had suppressed their symptoms like any other bloodletting. There was no suggestion that the blood had been rejected. 86.161.110.118 (talk) 18:20, 20 February 2011 (UTC)

Oh, I should also note that the article on bloodletting you link helpfully includes a few other diseases it is still considered useful for treating (although sometimes only in the absence of other treatment), including 'the fluid overload of heart failure' (?) and possibly high blood pressure. 86.161.110.118 (talk) 18:34, 20 February 2011 (UTC)

There's been some speculation that it might help prevent heart attacks in men. If I remember right, the hypothesis is that iron overload is the reason for the difference in heart attacks between men and women since women lose iron on a monthly basis, but no one's been convinced yet and honestly the whole gender difference there may be overblown (q.v. the red dress campaign here in the US and A). Charity in general tends to make people feel good about themselves, which is probably beneficial for health, but so far the only people that obviously benefit from blood donation are people with specific health issues, most frequently hemochromatosis and polycythemia. SDY (talk) 18:10, 20 February 2011 (UTC)

Does donating blood make you hungry ? (If you lose some portion of your blood sugar, it might.) If this results in you eating a steak, then you might replace all the lost iron, and then some.

Also, I would expect that donating blood could help with any excess nutrient in the blood, where that excess is harmful, and similarly could hurt where a deficiency is present. If both excesses and deficiencies exist, then determining whether it's a net benefit could be tricky. StuRat (talk) 20:45, 20 February 2011 (UTC)

Perhaps this is the case only because the evolutionary design for men assumes that men would be doing hard physical work every day like running for one hour. Count Iblis (talk) 15:05, 21 February 2011 (UTC)

What? If you think the traditionally female roles are not hard physical work, either in a hunter-gatherer society or an agrarian society or a herder society or a modern society, I strongly suggest you try them for a week. Make sure you are actually doing all the work properly, not just your idea of the work. In many hunter-gather societies, the men's only real work is hunting, and they don't even nearly do it every day. Have you ever ground grain to flour using a saddle quern? But then, women in hunter-gather societies aren't having periods every month. 86.161.110.118 (talk) 11:32, 22 February 2011 (UTC)

I don't think it was true to say that men did harder work, but the nature of the work did vary. Men primarily had to do sprinting-type activities (requiring brief surges of energy), like hunting, while women had to do more marathon-type activities, like carry babies as they migrated thousands of miles. So, as a result, men tended to have more muscle, which is good for short term power, and women had more fat, which is good for long term energy. I suspect that additional red blood cells are more important to sprinting-type activities. StuRat (talk) 21:05, 22 February 2011 (UTC)

The more cell divisions there are, the more mutations there are. Does that apply here? 66.108.223.179 (talk) 20:33, 22 February 2011 (UTC)

Radiation

What dosage of radiation (only radiation, not any explosion that might come with it) would be required to kill the stereotypically-used-as-an-example healthy adult male immediately or within several seconds? What might be the source of this (I'm interested in a small, very radioactive substance such as that involved in the Demon Core accident, rather than a bomb)? I do not plan to try this at home and I suggest any readers to similarly refrain.~ 72.128.95.0 (talk) 04:15, 20 February 2011 (UTC)

page 2 of this NASA document for kids says that 8,000 rems (80sv) will be fatal within an hour and 10,000 rems (100sv) will cause instant death. Radiation_poisoning#Exposure_levels has some figures in Sieverts. Nanonic (talk) 05:56, 20 February 2011 (UTC)

See also Polonium#Toxicity for other figures on what is regarded as the most radioactive element. It states that a tiny tiny amount ingested or inhaled would be fatal. Nanonic (talk) 06:15, 20 February 2011 (UTC)

Polonium-210, which I imagine is the one you're talking about, is pretty damn radioactive, but very far from "the most radioactive". Grosso modo, the shorter the half-life, the more radioactive the isotope (although there are other variables such as the decay type and energy; for example tritium is not nearly as dangerous as its short 12-year half-life might suggest). For Po-210, the half-life is 138 days. But for, say, francium, the longest-lived isotope has a half-life of 22 minutes. You could kill someone with much less francium, assuming you could somehow accumulate it and get it into the victim's system before it all decayed. --Trovatore (talk) 09:22, 20 February 2011 (UTC)

Indeed. Polonium has a reputation for dangerousness because it's just lasting enough to actually be useful (e.g. as a poison, or, more commonly, as a component to neutron initiators). --Mr.98 (talk) 14:01, 20 February 2011 (UTC)

Calculating Jerk in a BUS...

Hi I have observed that while travelling in a bus, when bus passes over a speed breaker passenger sitting in front side of bus feel less jerk than those sitting on rear side of bus. although I don't know the exact reason behind this but I assume that it is due to psychology of driver, as when front tyres passes over the speed breaker the driver slow down the vehicle and just after that he speedsup without caring about rear passenger........... If there is any other reason please tell me.. Now what I want to know that ... How can I calculate in such circumstenses that In bus where will be least jerk in the bus. —Preceding unsigned comment added by 220.225.96.217 (talk) 05:17, 20 February 2011 (UTC)

Yes, I've noticed the same effect in both buses and cars, and it happens even at constant speed. I think it is a feature of the rear suspension, perhaps combined with a coincidence of the rear wheels receiving an upwards impulse just as the rear of the bus is already rising with the rebound of the front-wheel impulse. The effect varies with speed, and the mathematics is not simple because " jerk" is the third derivative of distance with respect to time. I'm fairly certain that somewhere close to the center of the bus will be optimal for "least jerk", but I'll leave this for someone else to prove. Dbfirs 08:18, 20 February 2011 (UTC)

Setting aside speed and suspension, which obviously affect the bounciness of everything, the rider's position relative to both tires matters. A normal (at least that I see, similar to File:NYC Transit New Flyer 840.jpg:) bus has one set of tires very near the front (often ahead of the whole passenger area) then the seats extend back to a few rows behind the rear tires. When the front tire goes up and down on a bump, it's like moving the whole bus as a lever pivoted at the back tire: the further forward or backward of the rear tire you are, the greater your vertical motion. Nobody is very far behind that pivot and the driver is the furthest forward (so maybe he's minimizing his bounce?). When the rear tire goes up and down up a bump, it's like moving he whole bus a a lever pivoted at the front tire: the further forward or backward of the rear tire you are, the greater your vertical motion. The driver is right near the pivot, so mostly just tilts forward/backward rather than actually bouncing up/down. Moving towards the rear, the vertical motion gets greater. Those behind the rear tire move even more up/down than the tire itself (being beyond it on the lever), which is a type of motion that nobody else experiences. DMacks (talk) 10:32, 20 February 2011 (UTC)

Jerk is the rate of change in acceleration. If F = ma is considered, then jerk is also the rate of change in force (with respect to time) per unit of mass. Use spring mechanics and lever rules to develop an approach. Plasmic Physics (talk) 11:57, 20 February 2011 (UTC)

Jerk as a technical term is the rate of change of acceleration. The connection to what the OP calls the jerk that a passenger "feels" is less than clear. --Trovatore (talk) 20:15, 20 February 2011 (UTC)

Yes, the body is most sensitive to rate of change of acceleration, but the eyes notice the maximum displacement and the velocity of the oscillation. Dbfirs 22:04, 20 February 2011 (UTC)

I'm guessing (this is speculation) that the body is most sensitive to things like maximum displacement of internal organs, or the energy dissipated within the body as the organs bounce around. There is not going to be any simple relationship between those things and the rate of change of acceleration. --Trovatore (talk) 22:10, 20 February 2011 (UTC)

Yes, these effects can be felt, especially if some organs resonate, but the body is most sensitive to the third derivative of displacement, and cannot feel either displacement or velocity. Acceleration is felt, but is not immediately noticed unless its rate of change is large. "Bouncing around" produces a large rate of change of acceleration and this is the jerk that is most noticeable. Dbfirs 23:00, 20 February 2011 (UTC)

My point stands, though. The human body has no "jerkmeter" per se (in the sense of an instrument that measures the third time derivative of position), and jerk in the sense of the third derivative has no special physical significance, unlike the second derivative. The third derivative is being used as a proxy for other things; I don't know what specifically, but I made my best guess as to two of them. The natural-language word jerk should not be confused with the third time derivative; the latter is a technical sense and does not correspond directly to any felt quantity. --Trovatore (talk) 10:00, 21 February 2011 (UTC)

I strongly disagree. Rate of change of acceleration corresponds to rate of change of force, and this is exactly what the body is most sensitive to. To see this, try travelling in a high-speed lift (elevator in your country). The high quality elevators in tall buildings have smooth acceleration that is barely noticeable, whereas the cheap lifts in smaller buildings produce a very distinct "jerk" on starting and stopping because they change very abruptly from zero acceleration to a large acceleration and back to zero. The mathematical term jerk (also called jolt, surge or lurch) for rate of change of acceleration was chosen precisely because it corresponds to the perceived jerk that the body is sensitive to. An alternative experiment would be to apply a force very gradually to any part of the body. You will hardly notice the force if it is applied very gradually (assuming that it is not large enough to cause damage), but if the same force is applied suddenly (with a large third derivative) then you will notice it much more. Dbfirs 14:01, 21 February 2011 (UTC)

If you think the body is directly sensitive to rate of change of force, then I think you're just wrong. The reason you have more reaction to force that increases quickly is that you have different equilibrium positions for internal organs depending on how fast you're accelerating. If the acceleration comes on suddenly, then they have to move to different positions, and will therefore bounce around. It's that bouncing that you feel, not the rate of change per se. --Trovatore (talk) 17:49, 21 February 2011 (UTC)

I think that you are wrong, but I'll clarify. The body has sensors which directly detect pressure, but both body and brain are designed to detect changes in stimulus much more readily than a constant stimulus. Both the chemical signals and the brain's interpretation tend to gradually reduce in the case of a constant stimulus, but are renewed whenever the stimulus changes. Thus the body (and brain if you consider it separate from the body) is most sensitive to rate of change of force. This is equivalent to the third derivative of displacement. I am particularly attached to this idea (and thus biased) because, nearly fifty years ago, I asked a teacher what the third derivative represented (given that first derivative is velocity and second derivative is acceleration), and was disappointed to get just a strange look and no reply. You might also be interested to know that fairground rides have safety limits on second derivative (g forces) and also on third derivative (rate of change of g-force) because the latter can be more dangerous. Dbfirs 23:30, 22 February 2011 (UTC)

Another application similar to fare-ride design is the design of curves on railroads and highways. Instead of going straight then suddenly curving along a circular arc (higher centripetal acceleration than going straight) and then going straight again, the road gradually eases into the curve and then gradually straightens out again: even if you wind up turning the steering wheel a lot, you try to do it slowly). The gradual change in curvature is analogous to derivative of centripetal acceleration to stay in lane. And heck, this is wikipedia, so Track transition curve. DMacks (talk) 02:30, 23 February 2011 (UTC)

In addition to the comments above, another factor is the weight distribution on the bus, based mainly on the position of the engine. If the engine is in front, then the front will bounce slowly and the rear will move more quickly. With a rear engine design, this should be reversed. However, as people are often seated further behind the rear wheels than in front of the front wheels, the rear passengers might still have more of a bounce than those in front, even in this case. StuRat (talk) 20:32, 20 February 2011 (UTC)

Let me expand on my previous comment, based on what I said earlier, a "jerkometer" could be constructed using a spring with a mass attached, and a high speed video recorder, is what is needed. Record the spring as it jerks. Using spring mechanics, calculate the maximum force experienced ar the peak of displacement in the spring. Divide this force by the known mass attached to the spring, and divide the answer by the time ellasped between equilibrium and maximum displacement. Plasmic Physics (talk) 10:53, 21 February 2011 (UTC)

While we're on the topic, what is it that kills a person in a fall from great height? The high deceleration or the high (dejerk?)? Plasmic Physics (talk) 10:57, 21 February 2011 (UTC)

It's the high deceleration that kills. Jerk is irrelevant here since it is the force that kills, not its rate of change. I like the jerkometer, though it measures average jerk not instantaneous jerk, but with suitable spring constants it would be an excellent instrument for use at the back of a bus. Dbfirs 14:10, 21 February 2011 (UTC)

If you measure displacement (either maximum or time-average), don't you still only have a forceometer, not a jerkometer? As Dbfirs says, neither rapid acceleration nor large total acceleration over a long time are easily felt as a sudden change of acceleration. Mass on a spring is a pretty good approximation of a person though (organs and stuff suspended in all the goo and stringy stuff). If we don't easily feel constant forces, we wouldn't care about a constant string stretch either. Fighter pilots care about maximum force ("g"s) for safety, but that could still be a smooth acceleration in a tight turn rather than the jerk of an aircraft carrier take-off. The jerk would be the rate of motion of the mass on the spring (per the third-derivative)...if it's bouncing up and down, so's your lunch, if it's stretched stably, you more quickly become accustomed to it. DMacks (talk) 17:37, 21 February 2011 (UTC)

No, change in force divided by change in time divided by mass equals jerk; it is infact a jerkometer. Plasmic Physics (talk) 08:58, 22 February 2011 (UTC)

"maximum force experienced ar the peak of displacement in the spring [...] divide the answer by the time ellasped between equilibrium and maximum displacement". Shouldn't the starting point for each jerk be the previous mass position, not the unaccelerated ("just gravity") equilibrium natural spring length? For example, if going to use a rocket to propel myself, I feel a jerk when the rocket ignites, which would . I do that when I'm already falling off a building (accelerating down at 9.8m/s²), is the jerk when the rocket adds to my acceleration the same as if I were in a rocket-sled on a flat track? But when falling, the mass-on-spring displacement is greater (two downward forces add). Or another way, in a rising elevator that smoothly moves faster up and then rapidly stops, isn't that a jerk at the top but seen as a decrease in spring stretch compared to it being stretched from the the vertical acceleration (and unrelated to its maximum stretch)? Rather, the jerk would be seen as the mass on the perhaps-stably extended string move up again...jerk as an absolute change or force during that particular episode, not maximum in any one direction. DMacks (talk) 17:17, 22 February 2011 (UTC)

[removed some notes to self while composing preceding comment, redundant to what I wrote] DMacks (talk) 02:22, 23 February 2011 (UTC)

New suggestion: use the same jerkometer, but process the data differently - plot the acceleration (calculated from F = ma) against time. Using software, estimate the tangents at each of the plot point. Take the modulus of the steepest tangent as your peak jerk. Plasmic Physics (talk) 23:51, 22 February 2011 (UTC)

I like it! DMacks (talk) 02:30, 23 February 2011 (UTC)

Upper bound of the size of the universe

Tom Murphy says: "All we are prepared to say for now is that over 13.7 billion light year scales, the universe looks pretty flat: it doesn’t deviate by more than 2% from being flat. But, the possibility exists that the universe is still curved on much larger scales. It’s just like the fact that the earth looks flat locally, over small scales, but is curved on the whole. The universe could be closed into a sphere, but on a much larger scale than what we can see. A 2% limit translates to a factor of 50 (it takes 50 2%’s to make 100%), so we could say that if the universe is finite, it must be at least 50 times bigger than our 13.7 billion light year horizon."

If we use the same conditions (flat universe to a certainty of 2% and assuming the universe is finite), what is the upper bound on the volume of the universe? Leptictidium (mt) 09:05, 20 February 2011 (UTC)

Even allowing your assumption that it's finite, why should there be an upper bound? It would be rather odd if we could say "either the universe has radius at most 10^70 gigaparsecs, or else it's infinite". Not being an expert I can't rule out that cosmologists can make such a statement, but it seems very unlikely to me. --Trovatore (talk) 09:11, 20 February 2011 (UTC)

I was assuming that a maximum uncertainty of 2% also sets a maximum uncertainty for the size of the universe. Leptictidium (mt) 09:25, 20 February 2011 (UTC)

It doesn't deviate by more, but it could by less. So it's a minimum bound, not an upper - and the upper should include infinity. BTW I'm not convinced your math of: 2% deviation = 50 times for a circle is correct. I guess it depends on what he means by deviation. Ariel. (talk) 09:31, 20 February 2011 (UTC)

Lepictidium, are you aware that according to the simplest GR models, a universe that's flat or negatively curved is infinite, whereas one with positive curvature is finite? That means that "the universe looks pretty flat" means "the curvature is right on the borderline between predicting a finite universe and predicting an infinite one". As far as I know the question is still within experimental error.

I believe there are more sophisticated possibilities that don't observe this strict dichotomy. Certainly it's not a geometric necessity; for example the flat torus has zero curvature but finite size. Whether there are any GR solutions that look like a flat torus, or whether they are consistent with observations, I have no idea. --Trovatore (talk) 09:46, 20 February 2011 (UTC)

Murphy's argument is rather dubious. He converts a measurement of the geometry (curvature) to a limit on the topology (the size) of the Universe which is not possible. It is true that a positively curved Universe is necessarily closed (this refers to the curvature of 3D space) but the topology is, as far as I know, not unique, meaning that it does not necessarily have to be a (hyper-)sphere, as Murphy assumes. Flat and negatively curved spaces can be infinite but do not have to be (the flat torus is indeed a possibility). There are lower limits on the size of a finite Universe but they do not come from measurements of the curvature. One could also nitpick on the use of lookback time as the measure of the size of the horizon but we'll let that pass. --Wrongfilter (talk) 10:05, 20 February 2011 (UTC)

I don't think it is true that a positively curved universe is necessarily closed. I don't see why you couldn't take a Euclidean 3D space and assign a constant metric with positive curvature to every point. There is in general only a very weak connection between local geometry and global topology of a space. Looie496 (talk) 17:42, 20 February 2011 (UTC)

This is one of the strongest connections, see e.g. Page 12 of this article. The reason is that the hypersphere S³, the universal covering space of any positively curved space, is compact. Incidentally, "Euclidean" is "flat". --Wrongfilter (talk) 17:56, 20 February 2011 (UTC)

There are infinite spaces that are positively curved everywhere, such as a paraboloid, but there are no infinite spaces with constant positive curvature. The data can't rule out a universe with that shape (they can't rule out much of anything, unless it's so small that a significant fraction of it fits inside the visible universe). -- BenRG (talk) 19:57, 20 February 2011 (UTC)

"constant" is of course important. That is implied if the cosmological principle is assumed to hold. --Wrongfilter (talk) 20:02, 20 February 2011 (UTC)

Our article on this is Shape of the Universe. See also Doughnut theory of the universe. Red Act (talk) 15:39, 20 February 2011 (UTC)

With the doughnut theory, is space simply cut-and-pasted into a topological doughnut (like Asteroids, where when you go off one edge of the screen you arrive on the other), or is it physically doughnut-shaped, with some parts of space being longer to loop around than others when parallel lines are measured? Is spacetime is looped different ways in different directions, i.e. positive curvature around one "piece" of the doughnut, but negative around the hole? Does this imply that there's some experiment you can do on a tabletop that will give different results depending on which way the apparatus (or the Earth and the table holding it) is oriented relative to distant stars? Wnt (talk) 19:47, 22 February 2011 (UTC)

Decay constant and half-life

We did an experiment in which there were originally $N_{0}$ fair coins. Every minute, all the coins were tossed, and those that landed heads were removed. Then the half-life of this "decay" is clearly one minute, and so $\lambda ={\frac {ln2}{t_{1/2}}}$, roughly 0.7 per minute. However, if $\lambda$ is the probability that a nucleus decay in unit time, why is this not equal to 0.5? jftsang 12:07, 20 February 2011 (UTC)

One reaaaally hand-wavy way of explaining why there's the ln(2) term in there (i.e., the decay rate constant is greater than the half-life) is that a real exponential-decay isn't a process of "sit unchanged for the time of one half-life, then suddenly and intantly half decays", but rather half decays over the time increment. So some of the material that is observed to be gone after one half-life would actually have decayed before that full half-life time was reached. As a result, the rate of decay for all the decaying particles together is faster than just the ones in the "half that decays in the half-life" that manage to last to the end of the half-life before actually decaying. Our exponential decay article has a lot of gory math details proving the formula, if you prefer a "because the formula says so, but why is the formula correct and used at all?" type of answer. DMacks (talk) 14:15, 20 February 2011 (UTC)

Imagine that instead of tossing $N_{0}$ coins and than waiting a minute for the next round of tosses you were to toss a coin every (minute/$N_{0}$) for the first minute. Obviously that would lead to the same result - that is $N_{0}$ tosses after one minute. That would be a slightly more realistic representation of an actual decay process where half of the atoms don't simply decay all at once after one halflife. But then you would have to abruptly drop the rate of tossing to one toss every (minute/$(N_{0}/2)$). That abrupt change of pace shows that there was something wrong with what you were doing. You should have started with a slightly faster decay pace at the beginning (Because there were more coins to begin with and the pace of decay is proportional to the number of coins)

and than slowly easied that pace so there would be no discontinuity at the pace after the first minute. That means that the original decay rate must have been higher than the naive 0.5 per minute. It turns out to be about 0.7 per minute. Dauto (talk) 15:52, 20 February 2011 (UTC)

I hate to say this, but the previous answers will confuse you. From the half-life article, two equivalent formulas are

$N(t)=N_{0}\left({\frac {1}{2}}\right)^{t/t_{1/2}}$

$N(t)=N_{0}e^{-\lambda t}\,$

Note that the first formula uses only 1/2 as a number value and is very easy to work with, but you've used the second formula, which has certain appeal to mathematicians. Now why does your lambda value come to 0.7? Because ln 2 = 0.693147181 ! The 0.7 occurs in your formula solely to cancel out the "e" (the base of natural logarithms) which has been added into it, and convert it back to 1/2. Wnt (talk) 18:06, 21 February 2011 (UTC)

I thought my answer was pretty good. Dauto (talk) 19:26, 21 February 2011 (UTC)

Quark Gluon Plasma Experiments - trillions of Celsius degrees ?

I was reading article about Quark Gluon Plasma and I noted that this experiments indicates that the temperatures in this phenomenons reachs around 4 trillion degrees celsius. My question is how can our devices in acelerators support this extremly high temepratures without burn out everything ? How can we keep this heat / energy isolated inside acelerators ? — Preceding unsigned comment added by Futurengineer (talk • contribs) 12:29, 20 February 2011 (UTC)

That temperature is achieved in a collision between two ions. It occupies a microscopic volume and last for a very small amount of time, cooling down as it expands. Dauto (talk) 15:27, 20 February 2011 (UTC)

I do not think that is a big problem, Quark Gluon Plasma is created when two heavy atomic nucleus collide, the volume of the plasma should be of the same order of magnitude as the volume of a nucleus 10 fm= 10*10^-15 m =0.00 000 000 001 mm. This plasma will expand and cool down very quickly, since all particles have relativistic speeds I would assume that it only exist for something like 10^-21 s = 1 zs. This happens in the centre of a cooled vacuum pipe a few centimetres across. By the time the particles in the plasma (or the produced particles) reach the pipe wall they will be very sparse. If we scale up this so the plasma is the size of a atomic bomb (r= 0.1 m approx.) then the container would have a size of the same order of magnitude as the earth's orbit around the sun. Of course Quark Gluon Plasma is much much hotter than an atomic bomb and there are several million collisions per second. I am no expert on this so other editors can maybe give a more detailed answer. Gr8xoz (talk) 15:44, 20 February 2011 (UTC)

An addition to what was said above: at a certain point in physics, you have to separate your "idea" of temperature from the definition of temperature, just as with every other "idea" that gets weirder as physics gets larger/smaller/hotter/colder. Temperature is defined, at the level of first principles (the basic axioms of classical physics), as the amount of heat energy needed to increase entropy (disorderly messy behavior of particles) by a given amount. So at these trillion-degree temperatures, relatively-enormous amounts of energy are needed to change the arrangement of the system just a tiny bit, but the particles in the accelerator don't necessarily "feel" hot (also because they don't really hold their own heat in like objects that we see around us do). For educational fun, if you want to reverse the definition above such that adding heat lowers entropy, you get negative temperature. SamuelRiv (talk) 19:23, 21 February 2011 (UTC)

As others said, you have to distinguish heat and temperature. Here's an experiment. A candle can easily burn you and you will perceive it as very "hot". But if you tried to warm up a swimming pool or a bath by holding a candle under it, your neighbors would laugh - it's obviously futile. Only a tiny amount of matter is "hot" in the candle's case. The heat needed to warm up a bath or swimming pool is far more than that which a candle provides even though a candle is "hotter".

Take it a step further. Imagine you could create a temperature of a billion degrees, but just for one subatomic particle. (The higher the temperature the harder it is for structures to hold together, so by the time we have a billion degrees molecules and even atoms cannot hold together which is why we get a plasma of even smaller particles and why extreme temperature is useful in studying tiny subatomic particles). Would that be enough heat to warm up your swimming pool?

It would obviously destroy any physical structure it touched, but it is so small, that the heat it contains would be easily exhausted in far less than a microsecond of contact with non-heated matter (or due to cooling by radiated heat or other mechanisms), that you'd never really notice it on a macro scale. The few million atoms it first touched would be affected but the pool as a whole would not. In research such as this, the containment system is designed to prevent heat transfer, and as others have said, the basic classical methods that must be present in the first place are removal of other particles (hard vacuum) and preventing contact with the surrounding structure (typically via laser or magnetic containment).

So the answer to the question is more like "how can we protect the few high temperature subatomic particles we produce from being almost instantly quenched" rather than "how do we protect the surrounding equipment from the high temperature particles". The answer is that if you prevent heat transfer then however hot the particles in the plasma get, the heat stays in the plasma. When it's no longer contained it is almost instantly quenched. Because the total heat is modest (though the temperature is very high) the energy can be disposed of when the beam has done its job.

That said the beam energy is not that small. You might find the article on beam dumps interesting. The problem is that the energy from repeated "bunches" of particles is disposed of in a tiny amount of time - in the case of the Large Hadron Collider that's less than 0.0000009 seconds (90 μs).

(Disclaimer, not a physicist, those who are please feel free to elaborate or correct any of this) FT2 ^{(Talk | email)} 15:58, 22 February 2011 (UTC)

cell wall in plant cells

I would like to know what is the role of cell walls in lifespan of a plant cell? Does thicker cell wall mean longer life or it is not so? And can plant cells live for 100s of years? If yes then how do they stay alive for such long time and is there any role of cell walls there?? — Preceding unsigned comment added by Ptamhane (talk • contribs) 16:04, 20 February 2011 (UTC)

We know that plants can live thousands of years so it stands to logic that plant cells can live thousands of years (Unless you meant live that long without going through cell division in which case I don't know the answer). I don't know whether cell walls play a role at all. Dauto (talk) 17:21, 20 February 2011 (UTC)

That is not necessarily true. Many trees shed their leaves every year. The living part of the bark may expand outwards every year, with the inner part dying. So a tree could live a thousand years, but that need not imply that any cell within it lives for that time. In the same way, a Chinese dynasty may exist over a thousand years, but none of its individuals lives a thousand years. 92.28.245.90 (talk) 17:51, 20 February 2011 (UTC)

The disagreement here are over the definition of the age of a cell. When a cell divides in to two do you get two new cells or two cells that each are as old as the original cell. If I split a stone in halves the halves are as old as the original stone but if we calculate the age of cells in the same way then we can argue that all cells are 3.7 billion years old and that does not seems to be a useful definition. (It is unclear how such a definition applies to sexual reproduction.) The problem are that after the cell division there are no distinction between child and parent. Gr8xoz (talk) 19:10, 20 February 2011 (UTC)

If we say that all cells are the age of their parent, plus that cell's parent, etc., then aren't all cells the same age (of approximately 3-4 billion years), dating from the first cell ? Perhaps you had in mind an exception for the reproductive cells in sexual reproduction, but some plants reproduce asexually, so this doesn't seem like a sensible way to measure the age of cells, in such a case. StuRat (talk) 20:20, 20 February 2011 (UTC)

As I said "if we calculate the age of cells in the same way then we can argue that all cells are 3.7 billion years old and that does not seems to be a useful definition.". I just wanted to explain why the previous answers were contradicting. I do think it is more useful to measure from the latest cell division just as you seem to do. I do not how ever feel that this must be obvious for everybody so I think the age (lifespan) of a cell need to be defined before the question can be answered. Gr8xoz (talk) 21:10, 20 February 2011 (UTC)

The length of time a cell is viable is programmed into it. Programmed cell death in plant tissue. Look at very old trees and you'll see that they are hollow with just the outer layer still living. The wood itself is dead. --Aspro (talk) 17:36, 20 February 2011 (UTC)

Exactly. Only the layer between the bark and the wood is alive. And some of the cells in that layer may have been there for a long time. The question is how long? Also, How long do they live between cell division cycles? I'm not sure what exactly the OP has in mind. Dauto (talk) 18:04, 20 February 2011 (UTC)

The OP is simply asking if its the morphological characteristic (e.g. thickness of cell walls) or something else. --Aspro (talk) 18:22, 20 February 2011 (UTC)

I'm not sure about any direct relationships between cell wall thickness and longevity, but the trees with the longest living leaves tend to have thicker cell walls. This is because they live in dry (including cold) or nutrient-poor habitats where they are only able to photosynthesise very slowly, so to end up with a positive carbon gain over the lifetime of the leaf, they need to be kept for a long time. Thick cell walls help with this because they are harder for microbes to penetrate, deter herbivores as they are relatively inedible and in cold areas withstand ice blasting and allow water to be stored in the apoplast where it can freeze without damaging the inside of the cell. These papers discuss the inherent trade-offs that plants make when 'deciding' how to construct a leaf and sort of shows what I mean. I can't find any references, but I'm fairly sure such cells will be the plant cells with the longest life spans - most of a tree trunk is dead - as has been said, and the same applies to the large parts of the roots. I do have a paper reference showing that some leaves live for up to 12 years so this is probably the upper limit, unless you want to include embryos in seeds. SmartSE (talk) 23:46, 20 February 2011 (UTC)

Gholson, Mississippi

What is the population of Gholson, Mississippi, and what is the population? --Perseus8235 17:02, 20 February 2011 (UTC)

The United States Census Bureau does not recognize "Gholson" as a census designated place (nor as an incorporated town, city, or other municipality) in Mississippi. You can verify this at the official Census Factfinder website. There is a Gholson, Texas, nowhere near Mississippi. If an unofficial place informally called "Gholson" is located in Mississippi, no authoritative statistics exist for its population. Nimur (talk) 17:33, 20 February 2011 (UTC)

Before posting a request for help here, Perseus began an article about this community. I've found some sources and added them to the article, but a population source is not among them. As an unincorporated community, it has no official boundaries: as a result, it can't possibly have a specific population. Nyttend (talk) 21:19, 20 February 2011 (UTC)

Why Doesn't the recoil from Handguns hurt?

Although I have never been shot by a bullet, I've read anecdotes on the web that being shot with a handgun bullet while wearing a bulletproof best is similar to being struck by a sledgehammer or a baseball bat. From Newton's Third Law would this not mean that the recoil of the gun would feel equal or more than being hit by a sledgehammer/bat? Having fired handguns, I do not feel like I'm being hit by a sledgehammer each time I pull the trigger. Acceptable (talk) 02:50, 20 February 2011 (UTC)

This was asked on the misc desk, some of the regulars here might have some insight. CS Miller (talk) 17:40, 20 February 2011 (UTC)

Because the bullet is accelerated over the entire gun length, but decelerated over a much shorter distance (and hence in a much shorter time). 213.49.110.216 (talk) 18:04, 20 February 2011 (UTC)

And that's why powerful guns need longer barrels. Dauto (talk) 18:07, 20 February 2011 (UTC)

I do not think that is correct, I think the bullet leaves the barrel before any significant momentum is transferred to the body so the velocity that the gun hits you with are dependent on the generated momentum and the mass only. See [2]Gr8xoz (talk) 18:56, 20 February 2011 (UTC)

I believe longer barrels are for increased accuracy, since that ensures that it will leave the barrel at a more precise heading towards the target. Some rather high-powered guns, such as mortars, have rather short barrels, but aren't as accurate. StuRat (talk) 20:35, 20 February 2011 (UTC)

Newton's Third Law states that if two objects interacts directly then they will be affected by forces of opposite direction but the same magnitude. This apply to direct interactions with no time delay such as the force between the bulletproof vest and the bullet, it does not apply to indirect interactions such as between the gun and the bulletproof vest.

Newton's Second law is more interesting, it states that the same momentum but of opposite direction is needed to accelerate and to stop the bullet. (We will ignore air resistance and recoil from combustion gases.) Momentum can be calculated as mass multiplied by velocity or force multiplied by time. Since the bullet plus the moving part of the bulletproof vest has less mass than the gun it follows that the bulletproof vest will hit the body with a higher velocity than the gun. If we assume that the forces on the bodies in both cases is the same (not necessary to be true but hard to estimate and it illustrates the point.) then it will take the same time to stop the gun and to stop the bullet and the moving part of the vest. Since the vest hit the body at higher velocity it will move longer in that time so it will do more damage to the body. Gr8xoz (talk) 18:44, 20 February 2011 (UTC)

In addition to the force of the bullet strike being more compressed in area and time, I can think of two other factors:

1) The bullet may strike in an area less able to withstand high forces than the hand.

2) The victim is more likely to be surprised by the shot, and the lack of psychological preparation may make it seem worse than it would otherwise. StuRat (talk) 20:08, 20 February 2011 (UTC)

So you think that the human hand is more able to withstand a bullet strike than other parts of the body, and a gunshot will do less damage if one is psychologically prepared for it? I would not recommend putting either of those postulates to the test. SpinningSpark 21:01, 20 February 2011 (UTC)

I said it would "seem" worse, not "be" worse. Do you not know the difference ? As for the hand being more able to withstand forces than some other parts of the body, then that certainly is the case for some body parts, like the eyes. And I'm not talking about a direct bullet strike, but rather one felt through a Kevlar vest. StuRat (talk) 22:28, 20 February 2011 (UTC)

The forwards momentum of the bullet hitting the body is the same as the backward momentum of the gun (assuming no loss in speed from air resistance), so the impulse on the body is the same. What the questioner has not taken into account is that this impulse is really quite small, initially moving the body at less than an inch per second. The effect is noticeable only if it occurs as an impact over a small area, as with a bullet hitting a kevlar vest or a gun held lightly an inch or two from your face. Normally, a gun is held firmly so that there is no impact with a part of the body, and the recoil momentum is hardly noticeable as it is transferred to the body, though a shotgun recoil can be painful on the shoulder if not held firmly. If the bullet is caught in a thick pad that absorbs the energy over a sufficient distance and area, then the effect on the body will be similar, though perhaps more noticeable if unexpected. The film cliche of victims being thrown backwards by a bullet is entirely faked. It would take a hit from a missile launcher to produce the momentum effect depicted. Dbfirs 21:41, 20 February 2011 (UTC)

I thought of another reason:

3) The hand, being on the end of the arm, is more free to "recoil" than most of the body. Thus, the force is more gradually absorbed by the hand, wrist, elbow, and shoulder, instead of all being absorbed by the hand immediately. I suspect that if the hand was held firmly in a fixture while the gun was fired, it would hurt more and do more damage. As for shotguns held against the shoulder, the body likely still recoils somewhat, although not as much as the entire arm. StuRat (talk) 22:34, 20 February 2011 (UTC)

I've never shot a handgun, but I always thought that best advice was to hold it firmly, thus transferring momentum to the body rather than allowing the gun to recoil freely and hitting a part of the body. You would notice the recoil if you shot with an outstretched arm with the direction of firing at perpendicular to your arm, but you will normally only feel pain if you allow the recoiling gun to hit you or you allow your recoiling hand to hit a hard fixed object. Dbfirs 22:53, 20 February 2011 (UTC)

I have shot several handguns, including 9mm, 0.45 and a 50 Cal revolver which would better be described as a hand cannon then a hand gun:). My take is, what sturat mentions just above, that your arm actually makes for a very good shock absorber: when you fire, your arm, not having a lot of mass, recoils with the gun over a certain distance. With the 50 cal in particular, you are strongly advised to take your head out of the way before you fire because your hands are almost guaranteed to travel backwards past the point of your head/face, regardless of how firmly you are gripping. When you get hit by a bullet however, your body, having a lot more mass, doesn't travel back very far at all, so the distance over which the energy is transferred is much shorter. Vespine (talk) 23:19, 20 February 2011 (UTC)

I would look at it in terms of "kinetic energy transfer" (though the "Force/area" issue and other points mentioned above are also relevant, I agree). So wouldn't the following reasoning be valid?

Because the momentum of the handgun must equal the momentum of the bullet, the energy transferred by the handgun to the hand is much less than the energy transferred by the bullet to the target.

Consider the M1911 pistol which weighs about 1100 grams and fires .45 caliber slugs which weigh say about 11 grams:

$m_{slug}=11grams$

$m_{gun}=1100grams=100m_{slug}$

and since

$m_{gun}\mathbf {v} _{gun}=m_{slug}\mathbf {v} _{slug}$

we also have

$\mathbf {v} _{gun}={\frac {m_{slug}}{m_{gun}}}\mathbf {v} _{slug}={\frac {1}{100}}\mathbf {v} _{slug}$

The kinetic energy of the slug is

$E_{slug}={\tfrac {1}{2}}m_{slug}v_{slug}^{2}$

and that of the gun is then

$E_{gun}={\tfrac {1}{2}}m_{gun}v_{gun}^{2}={\tfrac {1}{2}}100m_{slug}({\frac {1}{100}}\mathbf {v} _{slug})^{2}={\frac {1}{100}}({\tfrac {1}{2}}m_{slug}v_{slug}^{2})={\frac {1}{100}}E_{slug}$

So, in this example, the kinetic energy of the handgun is only one-hundredth that of the slug. The kinetic energy absorbed by your hand is therefore far less than the kinetic energy absorbed by, say, a human body-part struck by the slug, which is why there is less tissue damage to your hand than there is to your target. WikiDao ☯ 00:35, 21 February 2011 (UTC)

I concur with WikiDao completely. The transfer of energy to the shooter yields the perceived recoil and the shooter must present adequate force (F=ma) to counter the effects...a product of body mass and resistive force (muscular). But who says they don't hurt?

• Nice slow-motion video of the transfer of momentum energy to the human body.

• Greater ballistic energy transferred to smaller masses yields increased recoil...or another way of saying it, the bigger the load, and the smaller the body, the bigger the kick.
  ⋙–Berean–Hunter—► ((⊕)) 01:49, 21 February 2011 (UTC)

Since the question was about the impact of the bulletproof vest on the body and not about the impact of the bullet the mass of the moving part of the vest should be added to the mass of the bullet in the calculation. So if $m_{slug}+m_{vest}=55grams$ then the vest would impact the body with 20 times the energy of the guns recoil. Gr8xoz (talk) 04:04, 21 February 2011 (UTC)

(... but much less than the energy of the bullet.) The vest is designed to absorb and spread energy, so probably much less than you calculate because energy is lost in the collision of the bullet with the vest. Dbfirs 07:53, 21 February 2011 (UTC)

My calculation is correct, I assume a fully inelastic collision between the vest and the bullet, this means that the vest absorbs as much energy as possible. After a inelastic collision with a stationary object the bullet and the object will have the same momentum as the bullet had before and this can be used to calculate the velocity and kinetic energy. The problem is that the west does not move as a solid object hit in the centre of gravity so "equivalent" mass of the part that move has to be calculated. To do this in detail you need a detailed simulation of the deformation of the west. I assumed 44 grams for a soft vest, a ceramic plate has of curse higher mass. I assume that the collision between the vest and the bullet is much faster than the collision between the vest and the body. Even if the vest is "designed to absorb and spread energy" it must follow the laws of physics.--Gr8xoz (talk) 12:22, 21 February 2011 (UTC)

Yes, I initially mis-read your reply, and I agree with your calculation if the bullet hits the vest and the vest then hits the body, but the interaction is, as you say, more complex, with the vest being, to some extent, molded to become part of the body. Energy transfer is diffused, but transfer of momentum is simple and small. Dbfirs 13:39, 21 February 2011 (UTC)

It seems like there is a hidden punch in every bullet. Because when one person punches another, both hand and ribs suffer the same force of impact; yet the hand usually suffers less. And of course, "punching" with the barrel of a pistol makes the inequality greater. Wnt (talk) 17:54, 21 February 2011 (UTC)

Something else that helps to visualize this is an ice pick. If you hit someone with the business end, it will do far more damage to them than to you, both because of the decreased area (and thus increased pressure) at that end, and because the force is all applied to them at once, versus over maybe a half second to you, if you gave it a full swing. StuRat (talk) 19:44, 21 February 2011 (UTC)

Basic Phsyics

If I am going to shoot a bullet from 5m directly at the ground. At the same time as I shoot the bullet, I will drop a bullet from 5m towards the ground. Assuming same weight bullets and the rifle won't slow down the bullet, will both bullets hit the ground at the same time? —Preceding unsigned comment added by 12.180.137.195 (talk) 19:13, 20 February 2011 (UTC)

Yes they should. Both the forward moving and horizontal moving bullet will arrive at the ground at the same time. *Ignoring air resistance* See http://media.pearsoncmg.com/aw/aw_0media_physics/hewittvideos/projectile_demo.html --Tyw7  (☎ Contact me! • Contributions)   Changing the world one edit at a time! 20:08, 20 February 2011 (UTC)

You've misread the question. It said the bullet is fired "directly at the ground", not parallel to it. StuRat (talk) 20:12, 20 February 2011 (UTC)

No, the bullet fired from the rifle will hit first, as it will leave the rifle going faster than the dropped bullet. The final velocity depends on both the acceleration due to gravity and the initial velocity. Note that I am assuming that "5m" means 5 meters. If you meant 5 miles, then both bullets would be going at terminal velocity by the time they hit (the dropped one having accelerated to that point and the fired one having decelerated). However, even in this case, the fired bullet would have moved faster initially, so would still hit first. StuRat (talk) 19:58, 20 February 2011 (UTC)

I think the OP has misunderstood a common thought experiment where the bullet is fired horizontally at the same moment as a other bullet is dropped. The air resistance can complicate things a little bit but essentially it is expected that the bullets hits the ground simultaneous as long as the ground can be approximated as flat. See [3] Gr8xoz (talk) 20:13, 20 February 2011 (UTC)

See the article External ballistics for more background.
⋙–Berean–Hunter—► ((⊕)) 01:53, 21 February 2011 (UTC)

If the bullet was dropped compared to when it was shot from a gun, they would arrive at the same time. However, if it was shot downwards, then the bullet that was shot downards would arrive first. --Tyw7  (☎ Contact me! • Contributions)   Changing the world one edit at a time! 05:46, 22 February 2011 (UTC)

Assuming you meant shot horizontally, then that is correct. However, if shot even slightly upward it would arrive later, and if even slightly downward it would arrive earlier. StuRat (talk) 08:02, 22 February 2011 (UTC)

Technically if the ground was absolutely smooth and the air absolutely still, and one bullet dropped and the other fired horizontally parallel to the ground, you'd still have to allow for the curvature of the earth. Microscopic effect though. FT2 ^{(Talk | email)} 14:39, 22 February 2011 (UTC)

See [4][5]. Air resistance has an effect, but the effect is surprisingly small. Wnt (talk) 19:37, 22 February 2011 (UTC)

Light waves/photons

Are light waves composed of photons moving in a wave light pattern? --Tyw7  (☎ Contact me! • Contributions)   Changing the world one edit at a time! 19:55, 20 February 2011 (UTC)

Unfortunately it is not that easy se Wave–particle duality Gr8xoz (talk) 20:03, 20 February 2011 (UTC)

(ec) No, "one photon" is just a certain amount of the energy in a light wave. It doesn't have a location in the wave. It's similar to asking whether your checking account is composed of one-cent/penny/yen coins. -- BenRG (talk) 20:10, 20 February 2011 (UTC)

That's not entirely true. For example, if you shoot one photon at a target, there will be exactly one collision. If you shoot half of two separate photons at it, there could be anywhere from zero to two collisions. The difference is more pronounced with fermions, which have a probability of zero of being in the same place, affecting the wave-form. — DanielLC 21:04, 20 February 2011 (UTC)

The easiest way to imagine wave-particle duality is that the wave is chopped into small sections, each with a fixed energy.
Actually, the wave is a probability wave, and the square of its amplitude is the probability. So, the concept of "photon" only comes in place only when you know where it is for sure, i.e., the probability at that point is 1. Otherwise, the photon is nowhere and everywhere at once, so its better to think of it as a wave. The same applied for matter. ManishEarth^{Talk • Stalk} 13:34, 21 February 2011 (UTC)

Faking cold fusion

On the JANUARY 15th FOCARDI AND ROSSI PRESS CONFERENCE an demonstration was done that supposedly shows cold fusion. The researchers them self present it on [6]. They have got some media attention such as [7], [8] and [9]. Many images can be found at [10]. They demonstrated 12 kW heat generation by boiling water in about one hour. This is so much energy that it seems to rule out a mistake. This leaves two possibilities:

1. A fraud.
2. An interesting nuclear process such as the fusion of nickel and hydrogen that the researchers suggests.

If this is true it will be truly revolutionary but I think experience show that this type of demonstration is probably some kind of fraud. It is also hard to explain theoretically how it can be possible to get cold fusion and why there are so low radiation levels.

Some "experts" have calculated that it would be hard or impossible to hide enough chemical energy in the device to fake it that way. (They calculated the amount of hydrogen and oxygen needed.) One suggestion on how to fake this that I have read in an article comment is that they did not feed it with water, instead they could have used hydrogen peroxide. Hydrogen peroxide decomposes in to oxygen and water vapour when it come in contact with a catalyst. Would the audience detect the difference between water and hydrogen peroxide by smell, viscosity, colour and so on??? Does anybody has any other suggestions in how they can have faked this? Gr8xoz (talk) 20:59, 20 February 2011 (UTC)

I saw mention of nickel powder - if they used aluminium or magnesium powder I doubt anyone could tell the difference and those would have provided plenty of energy to boil water. As a general point of view scientists are rather bad at detecting deliberate fraud. A professional magician would be much better at that. Ariel. (talk) 21:36, 20 February 2011 (UTC)

As I understand it it was claimed to be a rather small amount of nickel powder but since the inside of the device was not inspected nobody except the people that build the device knows if there are aluminium or magnesium powder in it and in what amount. Do you mean the heat generated when magnesium reacts with water? Would that really be enough energy? Would not the magnesium compounds be easily detected in the steam, a white powder? Gr8xoz (talk) 22:17, 20 February 2011 (UTC)

I was just saying that measuring the volume of hydrogen needed is pointless - there are plenty of other fuels you can hide that are much smaller. Also, what was in the compressed gas cylinder? It was labeled as nitrogen - but that doesn't mean that it's the truth. Plus did anyone check that the amount of steam really included ALL the water? Perhaps they drained some water and only made a small amount of steam? Ariel. (talk) 00:19, 21 February 2011 (UTC)

The device was connected to a hydrogen bottle during the experiment(suposedly fuel for the fusion proces H+Ni->Cu), it was weighted by supposedly independent persons before and after the test and lost about 1 g H2. The nitrogen was connected afterwards to shut-down or clean the device. I think the calculation assume that since the device have no air inlet or exhaust pipe it would need to contain any oxidizer used and that the exhaust would need to be undetectable in the steam. As I understand it the device is rather small and mounted so that a drainage pipe could not have been hidden. I have not checked all the details. Gr8xoz (talk) 03:37, 21 February 2011 (UTC)

I skimmed the 3 videos and I didn't see any of that. You should utterly ignore what they said, and only go by what you can actually verify. Ariel. (talk) 04:07, 21 February 2011 (UTC)

Thank for your answers. I can not verify anything, the videos could easily been faked. It could of curse be so that all journalists and "independent" observers there was in on it or that some was easily fouled and the rest was in on it. If we assume the observers were not in on it how cold it be faked? Given that respectable magazines like this [11] (Swedish) write about it I assume they could not just make up the names. So I think the report[12] from the observers should be given some credibility. (Not that I blindly trust it but I am much less suspicious on that than something Rossi him self has written.) --Gr8xoz (talk) 11:55, 21 February 2011 (UTC)

Maybe I missed it (I was skimming since I don't speak the language in the video), but weren't they not even in the room? It looks like they did the demo by video? Also, journalists do not attempt to detect fraud, they simply want to report what happened, and the more interesting the better. Please remember: I have NO idea if this is real or not, all I can tell you is that there were opportunities for fraud. But I can't tell if there actually was fraud. Ariel. (talk) 21:09, 21 February 2011 (UTC)

Since the observers report that they used there own power meters, thermometers, scale and radiation detectors I would assume that the demonstration was not just a video recording. Obviously some videos show a presentation and some show the actual demonstration. --Gr8xoz (talk) 22:42, 21 February 2011 (UTC)

I know that James Randy says that magicians are better than scientists in detecting fraud but I think that in cases like this you need both understandings of illusions and science to be qualified to detect any fraud. Gr8xoz (talk) 22:24, 20 February 2011 (UTC)

It's not so much that scientists can't detect fraud as that they don't evaluate it, preferring a different standard. Randi and the scientists would both agree that in order to start looking for fraud you'd have to be able to get there, look over the equipment, see where the hydrogen comes in from, test the amperage in the electric cables and so on. But scientists have the more exacting standard that someone else has to be able to simply read the report about the experiments, construct his own apparatus, and create cold fusion on his own. This clearly makes fraud difficult to perpetuate - though it is true, often "irreproducible results" are discounted by scientists without anyone ever knowing whether they were the result of fraud or simple contamination or error. Wnt (talk) 17:40, 21 February 2011 (UTC)

Since they don't want to release some details due to issues regarding the patent application it can not be independently reproduced. Supposedly independent researchers where there and did "see where the hydrogen comes in from, test the amperage in the electric cables and so on." They did so using their own instruments but they were not allowed to look inside the devise. There are three explanations: 1. The observers was in on it in this case the fraud is very simple and uninteresting. 2. They were fouled, I think it is interesting to think about how this could be done. 3. The device do actually work as advertised. --Gr8xoz (talk) 22:53, 21 February 2011 (UTC)

If they don't give independent labs the ability to reproduce the experiment, then it's very likely fraud. There are lots of ways they could enter into contracts with said labs that would render no threat to their patent applications, the same types of non-disclosure agreements that the FDA enters into with drug companies who have patent-pending medicines. They are either entirely non-confident in their method, or they are trying to pull a fast one (get a patent, use it to bark up funding, then get out of dodge before the number of non-confirmations becomes impossible to ignore). --Mr.98 (talk) 14:05, 22 February 2011 (UTC)

I can understand that someone with the Answer To The Energy Crisis could be paranoid about someone else taking their invention, but unfortunately, it's not science until it's reproduced. (I say someone could take their invention because, for example, there might be some component to their device that isn't optimized or has an error in it, and some company could jump in and patent the right way to do it, then hold the whole invention hostage, refusing to license it unless they get a large chunk of the profits - indeed, effectively mandating a high profit margin even if the inventors had been philanthropically minded toward a larger market at a lower price. Even if a NDA has been signed, no one can prove that the sudden rush of activity in improving the component was actually related to an illicit divulgence of the details) — Preceding unsigned comment added by Wnt (talk • contribs) 14:32, February 22, 2011

Collapsing pyramids?

In referring to ground stone used somewhat as mortar, Egyptian pyramid construction techniques says the following:

The filling has almost no binding properties, but it was necessary to stabilize the construction.

Why would filling be needed for the construction to be stable? How possibly could a pyramid collapse, since it's not top-heavy and it's made of solid limestone blocks with no holes larger than a burial chamber? Nyttend (talk) 21:23, 20 February 2011 (UTC)

If no filler was used between roughly cut stones, it would be impossible to get each to be level, and this would become more exaggerated with each new level. Thus, the size of gaps would increase near the top, and this would allow water infiltration, which, if it froze (does it get below freezing there ?) would tend to drive the stones loose, over time. There's also probably the "comfort factor", that the didn't want dripping water to ruin the pharaohs' afterlives. StuRat (talk) 22:19, 20 February 2011 (UTC)

The article states ...stones forming the core of the pyramids were roughly cut..., which means that any upper course of stone blocks would only have had a few points of loadbearing contact to the lower course. The resulting uneven stress (receive load from top / transfer load to bottom) in any given block would have caused it to split and crumble, resulting in the gradual collapse of the entire face of a pyramid (which you can clearly see on some of them). A filler (binding or non-binding) has the simple purpose of distributing this load equally.

Compare this with you sleeping on a hard surface. Your entire weight will be resting on a few "protruding" bits, the head, shoulder blades, pelvis, etc. If you were to sleep on a beach - on a mattress of sand - a bit of wriggling about would evenly distribute your weight to the delight of Morpheus. --Cookatoo.ergo.ZooM (talk) 23:06, 20 February 2011 (UTC)

"Collapsing" here doesn't mean that they unfold and pour out the edges; it is more like mine subsidence (wow, that's a red link?). Even a multi-ton block of stone can break if you lay it on an uneven surface with half a pyramid on top of it, and the same applies to the one on top of it and so forth. Wnt (talk) 17:22, 21 February 2011 (UTC)

We have Subsidence#Mining, I'm not sure how to create a redirect yet, if i learn in the next 30 minutes i might fix it. Vespine (talk) 23:01, 21 February 2011 (UTC)

Well, THAT was easy!:)Vespine (talk) 23:04, 21 February 2011 (UTC)

Actually, is that actually the OPPOSITE of what mine subsidence means? Vespine (talk) 23:04, 21 February 2011 (UTC)

General Relativity and Conservation of Energy

I've heard energy isn't conserved under General Relativity. For example, cosmic background radiation is being redshifted due to the expansion of space with the energy going nowhere. Does the stress–energy tensor act as pretty much the same thing, or would it be possible to take advantage of this? If you can take advantage of it, how hard would it be to build a perpetual motion machine? Would it require a solar system? A black hole? — DanielLC 21:35, 20 February 2011 (UTC)

Redshift is not due to a loss of energy, it is due to a change of reference frame. Imagine a car of mass m driving past you at speed v; you say, the car's kinetic energy is $1/2mv^{2}$. Now imagine you're sitting in the car. Now the kinetic energy is 0, because the speed v=0. Where did the energy go? Nowhere, you're just measuring it in a different reference frame. The fact that energy is not invariant to changes of reference frames is on the same level as time dilation and length contraction in special relativity. Conservation of energy always holds for local interactions like collisions or particle decay, if consistently described in the same reference frame. --Wrongfilter (talk) 22:01, 20 February 2011 (UTC)

I thought it's also partially due to the expansion of space. If a photon has a wavelength of 500nm, and space doubles in size, it will have a wavelength of 1000nm. — DanielLC 23:07, 20 February 2011 (UTC)

Change of reference frame in an expanding universe. Better? I should stress that I did not mean to imply that there is a one-to-one correspondence between the car example and cosmological redshift. I was just targetting the notion of energy not being conserved rather (that's what the example can do) than trying to explain what redshift is (it can't do that). --Wrongfilter (talk) 23:10, 20 February 2011 (UTC)

The problem with the expansion of space, is that it's the same everywhere. If you want to harness the energy, you have to have a difference in the expansion thus a harnessable "potential difference". Anyways, a perpetual motion machine built on this would be driven by an external force (cosmic expansion), and thus it's no longer a perpetual motion machine, rather a power cell rather like a solar cell. ManishEarth^{Talk • Stalk} 13:26, 21 February 2011 (UTC)

<bogus>I don't think redshifted light is losing energy. Remember, the light is redshifted because the universe expands. You may have light with 1/5 the frequency and 1/5 the energy as it had before, but meanwhile the universe has grown 5 times bigger and the same amount of light now covers a trail 5 times longer than before. I think...</bogus> there are some aspects that confuse me. (For example, if a single photon is emitted from a source moving away from you at great velocity, do you still receive a single photon? does the redshifted photon split somehow? is the existence of the single photon at the far end sort of blurred into a quantum haze for you, like a cat in the box?) Wnt (talk) 17:49, 21 February 2011 (UTC)

1st: Yes, the volume increases and the energy density decreases but those two effects don't cancel each other completely. Defining $a\,$ as the expansion factor, the volume scales as $V\sim a^{3}$ and the photonic energy density scales as $\rho _{E}\sim a^{-4}$, so the energy scales as $E=\rho _{E}V\sim a^{-1}$. 2nd: No, The photons don't split. What the OP is neglecting here is that since there is a change in volume and there is pressure, work is being performed and it is not surprising that there is a drop in energy. Dauto (talk) 18:12, 21 February 2011 (UTC)

Hmmm, so you're saying that the photons, indeed, lose energy according to the scaling factor, i.e. as their wavelength increases their energy decreases. And this energy somehow goes into expanding spacetime? Does this mean light's energy applies pressure and works like dark energy? I haven't seen this point often made, and it does hint at some more intimate relation between electromagnetic energy and spacetime than I realized... Wnt (talk) 02:40, 25 February 2011 (UTC)

Electrolysis of urine/Urine powered cars

I've read that performing electrolysis of urine produces far more hydrogen than water does.

http://www.nydailynews.com/lifestyle/health/2009/07/15/2009-07-15_urine_power_hydrogen_produced_from_urea_could_be_used_to_power_cars_houses.html http://www.wired.com/autopia/2009/07/pee-powered-cars/

These articles (among others) claim that one day we may be able to run our cars on hydrogen produced from the electrolysis of urine alone. Is this possible? I know that with water, you are putting more energy into electrolysis of water, than you are getting out of the hydrogen produced. But what about urine? ScienceApe (talk) 22:58, 20 February 2011 (UTC)

The electrolysis decompose urea (NH2)2CO to get hydrogen H2. Urine contains about 1% urea, a human would leave about 25 grams per day. I do not know the mileage but it would probably be worse than for gasoline so each person would be able to travel less than 250 meters per day. It can be a small contribution to the fuel supply but it will be very limited. Biogas from human feces would give a larger contribution. Gr8xoz (talk) 23:42, 20 February 2011 (UTC)

Let's make things clear. Hydrogen is a means to deliver energy, not a source of energy. Any method you chose to produce hydrogen is going to consume more energy than you get from the hydrogen at the end of the day. Dauto (talk) 02:11, 21 February 2011 (UTC)

Too bad. It would have been nice to start a long trip by buying a couple cases at BevMo rather than tanking up at Chevron. PhGustaf (talk) 02:41, 21 February 2011 (UTC)

It's not terribly uncommon for human waste to be turned into energy for use. The typical method is to let bacteria break poop down into biogas, and then using that for heating or cooking:[13], [14], [15]. Running a car on biogas is probably feasible, but I don't get the point of trying to break down urea for energy, a relatively small part of human waste, when we have all sorts of great fuel coming out our other hole. Buddy431 (talk) 03:03, 21 February 2011 (UTC)

In either case, I think the problem is that the total quantity of energy is low, relative to what cars need, so it wouldn't pay to set up an infrastructure to extract, store, and deliver such a fuel. It would make more sense to extract the energy at the waste treatment plant and use it directly there. StuRat (talk) 05:25, 21 February 2011 (UTC)

Of curse the processing will not be done in the car but biogas is already used in cars and buses on a rather large scale and hydrogen has often been proposed as a car fuel due to its high energy-content, it can also be used to produce liquid fuels. --Gr8xoz (talk) 12:01, 21 February 2011 (UTC)

LOL @ "Of curse".... yes it would be a curse to have your car storing manure while it decomposes into gas. I don't agree that bio-gas is used on a large scale now. There are a few experimental programs here and there, but you can't just drive up to your average gas station and expect to get bio-gas there. And if the source is as small as human (and even livestock) waste, it would never pay to set up such an infrastructure, at least in most places. A location that has lots of manure available, like a cattle feed lot, might want to use bio-gas for their tractors and trucks. StuRat (talk) 19:36, 21 February 2011 (UTC)

I do not know where you set the limit for "rather large scale", obviously it is not near the scale of gasoline but here in Sweden some cities run their city buses and garbage trucks on biogas. In each of the bigger cities there are a few gas stations with biogas that are open for private cars. I think the upgraded biogas and natural gas are feely mixed since they have nearly identical composition and the difference in buying the one or the other is only a difference in accounting. If you buy biogas someone have to put in the same amount of biogas in to the network, similarly to "green electricity". So the infrastructure are largely there already. There are more than 30000 gasdriven cars. I do think we use only a small fraction of all the usable biological waste to gas production so the potential are much bigger but it would not be nearly enough to provide all the fuel. See Tables of European biogas utilisation and Biogas --Gr8xoz (talk) 22:24, 21 February 2011 (UTC)

February 21

Conservation of (angular) momentum

I remember an argument Feynman used to show that the conservation of angular momentum implies the conservation of linear momentum. I've never seen this anywhere else, though, so is it possible that he was mistaken? The proof goes something like this:

Suppose we have a system of particles. Conservation of L doesn't depend on where the we choose the axis, so we can choose an axis that is very far from the all the particles. Then, all the particles will have the same position vector r relative to this axis, which doesn't vary with time. Then ${\frac {d}{dt}}\mathbf {L_{TOT}} ={\frac {d}{dt}}\sum \mathbf {L_{i}} ={\frac {d}{dt}}\sum \mathbf {r} \times \mathbf {p_{i}} =\mathbf {r} \times {\frac {d}{dt}}\sum \mathbf {p_{i}} =\mathbf {0}$ implies ${\frac {d}{dt}}\sum \mathbf {p_{i}} =\mathbf {0}$.

The only flaw with this argument that I can see is that r will be infinite, but I don't see how that invalidates the proof. Any insight? 74.15.137.130 (talk) 04:59, 21 February 2011 (UTC)

So basically he views the conservation of linear momentum as a special case of the conservation of angular momentum, where the radius is infinite. Sounds reasonable, to me. StuRat (talk) 05:20, 21 February 2011 (UTC)

Okay, after a little more thought, I realized that the reason that conservation of L w/ respect to one axis --> conservation of L w/ respect to all axes requires that the center of mass' velocity be constant, which is true iff momentum is conserved. So it's not all that surprising after all. 74.15.137.130 (talk) 06:42, 21 February 2011 (UTC)

If our choice of axes is one very far from all the particles, then r will be very large but it won't be infinite. It would be possible to do this with full mathematical rigor by taking the limit as r increases without bound. Some people would summarise this by saying the limit as r approaches infinity but increasing without bound is more rigorous than approaches infinity.

There is much in common between these two conservation laws. I assume Feynman was saying that if we have established the truth of one of these conservation laws then a consequence is that the other is also true. It doesn't matter whether we start by establishing the truth of the conservation of linear momentum, or angular momentum. Either way, it follows that the other is also true. Dolphin (t) 11:23, 21 February 2011 (UTC)

I don't think that angular momentum conservation follows from linear momentum conservation. For linear momentum conservation you just need "for every action there is an equal and opposite reaction", but for angular momentum conservation you also need to know that the action and reaction act along the same line. In terms of Noether's theorem, you can imagine a space that's translationally but not rotationally invariant (fill it with a constant vector or tensor field), but not the other way around. -- BenRG (talk) 11:37, 22 February 2011 (UTC)

Noether's theorem connects symmetries to conservation laws. I'm just parroting the article here, but a rotationally invariant/symmetric Lagrangian implies conservation of angular momentum, while symmetry under a continuous translation in space implies conservation of linear momentum. I guess that would be a good starting point if you want to derive this more rigorously. EverGreg (talk) 12:45, 21 February 2011 (UTC)

I think this is very easy: any translation is a composition of two rotations. -- BenRG (talk) 11:37, 22 February 2011 (UTC)

Infinite universe and probabilities

My friend told me a theory which he had heard from someone who had read it from somewhere. It's been bugging me for a week now: I believe it must be false, but I just can't tell him exactly why (except that the "from someone/where" part is pure snopes.com material). The theory goes something like this: if the universe is infinite, then there are infinite possibilities how matter is organized. Thus, the probability that in some place else in the universe the matter is organized exactly like here is 1. This would mean that there are at least two copies of me around, sitting in this universe typing this post, although the other guy might have some atoms missing.

My answer to him was that at least the amount of matter in the universe can't be infinite. But is there a more accurate / scientific answer available? --Albval (talk) 07:37, 21 February 2011 (UTC)

Yes, and there are several variations on the theory. See multiverse.--Shantavira|^{feed me} 08:00, 21 February 2011 (UTC)

This isn't actually about the multiverse - it applies equally well to a single universe operating under known laws of physics. Back to the question, while the amount of mass in the visible universe must be finite, there is no theoretical limit on the total mass of the entire Universe; it may well be infinite (see shape of the universe for speculation that touches on the potential size of the universe). Someguy1221 (talk) 08:34, 21 February 2011 (UTC)

The probability of something happening may be 0, (impossible) in which case it still does not happen even if there is an infinite chance of it. Graeme Bartlett (talk) 09:48, 21 February 2011 (UTC)

Graeme, I was just about to respond to Albval on this point; I thought you would know better. Probability zero does not mean impossible, and probability one does not mean certain. See our almost surely article. --Trovatore (talk) 09:50, 21 February 2011 (UTC)

Forgive me for being uncivil, but the "almost surely" article is patent nonsense. An event with probability exactly equal to 1 is identical to an event with probability $0.{\overline {9}}$ and is exactly guaranteed to occur. The "corner-cases" mentioned in the "almost surely" article deal exclusively with numerical models of statistical systems. I think some statisticians need a formal retraining in measure theory, basic mathematical limits, and the finite precision capabilities of numerical computers. The example case of the "dart" landing on an abstract concept (a "line") is merely a restatement of Zeno's paradox (something to the effect of never being able to get arbitrarily close to the abstraction of the imaginary line). See also, treatment of infinity in computers. This is a matter of axiomatic definition: an event with probability zero will not happen. Nimur (talk) 20:32, 21 February 2011 (UTC)

No, you're just wrong. Consider an infinite sequence of fair-coin flips. For any fixed ω-sequence of heads and tails, the probability that the actual sequence will match it is zero (so the probability that it will not match it is one). However, if every actual sequence were impossible, then it would be impossible for the sequence to come out any way at all, which is nonsense.

It is true that these concepts don't correspond to physical experiments that are easy to design, but they are absolutely fundamental to probability theory as it is modernly conceived. There are people who don't like them, but they have basically lost. You're the one who should learn measure theory. --Trovatore (talk) 22:58, 21 February 2011 (UTC)

I think the theory is correct. --Gr8xoz (talk) 12:27, 21 February 2011 (UTC)

The source: This exact theory, which relates to our universe under the assumption of inflation in an infinite universe, was put forward in 2001 by Alexander Vilenkin of Tufts University in Medford, Massachusetts and Jaume Garriga of the Independent University of Barcelona. Inflation guarantees that the different regions are isolated. The time elapsed since the big bang determine the size of each region (how far we can see in the universe).

As for "exactly like here", Vilenkin and Garriga argued that there are only a limited number of distinct universe configurations in the infinite universe. While say the distance between two atoms may be a continuous variable, 'quantum blurriness' means that two distances in two worlds can't be told apart if they'r too small. Vilenkin and Garriga estimated that there were only 10^10150 distinct histories in the universe.

Just because something is infinite, doesn't necessarily mean that every possible configuration has to happen in that infinity. A simple example is the decimal expansion of π which is known to be infinite, but it is not known whether it contains seven consecutive sevens (for example). See Trovatore's link above. Dbfirs 14:36, 21 February 2011 (UTC)

The decimal expansion of π is infinite, but it is not a random sequence of digits (because it can be encoded in a finite number of symbols, a series or 4 atan(1)). Whether that is true of the Universe is an interesting question but I guess it is metaphysical rather than physical. The probability of finding a copy of OP in the observable universe is 0 or virtually 0. --Wrongfilter (talk) 15:23, 21 February 2011 (UTC)

Er, Random#In mathematics says it's likely that π is random in some senses. I think you're taking an information-theory approach, not a math/statistics approach? But also, how would one write a finite expression to generate the numerical represenatation of pi? To get the digits, you have to start doing actual iterations of the expansion (and an infinite number of them to get the infinite-decimal-places string), not just say "in the theoretical limit, it is symbolically exactly equal to pi" I think the best you can say is that the whole value is something like non-arbitrary or that it's reproducible (I'm trying to pick words that don't have subtly different technical meanings than in lay-language), not that the string is a "non-random sequence of digits". DMacks (talk) 17:19, 21 February 2011 (UTC)

The digits of pi are absolutely not random. For example, the first one is always a 3. When people say "the digits are random", they (if you really pin them down) usually mean that pi is a normal number, which is only a conjecture. Staecker (talk) 01:10, 22 February 2011 (UTC)

It's "only a conjecture", but it would be astonishing if it were false. Mathematicians are extremely careful to separate what has actually been proved from what has not, which is a useful trait, but it sometimes causes outsiders to think that there is reasonable doubt about propositions when there really isn't. --Trovatore (talk) 01:59, 22 February 2011 (UTC)

In an generic infinite universe, not only would there be an infinite number of copies of you (actually of the entire visible part of the universe), but you also cannot identify yourself as any single copy. This is due to the fact that you have a limited amount of information about yourself. So, e.g., if you are not bald and assuming that you haven't counted exactly how many hairs you have on your head, there will be different versions of you, each with a different amount of hair and they all share the same personal identity. This despite the fact that the distance between the copies is astronomical, as Tegmark points out here. Count Iblis (talk) 16:35, 21 February 2011 (UTC)

One thought experiment that might help to understand this is "if there were an infinite number of monkeys randomly typing away on typewriters, they would eventually reproduce all the works of Shakespeare". You might think this is impossible too, but here would could actually calculate how many it would take. Let's say we just want to see the word "HI" (ignoring case) and that there are 100 keys on the typewriter (a typewriter with fewer extra keys would obviously give better odds, though). For two letters, it should take, on average, 100² or 10,000 monkeys (or one monkey with that many trials). For a 50 letter sentence, it would take 100⁵⁰ monkeys. If we assume Shakespeare's entire work is 10 million letters long (if anybody has a better figure than this guess, please let me know), that should take 100^10,000,000 monkeys. That's a very large number, but definitely less than infinity. StuRat (talk) 19:27, 21 February 2011 (UTC)

If anyone want to try this at home I really recommend implementing the standard RFC 2795. When running large experiments it is important to follow relevant open standards to avoid vendor lock in. I also think it would be a good idea to run it on RFC 2549 in in order to increase the biological diversity. Unfortunately RFC 2460 is way to limited to be used by it self. (Why do they never learn that you should not build standards with arbitrary limits) --Gr8xoz (talk) 21:35, 21 February 2011 (UTC)

I take it that site is just an elaborate hoax ? StuRat (talk) 22:35, 21 February 2011 (UTC)

*blink* Um, IETF? No, they are the standards-body for pretty much everything that lets your computer interact with wikipedia's servers. DMacks (talk) 22:41, 21 February 2011 (UTC)

Well, not the entire works but you need look no further then the Infinite monkeys article. Even if the observable universe were filled with monkeys typing from now until the heat death of the universe, their total probability to produce a single instance of Hamlet would still be less than one in 10^183,800.... but not zero! lol. Vespine (talk) 22:59, 21 February 2011 (UTC)

Ah, but that said the "observable universe", which is infinitely smaller than an infinite universe or infinite multi-verse. StuRat (talk) 07:57, 22 February 2011 (UTC)

You might want to take note of the publication dates for RFC 2795 and RFC 2549. Red Act (talk) 23:17, 21 February 2011 (UTC)

As far as I know, though, there's no RFC for IP over demographics. --Trovatore (talk) 23:27, 21 February 2011 (UTC)

No, they are actually the main standardization organisation for the internet protocols but they actually have humour, it is tradition to release a joke standard on the Aprils fools day. This is the case for RFC 2795 and RFC 2549. We even has an article about this tradition April Fools' Day RFC This started when the internet was very much smaller than to day, the very first joke standard was written when there was less than 300 computers on the Internet. RFC 2460 is a real serious standard for IPv6 the internet protocol that will replace IPv4. IPv4 has only 2^32=4.3*10^9 addresses and that will not be enough for all our devices in the future so IPv6 has improved this to 2^128=3.4*10^38. It is often joked about how huge this address space is and that it should be enough for all civilisations in the universe. --Gr8xoz (talk) 23:45, 21 February 2011 (UTC)

Thank you everyone for answers! Hopefully, somewhere in this universe there is a copy of me who understood them all:-) Albval (talk) 19:37, 24 February 2011 (UTC)

Diving: max. depth without minding

How deep can you dive without caring about your security? —Preceding unsigned comment added by 212.169.187.5 (talk) 13:44, 21 February 2011 (UTC)

Just to clarify, do you mean how deep should the water be in a swimming pool before you should dive into it, or how deep can you go when SCUBA diving? Googlemeister (talk) 14:19, 21 February 2011 (UTC)

Either way, the answer to the question depends on how stupid the diver is. Dauto (talk) 14:23, 21 February 2011 (UTC)

I meant SCUBA + free-diving. And no, the answer do not depend on how stupid the diver is. The question is how stupid the diver can be, without consequences. 212.169.177.33 (talk) 14:31, 21 February 2011 (UTC)

May be that's what he wanted to ask. But that's not what he asked. Dauto (talk) 14:51, 21 February 2011 (UTC)

Our article on Deep diving has some recommendations. Dbfirs 14:43, 21 February 2011 (UTC)

It's not just the depth, but how fast you reach this depth, how long you remain there and how fast you come back to the surface. The type of breathing gas you use is also relevant here. The answer for SCUBA is simply that you'll go through some certification program and keep the rules that you learned there. Quest09 (talk) 15:03, 21 February 2011 (UTC)

Free-diving is not limited by risk of bends (there is none) or breathing gas. Not sure what it is limited by, other than the ability to hold ones breath, but the no limits world record is 214 metres according to our article. SpinningSpark 18:09, 21 February 2011 (UTC)

As stated above, free-divers don't get bent. In SCUBA diving, this question is not answerable as asked, because you could get bent, theoretically, even diving to a 10 foot depth if you stay down long enough. Looking at the dive tables isn't actually helpful; the NAUI dive table doesn't start until 40 feet, but this is because people don't really do 20-foot dives, not because you're immune to the bends if you only do 20-foot dives. The question could be answerable if you included a time limit. Comet Tuttle (talk) 18:33, 21 February 2011 (UTC)

There have been cases as low as 6' deep getting the bends if the diver is there for hours and the dive was at a mountain lake. Googlemeister (talk) 22:15, 21 February 2011 (UTC)

The bends is not the only risk. Even in a short dive to 20 feet, it would be easy for a novice diver to accidentally inhale some seawater, panic, and shoot to the surface without remembering to breath out, which can do significant damage to the ears or lungs. There is really no type of scuba dive where the diver can be stupid without consequences. Looie496 (talk) 19:14, 21 February 2011 (UTC)

Agreed; I am assuming the original poster is asking about decompression sickness. Obviously a 1 inch dive could be fatal if you stupidly breathe in all the water possible. Comet Tuttle (talk) 19:43, 21 February 2011 (UTC)

"The bends" is one of many scuba-diving hazards. In addition to "the bends," decompression sickness covers more of the hazards related to re-gassification of dissolved nitrogen. Other risks for ordinary-air SCUBA diving include oxygen toxicity, and nitrogen narcosis - both of which can occur while still under water. Proper SCUBA training will teach you how to correctly account for each of these risks. As has been mentioned, the "maximum depth" is a complicated factor - it depends on how long you stay at each depth, and how long since your previous decompression or surface trip. Divers use a dive table or dive computer to assist them in calculating safe dive depths and dive profiles - because "safety" is not described by one single depth value. If you breathe a gas other than air, such as nitrox or trimix, or if you are a technical diver and breathe multi-gas out of various cylinders in sequence, other hazards can exist. If you are an underwater construction expert, you may dive a profile that is considered "unsafe" by recreational standards. If you are a Navy Seal, you may be ordered (or choose) to use a dive-profile that could be fatal (but Navy SEALs understand that risk of fatality is a part of their job). If you dive at altitude, such as in Lake Titicaca, you run additional interesting risks related to pressurization; divers typically undergo a specific training for "altitude diving" such as the NAUI Recreational Altitude Diver course. If you will be diving out of a submarine or hyperbaric chamber, your safe-dive profile is very different than a surface scuba dive. Nimur (talk) 20:08, 21 February 2011 (UTC)

In fact, we have an entire article: Maximum operating depth. Needless to say, "don't try this at home" - of course, technical dives require special training and equipment. Nimur (talk) 20:11, 21 February 2011 (UTC)

Has anyone tried to find a way to allow exchange of oxygen while free-diving, so that it would have the advantages of scuba? I mean, I think in theory you should be able to stick a pair of tubes down into someone's lungs, or even (maybe) provide oxygen via someone's dialysis shunt? A way to provide only the truly necessary amount of oxygen while leaving the lungs collapsed? Wnt (talk) 21:01, 21 February 2011 (UTC)

No, I'm not aware of any such technology. It would be incredibly hazardous to attempt to exchange gas at below the ambient water pressure. (Gas will not flow against the pressure gradient; you have to depressurize the lungs below ambient pressure to "suck" air in. That depressurization, which we normally call "inhaling", would be the same as "crushing" if ambient water pressure was very high; alternatively, your hose can "squirt" over-pressured air into the lungs, but the risk of hyperinflation is incredibly hazardous and defeats the point of a free-dive. For these reasons, a SCUBA regulator regulates the breathing gas pressure to ambient water-pressure. If for any reason you do not want the human to be exposed to ambient water pressure (such as when you dive to great depths), you use a pressure hull to isolate their atmosphere. A rigid diving suit can do this - but no such suit can withstand great depths. Most good pressure hulls look more like a submarine - sort of spherical or cylindrical, so there are no regions where force is concentrated (so the material won't crumple). Then, the "ambient" pressure inside the hull can be maintained far below ambient water-pressure. Submarine atmospheres are often kept higher than 1 atm, but still at a safer level than the water pressure. Take a look, for example, at bathyscaphe or bathysphere. I believe the crew of ALVIN breathes at atmospheric pressure - but they have two inches of solid titanium to keep the water out. Nimur (talk) 21:25, 21 February 2011 (UTC)

I don't know if anything like what I'm suggesting has been attempted, but I don't think it's that impossible. If you have a sack of oxygen in equilibrium with the external water pressure, then just a slight extra pressure should send it into the lungs; likewise the return tube should carry out air with just a slight pressure. Of course, tubes to the lungs (at any pressure) might have safety issues of their own, but I don't see why there has to be a hazardous amount of crushing or exploding involved. (The 'sack' might still have some minimum pressure it holds at the start, and use a regulator at such low pressures, to limit its buoyancy near the surface). Wnt (talk) 01:49, 22 February 2011 (UTC)

If you regulate the pressure, you have SCUBA! And if you don't regulate the pressure, it won't work. You might be underestimating the crushing-force that 10 meters of water can exert; you physically will not be able to inhale unpressurized air. It will be prohibitively difficult for you to "breathe in" air that is pressurized at 1 atm when your body is surrounded by 2 atm of pressure - your diaphragm musculature just isn't strong enough to work across that pressure gradient. (In other words - as you try to "breathe in" by expanding your thoracic cavity with your diaphragm muscle, the water will literally crush your lungs and your thoracic cavity back to the original shape). This is exactly the reason why you can't get away with a super-long snorkel: you need the air to be pressurized very close to ambient water-pressure so that the air pressure inside your lungs helps your diaphragm work, pushing back against the crushing-force of the water pressure. (Here's a "How Stuff Works" post explaining in more detail). Let me give you another example of exactly how much crushing you are subject to while diving: when you're down at even mild depth, say 50 feet, your weight belt (which was tightly fitted at the surface) will be very loose and often requires re-adjustment; and if you tighten it at the bottom, you'll find it uncomfortably tightens and squeezes your waistline pretty hard when you re-ascend. The weight-belt, usually made of nylon, barely changes size at all - but both your wetsuit and your body have been significantly and measurably "shrunk" by the massive compressive force of ambient water pressure. Nimur (talk) 03:26, 22 February 2011 (UTC)

I was talking about a flexible "sack" of air at the same level as your body, subject to the same force, under the same pressure. Wnt (talk) 19:21, 22 February 2011 (UTC)

If you could keep it from floating away, it might work; it'd be like SCUBA, but with an inconvenient air-tank. You would still have the exact same risks of SCUBA, though - you'd be breathing pressurized air, which means that if your downtime was long, nitrogen could supersaturate in your bloodstream, and you could get the bends; and you'd still have to worry about oxygen toxicity and nitrogen narcosis, just like regular compressed air dives. Nimur (talk) 21:34, 22 February 2011 (UTC)

Hmmm... if you don't add nitrogen, there should be no nitrogen narcosis. Having the air in the lungs be pure oxygen might be a problem, yet I'd think that with the lungs collapsed that less oxygen would be entering the bloodstream than with them expanded. And I wonder if the amount of nitrogen can be optimized more precisely. Wnt (talk) 02:48, 25 February 2011 (UTC)

I was snorkeling once and saw a shell, what must have only been 3 meters below me. I'm a pretty good swimmer but I've never had any diving training and never been deeper then the bottom of a pool. It took about 5 attempts to get down that deep and I regretted it when I got back up, my ears hurt quite badly for about an hour. It was a lot more taxing and dangerous then I suspected, so I don't recommend it unless you have an experienced diver give you some training. Vespine (talk) 22:03, 21 February 2011 (UTC)

On the ear thing, the Valsalva maneuver is what divers use as they descend in order to prevent their ears from feeling like someone is stabbing knives into them. I'm a little surprised to read that article talks about blowing the air with "moderate" pressure; I would never do it with anything stronger than "gentle" pressure. Comet Tuttle (talk) 18:41, 25 February 2011 (UTC)

Do not hyperventillate, even in shallow water, as it may lead to a shallow water blackout. ~AH1^(TCU) 21:26, 26 February 2011 (UTC)

DNA and RNA

Would the complimentary messenger RNA strand of a coding DNA strand, be an exact copy of the coding DNA strand, only with the execption of replacing T(thymine) with U(uracil)? Thanks.

for example

ATCGAATT dna coding strand

AUCGAAUU RNA messenger —Preceding unsigned comment added by 99.146.124.35 (talk) 14:58, 21 February 2011 (UTC)

Yes, as explained in Coding strand. Assuming no mutations occur, the RNA messenger will be the same except for the Uracil/Thymine. Chipmunkdavis (talk) 15:06, 21 February 2011 (UTC)

If an extra A is added after ATC in the DNA messenger strand, It would change many of the animo acids in the sequence. What exact processes would cause this mutation to happen? —Preceding unsigned comment added by 99.146.124.35 (talk) 15:26, 21 February 2011 (UTC)

Non sense mutation

Would

ATC-GAA-TTC-CGA-CCA-TGC... (non mutated dna)

ATC-AGA-ATT-CCG-ACC-ATG-C... (mutated added extra A)

Would it be consitered a nonsense mutation, Frameshift mutation or a missense mutation? —Preceding unsigned comment added by 99.146.124.35 (talk) 16:25, 21 February 2011 (UTC)

The example given is a frameshift mutation. You could get a frameshift in the RNA relative to the DNA by translational frameshift or RNA editing. Wnt (talk) 17:27, 21 February 2011 (UTC)

how much of an effect do daily multivitamins really have?

how much of an effect does taking a daily multivitamin supplement (like Centrum or whatever) every day really have on one's health? How would it compare to, say, cutting out one bad thing you do (go to mcdonald's, or not exercise enough). I mean, is it really a drastic effect, if you take those properly for the rest of your life, or more like accredited homeopathy. nobody really knows if it does anything, but why not? 109.128.213.73 (talk) 19:18, 21 February 2011 (UTC)

It's not correct to characterize homeopathy as "nobody really knows if it does anything". Homeopathy is known scientifically to be useless, and there are a lot of references in the homeopathy article to back that statement up. People buy that stuff because of cultural, psychological and marketing reasons, not because there's any scientific uncertainty about it being nonsense. Red Act (talk) 20:54, 21 February 2011 (UTC)

Forget something? Thank you for understanding my analogy. I've marked where you forgot to answer my question by applying the analogy to what I'm asking about. 109.128.213.73 (talk) 21:35, 21 February 2011 (UTC)

It depends drastically on the individual. There's no question that there are people with serious vitamin deficiencies, and also no question that many people have none at all. Wnt (talk) 19:25, 21 February 2011 (UTC)

Though the benefits of daily multi-vitamin pills can be (and are often) exaggerated, please do not confuse it with homeopathy. Unlike the homeopathy, the effects of vitamins, particularly vitamin_deficiency are well documented by the scientific community. SemanticMantis (talk) 19:40, 21 February 2011 (UTC)

Well, see our Multivitamin article; there are "evidence for" and "evidence against" sections. Comet Tuttle (talk) 19:42, 21 February 2011 (UTC)

My understanding is that the multi billion dollar "supplement" industry is mostly, quite literally "pissed" away (excuse the profanity). Unless you have a pregnant or have a deficiency, which most people in the developed world don't, there is not much evidence that taking supplements has any benefit. In fact, there is some evidence to suggest some supplements, specifically anti-oxidants, might actually have a negative effect. Vespine (talk) 22:04, 21 February 2011 (UTC)

A multi-vitamin is no substitute for a healthy diet. They only contain a portion of the healthy nutrients which we know about.

Also, some of the nutrients they contain may be slightly different formulations which may not be bioavailable, or may need to be taken in conjunction with certain foods to be absorbed. Furthermore, you can potentially overdose on some vitamins, and having them in pill form can make that easier to do.

Finally, if you use your multi-vitamin as a justification for continuing to eat junk food, keep in mind that the pills do nothing about the excess calories, sugars, animal fats, trans fats, bad cholesterol, and sodium you are getting.

So, eating a good diet will provide you with all the nutrients you need, in a form you can use, including things like fiber, which are difficult to give in pill form, due to their bulk. StuRat (talk) 22:23, 21 February 2011 (UTC)

Some dietary supplements are critical for optimum health. Count Iblis (talk) 00:18, 22 February 2011 (UTC)

I'm not sure how you reach that conclusion from the linked reference. Vespine (talk) 00:26, 22 February 2011 (UTC)

Last line of first paragraph:

As director of LPI, I am often asked what supplements I take—after all, thinking about and researching micronutrients every day, I should know what dietary supplements are most important. While I think eating a healthy diet, exercising regularly, maintaining a healthy body weight, and avoiding tobacco are of utmost importance to maintain good health, I also think that some dietary supplements are critical for optimum health.

Count Iblis (talk) 00:46, 22 February 2011 (UTC)

Sorry, are you kidding or? Just because someone thinks it, doesn't make it so. Linus Pauling is well known for his controversial views on vitamins and megadosing so you might want to read the claims coming from his eponymous institute with a low dose of sodium. Vespine (talk) 01:34, 22 February 2011 (UTC)

Linus Pauling is long dead, the LPI is named in his honor. The statement is made by the current director of the LPI: Balz Frei, Ph.D.

LPI Director and Endowed Chair Distinguished Professor of Biochemistry and Biophysics. Count Iblis (talk) 14:00, 22 February 2011 (UTC)

Taking regular exercise and avoiding McDonalds style junk food would have a much bigger effect than taking a multi-vitamin. I used to take multi-vitamins, but after reading scientific research about it I concluded that it was better to get your vitmains and minerals from a healthy diet including plenty of fruit and veg. Although I do take Vitamin D in the winter, as I live in the north. 92.15.8.100 (talk) 12:40, 22 February 2011 (UTC)

I also only take vitamin D supplements and I'm thinking of taking fish oil supplements. Count Iblis (talk) 14:03, 22 February 2011 (UTC)

I prefer eating sardines to taking fish-oil supplements. Being young fish, sardines have not accumulated any significant mercury as other fish do. They will contain many more nutrients than fish oil pills, including for example high levels of B12. I also eat tinned wild Alaskan salmon, which has plenty of fish oil in it. 92.28.241.15 (talk) 16:07, 22 February 2011 (UTC)

I also preferred these things ... but the damned gout says otherwise. Whether it's just the purines or the heavy metals also, fish oil offers an alternative for odd situations. Not that I ever remember to take it... Wnt (talk) 19:24, 22 February 2011 (UTC)

Cold

Cold slows down chemical processes, so why would that not apply to my own body? Even if it slows down only some of my cells some of the time, I'm sure that adds up to save my body some body life-cycle units (I'm not a biologist) in the long term. I don't know how it works, sorry, that is why I'm asking. I looked up stats and northern countries seem to have longer average life spans, and northern US states seem to have longer average lifespans. I realize there are many factors, such as availability and quality of health-care to be factored in, so those numbers alone are not good enough. —Preceding unsigned comment added by 88.160.92.233 (talk) 21:34, 21 February 2011 (UTC)

Your body is very good at regulating core temperature. In most of the places where it counts (i.e. internal organs, brain) your temperature does not vary much at all (just a few degrees) due to external temperature.Vespine (talk) 21:54, 21 February 2011 (UTC)

More like a few tenths of a degree. StuRat (talk) 22:08, 21 February 2011 (UTC)

If you could keep someone alive while their core body temperature is lowered, then you could theoretically extend life. (However, when you hit freezing, their cells all explode, and this certainly doesn't help extend life. ) This might actually be a reasonable medical treatment at some point in the future, if we can figure out how to make people hibernate, like many other mammals. This could keep someone alive, say, until a cadaver heart could be delivered from a pro-democracy demonstrator in China. :-) StuRat (talk) 22:08, 21 February 2011 (UTC)

Hypothermic people have been able to in some cases go without breathing for much, much longer then usual, our article suggests as long as 1 hour, but as the mortality from that particular situation runs 38-75%, I wouldn't want to try it. Googlemeister (talk) 22:12, 21 February 2011 (UTC)

Our Normal human body temperature article actually states In adult men and women the normal range for oral temperature is 33.2–38.2 °C so normal variation is about 5 degrees, but yes in a individual the normal range appears to be about 0.7 of a degree. I'm not sure if at all that correlates with the climate a person lives in. Actually, that reminded me that this "slowing down" is done during some surgeries, like Cardiopulmonary bypass surgery: hypothermia is maintained; body temperature is usually kept at 28ºC to 32ºC (82.4–89.6ºF). This gives the surgeons a lot longer to perform the surgery while lowering the risk of damage to other organs due to lower blood pressure or oxygen availability. But it's certainly not a state you would like to be in while conscious. Vespine (talk) 22:50, 21 February 2011 (UTC)

I can say from personal experience that this concept works well in nematodes (cold-blooded, if you count them as having blood at all); their life span extends significantly from lowering the ambient temperature from, say, 15^oC to 4^oC. I believe the same is true of flies (certainly, their life span takes longer), but I haven't bothered to deconvolute their longer lifespan from the increased availability of food (they eat less at low temperatures).

However, the important distinction humans have in this case is we're warm blooded. We've evolved as warm blooded for tens of millions of years, and as such, everything in our bodies is designed to operate at ~37^oC. Lower that, and things shut down. Maybe you could keep someone alive on life support at relatively low temperatures, but you're not going to eek out a meaningful existence at a persistent state of hypothermia. Nematodes and flies are used to massive fluctuations in their body temperature, so they can handle it. Someguy1221 (talk) 21:04, 22 February 2011 (UTC)

"Personal experience" ? So you're a nematode ? Welcome to Wikipedia, you may well be our first invertebrate member ! :-) StuRat (talk) 22:29, 22 February 2011 (UTC)

Something else to note is that even if we could get people to survive at lower body temperatures, and thus extend their life, and we were able to have them conscious, as well, it still wouldn't seem any longer, since their brain processes would be slowed down along with the rest. Everyone else might appear to go zipping by and speak too fast for them to understand. StuRat (talk) 22:18, 22 February 2011 (UTC)

This effect can be quite dramatic in terms of developmental biology - to quote some unpublished data, if you take buckeye butterfly pupae and inject them with certain drugs you can change the pattern that forms on their wings. If you put them at 4 C for a week and inject them just before or just afterward, you get the same patterns as if they had not developed more than a few hours in the whole time. How this works I have no idea, though I'm suspicious that c-myc is involved. Wnt (talk) 06:39, 23 February 2011 (UTC)

See also North-South divide and multiple sclerosis. Life expectancy is dependant on many demographic factors and not likely directly on average temperature and climate. ~AH1^(TCU) 21:18, 26 February 2011 (UTC)

Plants

Are there any plants that photosynthesis in moonlight, or entirely in moonlight and never sunlight? — Preceding unsigned comment added by K4t84g (talk • contribs) 21:37, 21 February 2011 (UTC)

I doubt it. First, the visible light from the Moon is orders of magnitude less. But, more importantly Also, there is virtually no proportionately less UV in moonlight, which is what some plants tend to may also use for photosynthesis. StuRat (talk) 22:03, 21 February 2011 (UTC)

Plants don't tend to use UV for photosynthesis, see for example this graph and Simarouba amara#Physiology where there is some data on the wavelengths that chlorophyll and leaves absorb - most of it is the visible spectrum. CAM plants (like cacti) are as close as you'll get to photosynthesis in the dark, but they only temporary fix CO₂ to use in photosynthesis the next day. SmartSE (talk) 00:48, 22 February 2011 (UTC)

While there is an absorption tail on many of the photosynthetic pigments that extends into the ultraviolet, absorption at wavelengths shorter than 400 nm accounts for a very small fraction of total photosynthesis. (See also photosynthesis, action spectrum, absorption spectra of plant pigments.) Elevated UVB exposure – caused by the ozone hole over Antarctic waters – has been shown to decrease the activity of marine phytoplankton; such exposure has also been shown to have variable but generally negative effects on most plants tested: [16]. It's also worth bearing in mind that many plants are grown quite successfully in greenhouses, whose glass panes reduce UV exposure. TenOfAllTrades(talk) 14:57, 22 February 2011 (UTC)

One more point—while the moon's reflectivity is lower at near-ultraviolet wavelengths, it's substantially better than zero. This report (figure 4.2, middle row) shows spectra of sunlight (left) and reflected moonlight (right). While the UV 'shoulder' below 400 nm is somewhat smaller in moonlight (relative to the visible portion of the spectrum) than it is in sunlight, it's definitely present in respectable relative amounts. It is the deficiency in absolute quantity of light that is most problematic here, not a spectral issue. TenOfAllTrades(talk) 15:10, 22 February 2011 (UTC)

OK, I modified my answer to de-emphasize UV light. StuRat (talk) 20:56, 22 February 2011 (UTC)

According to Howard Griffiths of Cambridge University, "probably no". In fact, he states that plants avoid moonlight, speculating that it might disrupt their circadian rhythm. Clarityfiend (talk) 22:08, 21 February 2011 (UTC)

Deeply ingrained superstitions...

I'm sure that I'm not the only person who finds himself in this situation sometimes - that is, knowing in my head that a certain superstitious behaviour/reaction is pure bunk, but still getting a deep-seated feeling of 'wrongness' in the gut if one does not act in the accepted manner in a situation where folklore suggests that bad luck may result.

For example. I see one magpie, I look around for another one. If there is only one magpie present, I salute the bird and verbally acknowledge its presence.

Or never, ever, under any circumstances walking under a ladder. Even if it means going out of my way.

Or feeling slightly uneasy if the numbers 13 or 666 come up in random situations and taking steps to avoid prolonged exposure to the number.

Just wondering if there is a name for this conflicted feeling? Thanks. --95.148.108.189 (talk) 21:56, 21 February 2011 (UTC)

If you want fancy words to describe this scenario, you might say that you "disbelieve the superstitions, but the prevalence of these cultural presuppositions amongst your peer group has had a normative social influence on you, and you willingly suspend rational analysis in certain circumstances." You could also use the term "cognitive dissonance" to describe the conflict between rational and irrational thought. Nimur (talk) 22:02, 21 February 2011 (UTC)

Sometimes conscious understanding of an illusion "breaks the spell" but there are lot's of illusions that are not dissipated with conscious knowledge. Sometimes you have to try really long and hard to break the habits you've been raised with and sometimes you will never truly rid yourself of them. I heard once that there have been "superstition parties" where people go to break mirrors and walk under ladders etc.. to try to clear them selves of silly beliefs / habits, but i can't find any reference off hand. Vespine (talk) 22:39, 21 February 2011 (UTC)

Having been brought up without these superstitions, I have no problems with 13 or broken mirrors or ladders (though I do check that nothing is likely to fall on my head before I walk underneath). In the past, I have had to fight (mentally) to avoid being "infected" with these superstitions from friends who do seem to believe in them. The cultural transfer seems to be surprisingly strong. Dbfirs 00:12, 22 February 2011 (UTC)

Could this be because the alleged 'punishment' for not warding off the superstition (be it death, a curse, eternal damnation, etc.) sounds so much worse than the minor little act (such as politely greeting a magpie - I know a lot of otherwise rational people who are terrified of a lone magpie) required to ward it off? Sort of a 'just in case I'm being too smart for my own good and it turns out to be real' thing... --Kurt Shaped Box (talk) 14:03, 22 February 2011 (UTC)

There actually is a word for it: compulsive. In obsessive–compulsive disorder these feelings become so strong that they can make life miserable, but a great many people show milder forms of obsession or compulsion. Looie496 (talk) 00:52, 22 February 2011 (UTC)

• This brings to mind Niels Bohr's (alleged) response to being asked if he really believed a horseshoe over his door would bring him luck - "Of course not ... but I am told it works even if you don't believe in it". DuncanHill (talk) 18:07, 22 February 2011 (UTC)

Could someone expand on the references above to saluting/greeting a lone magpie? The article doesn't seem to have anything and I've never heard of such a thing. Matt Deres (talk) 20:05, 22 February 2011 (UTC)

There's some information at European Magpie#Magpie in culture, referenced to Brewer's Dictionary of Phrase and Fable. Karenjc 20:15, 22 February 2011 (UTC)

Magpies seem to have evolved a powerful survival mechanism in relation to humans. Much like dogs and cats. 20:19, 22 February 2011 (UTC) — Preceding unsigned comment added by Bus stop (talk • contribs)

The name for this is "being human". thx1138 (talk) 22:46, 22 February 2011 (UTC)

Don't you mean "being unknowingly conditioned by the culture in which you live or were raised"? Dbfirs 09:36, 24 February 2011 (UTC)

Gasoline Octane - can you FEEL the difference when you drive?

So my buddy fills up the tank of his crappy, old Chinese-brand automobile yesterday and promptly declares that switching from 94 to 97 octane fuel has made a tremendous difference in the way it runs. I find this hard to believe and suggested it was a bit of bias due to the higher price he paid. He insisted he was correct. He's agreed to blind test octanes over several tanks of gas, but that's going to take awhile. In the meantime, I'm asking here: can you feel the difference between octanes when you drive, particularly between such a small change as 94 to 97? I hypothesized that a high performance vehicle might run differently, but a beater like his wouldn't know 97 octane if it bit it in the bumper. There's just too many other inefficiencies in the engine for that to bubble to the top as a defining quality. What says the RefDesk? The Masked Booby (talk) 22:05, 21 February 2011 (UTC)

Higher octane is not "better" fuel - it is "chemically different fuel that burns with different parameters." If your engine is not designed for high octane, you will obtain worse performance by filling up with it. A common symptom is engine knocking. See our article's section: effects of octane rating. Nimur (talk) 22:07, 21 February 2011 (UTC)

You won't get any knocking by using higher octane gasoline. It is just a waste of money. Dauto (talk) 22:25, 21 February 2011 (UTC)

High octane gasoline typically has a lower calorific value than the normal stuff. So it is possible the he would see an unmeasurably slight deterioration in performance and/or economy. However there are all sorts of possibilities. If he's happy paying for 97 octane, why not? In my own car using premium gets rid of a slight knock on acceleration, makin the engine seem smoother. Frankly i am not prepared to pay extra get rid of it. Greglocock (talk) 22:37, 21 February 2011 (UTC)

I once fell victim to the claim that high octane gasoline "helps clean out an engine". What it actually does is make the CHECK ENGINE light come on. Wnt (talk) 22:53, 21 February 2011 (UTC)

The octane rating has to do with the requirements of high-performance, high-compression engines or turbocharged engines to fuel that isn't subject to pre-detonation ("knock") in the compression stroke, which can damage the engine. The fuel is less volatile (in a different sense from its tendency to evaporate) under compression than "regular" fuel, and, as pointed out, can have less energy content. It has no value at all in an engine that doesn't specifically require the higher rating. It isn't better, regardless of the silly "premium" marketing, it's just suited for specific engines that need it to perform their best. If the engine isn't designed to demand high octane ratings, you're wasting money and buying lower-performing fuel at a higer price. Acroterion _(talk) 02:08, 22 February 2011 (UTC)

One kind of obvious point that no one seems to have mentioned is that it could be that your friend's car pings at 94 octane, but not at 97. Yes, he would notice that difference, for sure. It seems a little unlikely because it's usually very high-compression engines that want the very high octanes. On the other hand, I believe that old, dirty engines are more prone to pinging, so it's not totally beyond the realm of belief. --Trovatore (talk) 02:14, 22 February 2011 (UTC)

Here's a fun summary : The Straight Dope : What's the difference between premium and regular gas?

APL (talk) 06:00, 22 February 2011 (UTC)

February 22

checks

is there a way to do a free background check on someone? most sites want money example http://www.publicbackgroundchecks.com --Tomjohnson357 (talk) 01:23, 22 February 2011 (UTC)

yes, if you put quotation marks around the peron's name "like this" then only people with that exact name will be returned. if it's a common name, try adding, one at a time, anything you know about the person, such as a city he or she has lived in, a school they have gone to, etc. If you really want to go into detective mode, add them as a friend on facebook. 109.128.213.73 (talk) 01:43, 22 February 2011 (UTC)

This is a terrible answer. APL (talk) 05:47, 22 February 2011 (UTC)

im referring to a criminal background check — Preceding unsigned comment added by Tomjohnson357 (talk • contribs) 04:52, 22 February 2011 (UTC)

Why is this on the science desk? I thought that Nimur had deleted it by mistake when removing something else, but now I'm not so sure. --Trovatore (talk) 05:18, 22 February 2011 (UTC)

I have to say that I doubt it. Doing a background check is not a purely automated process, so at some point a human will be involved. I don't know, but I assume that's got to cost something. APL (talk) 05:47, 22 February 2011 (UTC)

Convolutor worm?

I was just watching a documentary which mentioned an unusual worm (I think they called it a "convolutor worm") that lives in symbiosis with algae and whose nutrition is provided by the algae's photosynthetic activities rather than by digestion.

I can't find an article on such a species, nor google hits. Any hints? FT2 ^{(Talk | email)} 14:17, 22 February 2011 (UTC)

Try Convolutriloba in Google, I think you'll be pleased! Richard Avery (talk) 14:30, 22 February 2011 (UTC)

Symsagittifera roscoffensis perhaps? Nanonic (talk) 14:40, 22 February 2011 (UTC)

Running an ordinary gasoline powered car on ethanol

My car runs on normal unleaded gasoline. The gas station near by house sells gas pretty cheap, but it has 10% ethanol in it. Is this safe for my car? Would it be alright if my car ran on 100% ethanol fuel? ScienceApe (talk) 15:18, 22 February 2011 (UTC)

Here's a Popular Mechanics article that answers your questions. The specific quotes: "The engineers at [Ricardo] also disassembled some old fuel systems to examine them for the effects of prolonged E10 exposure. They didn't see any." and "Pure ethanol can damage cars that aren't designed to handle it. On its own, the grain alcohol can cause galvanic corrosion, swelling of rubber components and rust or corrosion on certain metals, because of its ability to hold water." Andrew Jameson (talk) 15:40, 22 February 2011 (UTC)

Yes, 10% ethanol is safe. That's why most gas include some ethanol. No you could not use 100% ethanol in your car. 100% ethanol would require an engine with a different compression ratio.Dauto (talk) 15:42, 22 February 2011 (UTC)

That's incorrect. Using 100% ethanol in a spark-ignited does not require a different compression ratio. Ethanol has a higher octane rating than gasoline, so you could run a higher compression ratio if you desired to, but it's in no way required. Andrew Jameson (talk) 16:25, 22 February 2011 (UTC)

That's right. It's not strictly required. But without a change of compression ratio and/or adjustment of spark timing it would run very inefficiently and with low power. Dauto (talk) 02:21, 23 February 2011 (UTC)

As I understand it, the key differences in flexible-fuel vehicles is the seals, gaskets, tubing, etc. Ethanol has a different reactivity profile than gasoline, so plastics and rubbers that are perfectly stable in contact with gasoline can become brittle/break down in contact with ethanol (and vice versa). Vehicles designed with only gasoline in mind may have components which aren't stable to high ethanol concentrations. My understanding is that all modern cars can handle ethanol up to about 10%, it's E85 and the like which require special attention to materials. My understanding is that the reports you sometimes see of people having problems with 10% ethanol is due to fouling - since ethanol can dissolve different compounds than gasoline can, "gunk" that has accumulated in the fuel lines/fuel filter might be solubilized and brought to the cylinders; it's not that 10% ethanol is bad for the car, it's more that the fuel lines were dirty. -- 174.21.250.120 (talk) 17:01, 22 February 2011 (UTC)

It depends on the definition of "run". I ran a normal car engine on diesel fuel once. It would "run" if you mean "it made the car move forward when pressing on the accelerator". However, it didn't "run" if you mean "made the car travel at a reasonable speed without causing damage to the car and passengers". I expect running on ethanol without conversion to result in very very poor miles per gallon and low speeds. -- kainaw™ 17:05, 22 February 2011 (UTC)

"Diesel fuel" and "ethanol" are two different things. Diesel fuel has a lower octane rating than gasoline, while ethanol has a higher octane rating than gasoline. Diesel is also much harder to vaporize, which is why putting diesel into a gasoline engine will make it run poorly or not at all. Andrew Jameson (talk) 17:39, 22 February 2011 (UTC)

Complementing the above. I think it is also possible to get around the different compression ratio requirements by changing the timing and the behavior of the electronic injection system. Either way, you car wouldn't be able to deal with ethanol unless it is a flex-fuel design. Dauto (talk) 16:12, 22 February 2011 (UTC)

At least E10 scrounges up bits of water out of your fuel ... though apparently those using boats have many nasty things to say about it! [17]

E10 is not supposed to be used on boats or planes. Graeme Bartlett (talk) 04:12, 23 February 2011 (UTC)

Tinkerers in the 1970s (or earlier) modified automobile carbs to allow the vehicles to run on ethanol. Thee wee some limitations, but I do not recall exactly what. Low temperature operation, perhaps? Edison (talk) 05:20, 23 February 2011 (UTC)

World oil consumption

How much oil is burned in the world every day? This does count plastics, lubricants, asphalts etc. If you answer, can you please give a reference? --T H F S W (T · C · E) 18:22, 22 February 2011 (UTC)

According to Peak oil#Worldwide_production_trends (which references this report by the International Energy Agency), production has been 85.24 million barrels/day for the average of 2005-2008. Since we don't build or deplete large artificial oceans of oil, production and consumption are about equal over the medium term. --Stephan Schulz (talk) 18:38, 22 February 2011 (UTC)

I already know how much is consumed. I'm asking how much is burned. --T H F S W (T · C · E) 18:42, 22 February 2011 (UTC)

Sorry, I was confused because you listed all the other uses of oil (that only eventually may be burned). As a rough estimate, I'd still say that nearly all produced oil is burned eventually. Only a small part is made into plastics, and even those are often burned in thermal recycling. According to the US EIA, in 2009 about 94% of oil ended up in products for immediate burning. --Stephan Schulz (talk) 18:56, 22 February 2011 (UTC)

Oh, whoops, I forgot the "not". Thanks though. --T H F S W (T · C · E) 19:25, 22 February 2011 (UTC)

masonary

1.what are the composite wall?

2.please draw me 1 if any bricks in english bond with a turning corner as the blocks — Preceding unsigned comment added by Bruce sarinke (talk • contribs) 18:43, 22 February 2011 (UTC)

Here's a pic of English Bond, but there's no turning corner: [18]. StuRat (talk) 20:00, 22 February 2011 (UTC)

atom

In a previous post it was stated that an electron in an atom can be a kilometer from the nucleus of the atom but still be "part" of that atom. What force is maintaining that "connection?" —Preceding unsigned comment added by 165.212.189.187 (talk) 18:47, 22 February 2011 (UTC)

This is true only in the sense that there is nothing else perturbing the system - if you have miles and miles of perfect vacuum, and nothing affects the movement of the electron but its ever so gradual revolution about one proton. Otherwise this doesn't happen, and the atom is just plain ionized. (Though I suppose that there is probably some way you can analyze the interactions of non-bonded atoms as if each electron is also considered as trying to 'orbit' the other nucleus at great range, maybe under the rubric of "Van der Waals forces"?) Wnt (talk) 19:18, 22 February 2011 (UTC)

The force is electrostatic attraction, though in the case of an electron orbiting an atomic nucleus, the force must be described using quantum mechanics. When an electron's effective radius is very large, it requires very little energy to totally ionize it - in other words, the electrostatic potential energy well is much smaller than, say, the average thermal energy of a random collision with another object. So, to be stable, you have to find a way to thermally isolate the electron - which is pretty impossible in a real set-up; but hypothetically, in a perfectly isolated vacuum, with no incident energy or particles of any kind, an energized electron can stably orbit indefinitely at macroscopic radii. Nimur (talk) 21:29, 22 February 2011 (UTC)

The orbit wouldn't be stable. The electron is accelerated and would emit soft photons and cascade down to a low orbit. Dauto (talk) 02:12, 23 February 2011 (UTC)

For a thorough introduction to these highly-excited-but-not-quite-fully-ionized creatures, Rydberg atom is worth a read. TenOfAllTrades(talk) 23:42, 22 February 2011 (UTC)

An interesting feature is that these expanded atoms have a longer lifetime than the smaller excited atoms. Graeme Bartlett (talk) 04:10, 23 February 2011 (UTC)

I think you're missing the point of the question. The original question was about the atom in the ground (or indeed any other) state. And yes, even in the ground state there is a small but finite probability that the electron can be found at arbitrarily large distances from the nucleus. This is the result of the quantum mechanical treatment of the atom, with electrostatic potential/forces holding the thing together. This has no effect on the energy needed to ionize the atom (barring lots of tricky questions of how to measure or observe any of this). --Wrongfilter (talk) 09:00, 23 February 2011 (UTC)

What would be the volume of this atom? —Preceding unsigned comment added by 165.212.189.187 (talk) 15:40, 23 February 2011 (UTC)

That was actually part of the point I was trying to make with the comment "simultaneously infinitesimally close to the nucleus and infinitely far from the nucleus" [19], you can't give a definitive radius or volume to the electron density of an atom, because the electron density never really ends, it just sort of tapers off. You can set arbitrary cutoffs (say, "the radius within which 95% of the electron density is contained") but those are just that: arbitrary. - However, if you're looking for the radius/volume of the atom rather than that of the electron density, you can resort to phenomenological definitions. For example, the van der Waals radius can be defined by the the behavior of gasses, or by one half the minimal non-bonded atom-atom distances in crystals. That's the radius of a solid sphere that the atom "behaves like" under the conditions where you measured it. This tends to be rather consistent between different types of measurements for each element, somewhere in the 1-2 Å range for the common elements. The van der Waals volume is then defined as just the volume of the hard sphere with the van der Waals radius. -- 174.21.231.113 (talk) 17:53, 23 February 2011 (UTC)

Is there any evidence of atoms like this existing in the vacuum of space? —Preceding unsigned comment added by 165.212.189.187 (talk) 18:14, 23 February 2011 (UTC)

We say the electron has a certain probability of being very far from the nucleus. If you check a billion hydrogen atoms you'll maybe find one where the electron is very far from the nucleus. If you watch a single hydrogen atom, the electron will be close to the nucleus most of the time but might be very far for short periods of time. This is how nature works on the atomic level: we can only talk about probabilities of finding an electron here or there. In fact, it is impossible (by Heisenberg's uncertainty principle) to observe a trajectory of the electron. Quantum mechanics is weird and it requires quite a bit of effort to get one's head around it. There are also atoms in a special, highly excited state where the electron is even on average rather far from the nucleus (most of the responses here were about these Rydberg atoms), and yes, these have been observed. --Wrongfilter (talk) 20:19, 23 February 2011 (UTC)

Internal state

http://books.google.com/books?id=OAsQ_hFjhrAC&pg=PA216&hl=en#v=onepage&q&f=false

At the bottom of page 216, the author says that the internal state of a body will be the same regardless of whether it is moving or not. But if the internal state takes into consideration the relative positions of the particles in the body, wouldn't (by Lorentz contraction) it be different in two inertial frames? 74.15.137.130 (talk) 19:32, 22 February 2011 (UTC)

A moving body would appear different in two different inertial frames, yes. But in the frame containing the body, the internal state when in motion isn't any different from when at rest, if the motion is uniform. If it was different, then you'd be able to tell when the body is in motion. That's what I gather from that page. ~Amatulić (talk) 21:13, 22 February 2011 (UTC)

... and remember that the Lorentz contraction is symmetric i.e. the moving observer measures the "stationary" body's dimensions as contracted. (so, in one sense, the contraction is not "real" but just a feature of the observation, though in other senses it is measurably real) Dbfirs 23:44, 22 February 2011 (UTC)

Etymologies of methane, ethane, propane, and butane

What are the name origins for methane, ethane, propane, and butane? Unlike the later pentane, hexane, etc., they don't derive from the Greek or Latin words for their number. --75.15.161.185 (talk) 22:10, 22 February 2011 (UTC)

Methane comes from Methyl, and see the bottom of that article for where that came from. Ethane is from Ethyl, and butane is from Butter. See this site for any others you have an interest in. Ariel. (talk) 00:32, 23 February 2011 (UTC)

This says that methyl was defined as "univalent hydrocarbon radical" by Jöns Jakob Berzelius in 1844. Ethyl apparently comes from ether, which itself was coined in c.1730. Here's propane and butane for your perusal as well. In reality, it looks like we should ask why pentane etc. have got such boring names. SmartSE (talk)

from etymonline

1873, from ethyl + -ane.

1868, coined from meth(yl) + chemical suffix -ane.

1866, with chemical suffix -ane, from prop(ionic acid) (1850)

1875, from but(yl),

Vespine (talk) 00:35, 23 February 2011 (UTC)

Methane from methyl from methylene: spirit of wood

Ethane from aether: ether

Propane from propionic: of first fat

Butane from butyric: of butter

Plasmic Physics (talk) 03:07, 23 February 2011 (UTC)

Bacillus cereus and alcohol prep pad contamination

I went to the Wikipedia article on Bacillus cereus when I learned that widely used alcohol prep pads(70% Isopropyl alcohol) made by the Triad Group and sold under many store names have been recalled in the US by the FDA due to contamination with Bacillus cereus. See [20]. Parents whose 2 year old died from meningitis have sued the company with the claim that the contamination caused the fatal infection: [21].

• 1)I am puzzled by the Wikipedia article on the bacillus, because it talks about it as a common inhabitant of the human intestine, and says little about it being dangerous when it enters broken skin, where a contaminated alcohol prep might introduce it, or a noncontaminated one might fail to kill it, such as before an injection or other skin penetration. There are only 2 brief sentences about its effect in skin infections,and nothing about systemic infection such as meningitis. The article actually says some positive things about it: "It recommended as pathogenic microflora in pharmaceutical oral products in Brazilian Phamacopaeia." Could someone add a bit about the risk from it being on the skin when alcohol preps are used before an injection or incision?
• 2)Also, if the purpose of an alcohol wipe is to kill off harmful bacteria, why doesn't it kill the b. cereus? Is it becasue the b. cereus as a spore is resistant to alcohol? The Wikipedia article only discusses the bacillus's sensitivity to heat. Some expert editing of the article is badly needed. In addition, the article on Alcohol says almost nothing about its use as a skin prep before injections, or for cleaning medicine vials before a hypo is introduced.

Bacillus cereus endospores are highly resistant to ethanol, although it does have some activity against them.[22] Also, according to a cute little wiki (alas, "noncommercial use only" and not snarfable) [23] the bacillus creates ethanol by fermentation, perhaps explaining some of its resistance. Wnt (talk) 23:00, 22 February 2011 (UTC)

• 3)Rubbing alcohol only vaguely says "good against vegetative bacteria and fair against fungi and viruses, but is ineffective against spores" without any links to see what is included in "vegetative bacteria." What is and is it not effective against? Strep? Staph? Pseudomonas? Anthrax? The article on Bacterial infection lists many common infectious bacteria, but again there is no mention of what antiseptics are effective against them. The article Antiseptic says that alcohol is "commonly used" but again with no note as to its failings such as the apparent one in the case of b. cereus.
• 4)Betadine (Povidone-iodine is a more modern antiseptic. Is it a more effective prep than alcohol pads, and if so why isn't it or Tincture of iodine used instead of alcohol (if the skin discoloration due to Betadine is not a problem)? Edison (talk) 22:24, 22 February 2011 (UTC)

You make excellent points about our articles. Regarding point #4, though, povidone-iodine can cause significant reactions (e.g. PMID 17064209), and one can sometimes grow pathogenic organisms from povidone-iodine (e.g. PMID 19815282) illustrating its slow antiseptic action when in liquid form (it's activated by drying); in fact, there have been infection outbreaks associated with povidone-iodine (PMID 1376156). So, no antiseptic is perfect. -- Scray (talk) 23:34, 22 February 2011 (UTC)

So how about tincture of iodine as an alternative to practically useless 70% isopropyl alcohol prep pads?Edison (talk) 05:16, 23 February 2011 (UTC)

Tincture of iodine (elemental iodine solubilized by the presence of iodide through formation of triiodide ion) is far more prone to hypersensitivity reaction than is povidone-iodine, due to the large amount of free iodine (this is discussed in the povidone-iodine article, I believe). Thus, it is used when rapid killing is particularly important and a reactions are not (i.e. its use is relatively limited). This episode of contamination of alcohol prep pads should not, IMHO, overshadow their demonstrated utility (when used appropriately and not contaminated). Anything can be contaminated, even adhesive bandages and tongue depressors (PMID 19754749). This is why we need effective regulation of the manufacture of medical equipment, and each outbreak teaches new lessons. -- Scray (talk) 13:03, 23 February 2011 (UTC)

This news coverage was the first mention I had seen of "nonsterilized alcohol pads." I had assumed that the alcohol was an effecetive antiseptic. There is a contrast between an antiseptic which contains dangerous microbes, and their presence on things not expected to kill germs, such as gauze, cotton pads, tongue depressors, or adhesive tape. The bottle of tincture of iodine in the medicine cabinet has a little glass applicator which is applied to the cut or scrape. Presumably the iodine is supposed to kill any pathogens picked up by the applicator. An weakly effective antiseptic or antibiotic ointment applicator/dispensor seems like a route for spreading microbes from one wound or person to another. Edison (talk) 17:14, 23 February 2011 (UTC)

Agreed. Medical practice is complex, so in addition to good equipment it requires common sense, beneficence, excellent training, adherence to proper technique, and vigilance. -- Scray (talk) 18:30, 23 February 2011 (UTC)

is there a link between gender orientation to sexual orientation?

ya know, in a superficial way it seems that indeed there is, i myself don't agree and i think that even if there is, it ain't that robust...

but, what does the research shows to us?, is there a link?, and if there is, how significant it is by this stage of the research?

thanks and blessings,

beni. —Preceding unsigned comment added by 79.176.18.237 (talk) 23:16, 22 February 2011 (UTC)

What do you mean by "gender orientation"? Are you possibly referring to gender identity, or simply to which gender a person prefers the non-sexual company of? Or maybe something else entirely? Someguy1221 (talk) 00:18, 23 February 2011 (UTC)

I was going to ask the same. The term "gender" tends to be used in terms of the gender one is, or perceives oneself to be. The term orientation tends to be used to mean what one is attracted to. So you wouldn't usually talk about gender orientation. You might talk about the gender a person is or the gender they believe themselves to be, and the gender they are sexually attracted to, but you probably need to clarify the question before it can be responded any better. FT2 ^{(Talk | email)} 03:18, 23 February 2011 (UTC)

Hey, indeed, i speak about Gender Identity.

thanks! —Preceding unsigned comment added by 79.176.18.237 (talk) 11:33, 23 February 2011 (UTC)

February 23

An interesting physics problem

Dear Wikipedians:

I am working on this very interesting physics problem:

A spring attached to a ceiling has mass of 500 g suspended from it such that the spring stretches 4.0 cm. Calculate the spring constant.

I have taken two approaches to this question, but they seem to yield different answers!

The first, dynamic approach:

kx = mg

k = mg / x = 0.5*9.8/0.04 = 122.5 N/m

The second, energy approach:

½kx² = mgx

k = 2mg / x = 2*0.5*9.8/0.04 = 245 N/m

I find it very mind-boggling: how can there be two answers???

Thanks,

174.88.35.131 (talk) 02:48, 23 February 2011 (UTC)

Somewhat amusing. You might like to consider the spring deflection a little later, in the second case. In mechanical engineering this is why 'suddenly applied loads' are multiplied by two, but not many people know that. Greglocock (talk) 02:54, 23 February 2011 (UTC)

The first approach is straightforward Hooke's law - you've calculated the stable applied force mg from the weight, and set the spring to be the right length to be supporting it. But the second approach assumes that the spring is extended solely because the weight was dropped from the point where the spring's length is zero - which isn't necessarily true. It assumes that the kinetic energy is zero, but this is true when the spring is maximally extended and about to rebound. In any case, there are two answers because the spring can be extended to that distance for more than one reason - it could be because it is hanging at a constant distance, or because it is bouncing up and down after having been released. Wnt (talk) 03:31, 23 February 2011 (UTC)

I agree with Wnt. Considering the wording of your question, your first equation is correct and 122.5 N.m^-1 is the answer. You have called this a dynamic approach but it would be better described as a static approach. You have called your second equation the energy approach and this is reasonable. Alternatively it could be described as a dynamic approach. Dolphin (t) 06:40, 23 February 2011 (UTC)

You can solve the equilibrium problem using energy by equating the first derivative of ½kx² − mgx to zero, since an equilibrium configuration is a local minimum of the potential energy. That gives you kx = mg. What you did is equate ½kx² − mgx itself to zero. Since energy is only defined up to an arbitrary additive constant, that will give you different answers depending on, in this case, the point you happened to choose as the origin of x. -- BenRG (talk) 10:08, 23 February 2011 (UTC)

Basically, you have a system in equilibrium. With springs, being in equilibrium and being at rest are two different things. If the body was drooped from the unextended state as above, the spring will reach an extreme position, where energy is zero and it is temporarily at rest. But, the spring force still acts on it. Thus, this is not your equilibrium position. Springs have a tendency to oscillate in simple harmonic motion, so you have to look carefully at the wording. ManishEarth^{Talk • Stalk} 10:31, 23 February 2011 (UTC)

Thanks all for your response! Now I fully understand the difference between the two approaches! 174.88.35.131 (talk) 22:12, 23 February 2011 (UTC)

Resolved

Pepper seed

I have heard that pepper seed (from bell peppers and hotter varities alike) are mildly poisonous and shouldn't be eaten. This true? 212.68.15.66 (talk) 06:55, 23 February 2011 (UTC)

They're not poisonous. However, chefs do tend to remove them, because they're a bit bitter when bitten into. Sophus Bie ^(talk) 10:36, 23 February 2011 (UTC)

The seeds of a pepper also tend to be spicier than the flesh. While pepper seeds are not 'poisonous' in any meaningful real-world sense, note that capsaicin is technically toxic at high doses. See e.g Capsaicin#Toxicity. However, there's no real reason to avoid eating pepper seeds; I use them in things like gumbo. SemanticMantis (talk) 15:46, 23 February 2011 (UTC)

You may be thinking of apple pips or peach pits, which contain cyanide, though poisoning yourself with those would involve eating the seeds of hundreds of fruits. The article about apple seed oil says it smells of almonds (a giveaway that it contains cyanide) and says vaguely that you shouldn't consume too much of it (however much that is). One food you could feasibly poison yourself with is the bitter almond, but for this reason they don't get much use as an ingredient. 213.122.19.152 (talk) 17:47, 23 February 2011 (UTC)

Keeping soda carbonated longer

Resolved

 – –Kerαunoςcopia^◁_galaxies 09:22, 23 February 2011 (UTC)

I have a 2 liter bottle of regular soda. After opening it, it gets flatter with each day (or rather, with each time) that I open the bottle. Would squeezing the bottle to the point where the liquid was near the top of the spout, and then sealing the bottle closed with the cap, help slow down the rate at which it gets flat? Or would the bottle still expand as the carbonation leaked from the liquid, and this would just be futile? I refrigerate the soda before and after each opening. Also, I did check the archives for similar questions and didn't find anything about this. – Kerαunoςcopia^◁_galaxies 08:56, 23 February 2011 (UTC)

No. By squeezing out the air and hence lowering the pressure above the drink, you encourage the CO2 to come out of the drink. If you wanted it to stay carbonated it would be better to pump CO2 (or just air) into the bottle.--Shantavira|^{feed me} 09:07, 23 February 2011 (UTC)

Ah, that makes perfect sense!! Excuse me while I go fix my bottle lol. – Kerαunoςcopia^◁_galaxies 09:19, 23 February 2011 (UTC)

So here I think we landed on a million dollar idea, but there are a half-dozen variations on this invention already (fizz-keeper bottle caps). So much for quitting my day job. – Kerαunoςcopia^◁_galaxies 09:24, 23 February 2011 (UTC)

Fizz keeper caps, appear effective just hours not days. Making your own in a soda siphon would be better. See Sodastream--Aspro (talk) 09:45, 23 February 2011 (UTC)

The fizz keeper article consists of an obsessively detailed answer to this vexing question. The best answer is to buy smaller bottles.--Shantavira|^{feed me} 10:09, 23 February 2011 (UTC)

These answers are great, thank you! – Kerαunoςcopia^◁_galaxies 10:42, 23 February 2011 (UTC)

Why would squeezing the bottle significantly affect the pressure of air above the drink? You aren't pumping out air while keeping the volume the same, you are reducing the volume of air keeping it at atmospheric pressure. I believe this would actually have a slight improvement although as with fizz keeper it may not be significant. A slightly better solution would be to pour the drink in to smaller bottles, this way the volume of air is reduced and you are only opening one so the others aren't affected every time you open one. Of course buying smaller bottles in the first place is best but that tends to be expensive. Nil Einne (talk) 14:20, 23 February 2011 (UTC)

It depends on the bottle. If the plastic resists deformation and exerts a force to rebound to its original shape after it is sealed, then it will generate a (admittedly very weak) partial vacuum above the liquid as the bottle itself counters some of the local atmospheric pressure. The bottle will tend to expand until it returns to its original volume; the entire headspace in the bottle will then be filled by carbon dioxide lost from the liquid solution.

The amount of carbon dioxide that ends up in this headspace is an interesting question. In principle, the final partial pressure of carbon dioxide above the soda should depend solely on the concentration of CO₂ in the liquid (Henry's law), so squeezing the air out shouldn't make a difference; in practice, I imagine that there is at least a small departure from ideality which will depend on the partial pressure of air already present. The best ways to reduce the amount of CO₂ lost are (as Nil Einne says) to reduce the amount of headspace in the bottle by transfer to smaller containers, and to keep the temperature of the liquid low when the bottle is open (increasing the solubility of carbon dioxide and decreasing its rate of loss from solution). TenOfAllTrades(talk) 14:59, 23 February 2011 (UTC)

Unfortunately, though, pouring the carbonated drink from the larger bottle into smaller bottles would increase the loss of CO₂ during that very action, probably making the action futile. However, being that I currently reside at high altitude (9,000+ feet), I wonder if the expansion of the air in the headspace during shipment would help prevent more loss of CO₂ than for a similar bottle at sea level? When I open my bottle, am I starting off with a higher amount of carbonation than a sea-level bottle would have? – Kerαunoςcopia^◁_galaxies 22:18, 23 February 2011 (UTC)

9000+ feet ? Where's that ? Do you need an oxygen mask ? StuRat (talk) 23:24, 23 February 2011 (UTC)

I'm in the American West, so I guess that makes me some sort of space cowboy? –Kerαunoςcopia^◁_galaxies 00:43, 24 February 2011 (UTC)

I'd just stop buying 2 liter bottles. They really are only good for parties. For individual consumption, you need something smaller. StuRat (talk) 23:25, 23 February 2011 (UTC)

But a generic brand of 2 liter soda is 1 cent per ounce, far cheaper than cans (of any brand!). – Kerαunoςcopia^◁_galaxies 00:43, 24 February 2011 (UTC)

This conversation surprises me. I remember that 20-30 years ago, soda pop seemed like it got flat easily, but then it stopped seeming to do so. I'd assumed that they'd figured out some technical trick to improve the CO2 solubility or something - nowadays, it seems like even open cans of soda seem reasonably fizzy after being left a day, and two liter bottles can be opened many times without trouble. Or have I just lost the ability to taste the "flatness"? Wnt (talk) 23:34, 23 February 2011 (UTC)

I don't know the science behind it, but I do know I can put an open glass of coke in the fridge at night and the next day it's still bubbly. – Kerαunoςcopia^◁_galaxies 00:43, 24 February 2011 (UTC)

Squeezing plastic bottles does make the carbonation last longer. I do this all the time. My guess is that once you close the cap, because there is a smaller volume of air, less CO₂ is drawn out of the pop each time to achieve equilibrium, so there's less gas available to escape into the atmosphere during the time you have it open again (hmmm ... partial pressure doesn't do a good job of explaining the phenomenon). Clarityfiend (talk) 02:04, 24 February 2011 (UTC)

Buy one of those computer dusters that shoots CO2 and shoot it into the bottle right befor closing it. this will saturate the volume above the soda with CO2. —Preceding unsigned comment added by 165.212.189.187 (talk) 13:51, 24 February 2011 (UTC)

This is a start, but pressure would still need to build up to a higher than atmospheric level. Now if you really want to fix the problem, you need a good supply of dry ice... ;) Wnt (talk) 17:20, 24 February 2011 (UTC)

Stripping all electrons from metal

Has anyone ever taken a very small peice of metal (such as copper) and applied enough electric force to pull all electrons from it? Would this make the metal invisible? Would the metal still stay as a solid? - Anon. —Preceding unsigned comment added by 114.77.191.84 (talk) 09:02, 23 February 2011 (UTC)

Any macroscopic piece of metal would explode due to the repulsion between the remaining positively charged nuclei. If you have just a single nucleus, you just get a highly ionized atom which is invisible anyway by virtue of its extremely small size. 157.193.175.207 (talk) 09:30, 23 February 2011 (UTC)

Yeah. You first keep in mind that the nature of being a metal requires the presence of electrons (see sea of electrons). Then just think of the forces involved. If you remove just 1% of the electrons from each of two humans standing one meter apart, the force between them would be strong enough propell the planet Earth at many times g. And the force increases asymptotically toward infinity as the distance decreases, and my two human subjects were at 1 meter. A metal would explode with a relatively tiny portion of electrons removed. Someguy1221 (talk) 10:37, 23 February 2011 (UTC)

Indeed; the energy cost of stripping all the electrons from an atom (of, say, copper) is much, much, much greater than the binding energy that holds a single copper atom in place in a lump of metal. As you get down to tightly-bound inner-shell electrons, it becomes progressively more difficult to remove them from the correspondingly less-shielded nucleus. Check out Ionization energies of the elements (data page) for some numbers. Consider a neutral copper atom with 29 electrons. Removing the first one costs 7.7 eV; removing the twenty-ninth electron (from Cu²⁸⁺) costs 11568 eV. While it is possible to make fully-stripped nuclei, it's something that's done to single nuclei in particle accelerators, not something achievable in bulk materials. TenOfAllTrades(talk) 14:41, 23 February 2011 (UTC)

Metal without electrons is in a plasma state, see article. Cuddlyable3 (talk) 14:47, 23 February 2011 (UTC)

In a plasma state, a metal is no longer a metal (it loses its sea of electrons). ManishEarth^{Talk • Stalk} 12:33, 24 February 2011 (UTC)

Effect of atmospheric pressure on reaction?

Our Chemistry class came along a question in a textbook regarding Le Chatelier's Principle, stating roughly: NO2(g) reacts with itself: 2NO2(g) «» N2O4(g) + heat. As the reaction proceeds to the right, 2 moles of gas form 1 mole of gas, so in a fixed volume, pressure would drop as the reaction proceeds to the right. If the total gas pressure is decreased, would the reaction proceed to the right or left?

We cannot decide whether the reaction would shift to the left, favouring more gas, or not shift at all, as a gas fills the container regardless of pressure :/

Thanks! 110.175.208.144 (talk) 10:26, 23 February 2011 (UTC)

For background reading, Wikipedia has an article about Nitrogen dioxide NO2 that mentions its existence in equilibrium with Dinitrogen tetroxide N2O4, called a dissociating gas. Le Chatelier's principle says If a chemical system at equilibrium experiences a change in concentration, temperature, volume, or partial pressure, then the equilibrium shifts to counteract the imposed change and a new equilibrium is established. If equilibrium is reached and the container volume is then increased, the equilibrium swings to the left (more gas molecules) to counteract. Cuddlyable3 (talk) 14:14, 23 February 2011 (UTC)

I had read the above articles, but it wasn't quite clear exactly what would happen if pressure decreased... thanks for your answer! 110.175.208.144 (talk) 20:14, 23 February 2011 (UTC)

Sometimes equilibrium reactions are easier to see one way, and then just recognize that because it's an equiliubrium, it goes the other way in opposite circumstances. For example, it makes sense to me that when you increase the pressure, pairs of molecules combine to make single molecules (in essence, absorbing that added pressure). So "therefore, reducing pressure allows or promotes them to split apart, as the opposite of what happens when the pressure is increased". Not always, but often, one direction is an actual effect, and the other is the absence of the effect (sometimes it sounds weird to say you are "pulling them apart" in this case, because there still is some pressure). Maybe useful, maybe more confusing if you already understand it. DMacks (talk) 22:58, 23 February 2011 (UTC)

Electrical Circuit Analysis

Why we use resistor , capacitor, Inductors in the electrical circuits what actually these passive elements doing in the circuits and how to choose those ranges i don't know why when where which what element should be used ? ex-: http://www.circuitstoday.com/wp-content/uploads/2008/03/max2606_single_chip_transmitter.JPG they used 5 capacitors and 8 resistors and one inductor how they know those elements and their values — Preceding unsigned comment added by Kanniyappan (talk • contribs) 13:06, 23 February 2011 (UTC)

Most of the circuit complexity is inside the integrated circuit MAX2606 that is connected to external components (resistors, capacitors and an inductor) that are recommended by the manufacturer. You will find a description in the MAX2606 data sheet. Here are simple articles about components: resistor, capacitor and [inductor].Cuddlyable3 (talk) 14:38, 23 February 2011 (UTC)

As a young and naive reader of science articles, I learned what the individual components did (caps store and release charge, resistors oppose the flow of electricity, etc) and tried miserably to understand a diagram of a radio, in terms of "The antenna brings voltage and current to the capacitor, which stores then releases charge". That analysis was flawed, because the components act in combination to do certain standard functions (a tuning circuit, an RC filter, a blocking cap, a dropping resistor, a transformer). It helps to understand what larger building blocks do. Edison (talk) 17:03, 23 February 2011 (UTC)

May I ask (at the risk of committing the sin of chatting), did you gain this understanding in a way compatible with the first principles (about capacitors, etc.), telling yourself things like "a flip-flop works because the first transistor ..." or did you let go of the need to make this conceptual connection, and simply accept that the standard circuits do what they do? I would find such letting-go incredibly difficult, and the thought occurs that maybe this is why I have never got beyond the stage of trying miserably to understand circuit diagrams in a reductionist way. 213.122.19.152 (talk) 17:58, 23 February 2011 (UTC)

I don't know this stuff, but I'll take a stab at it for fun. Let me know if I'm wrong.

• the 0.47 uF capacitor blocks a net current flow from the input into the chip. It's a fairly big capacitor, and it allows sinusoidal current flows, but (with resistor) it should block very low frequency sound from getting in (high pass filter).
• the 2200 pF capacitor should cancel out very short-term changes in current going into TUNE and thus (with resistor) act as a low pass filter.
• the largest 10 uF capacitor should compensate for changes in the power supply - I assume it helps prevent the chip from being damaged when you fool around with the battery.
• the large variable resistor R1 controls the voltage going into TUNE, except for short-term oscillations coming from the audio input. A volume control, in other words.Doh, it's FM! Note that its value controls how much power is being wasted whenever the circuit is on.

I think I'd better stop and see if someone corrects me... ;) Wnt (talk) 18:46, 23 February 2011 (UTC)

• Potentiometer R1 tunes the transmitter frequency by sending a steady voltage into pin 3 of the integrated circuit. It is NOT the volume control.
• Potentiometer R2 varies the strength of audio (an alternating current) signal into pin 3, causing the frequency to be modulated by the audio. It's an FM transmitter and this is its volume control.
• The two audio inputs left+right should not make you think this transmits a stereo signal. The two inputs get combined at R2 and transmitted in mono..
• The 0.47 uF capacitor allows an audio signal in but blocks the voltage from R1 from leaking away
• The 10 uF capacitor with the 4.7k resistor form a low-pass filter that prevents any ripple on the VCC supply adding noise to the sound. If VCC comes from a battery they can probably be omitted. However the VCC voltage has to be steady otherwise R1 will need continual adjustment to stay on frequency.
• Note that the 10uF and 1uF capacitors are electrolytic types that have to be connected a particular way round shown by the + symbols. Cuddlyable3 (talk) 21:24, 23 February 2011 (UTC)

Hey, thanks - I knew this wasn't my thing, but at least I got someone to answer. ;) Wnt (talk) 23:17, 23 February 2011 (UTC)

February 24

What is the effect of alcohol and drugs on non-human animals?

If you were to give a dog or a grizzly bear vodka, or LSD, what would happen?--X sprainpraxisL (talk) 01:01, 24 February 2011 (UTC)

We have an article on the Effect of psychoactive drugs on animals, though it doesn't look like it has much on the specific animal/drug combinations you mention.

Or, here's an article from 2000 in Alcohol Research & Health: Animal Models in Alcohol Research, if that's more the sort of thing you're interested in. WikiDao ☯ 02:08, 24 February 2011 (UTC)

There was a great part in The gods must be crazy I remember from my childhood where some animals got drunk on the fermenting fruit of some African tree. You can find the segment on youtube. Vespine (talk) 07:32, 24 February 2011 (UTC)

I think that was in Animals Are Beautiful People (by the same writer-director). -- BenRG (talk) 09:23, 24 February 2011 (UTC)

From personal experience, if you give a dog a couple of beers, he pretty much gets sleepy. Not especially entertaining. Googlemeister (talk) 14:18, 24 February 2011 (UTC)\

This article [24] is about a serious case of drunk monkey business. 10draftsdeep (talk) 17:49, 24 February 2011 (UTC)

Question on collision between hydrogen atoms (momentum / energy question)

Hi all,

I was doing my physics mid term exam today and I came across a question that I couldn't solve. Turns out it's from the textbook. Here it is:

When two hydrogen atoms of mass m combine to form a diatomic hydrogen molecule (H2), the potential energy of the system after they combine is -Δ, where Δ is a positive quantity called the binding energy of the molecule.

(a) Show that in a collision that involves only two hydrogen atoms, it is impossible to form an H2 molecule because momentum and energy cannot be simultaneously be conserved. (Hint: If you can show this to be true in one frame of reference, then it is true in all frames of reference.)

(b) An H2 molecule can be formed in a collision that involves three hydrogen atoms. Suppose that before such a collision, each of the three atoms has speed 1.00 * 10^3 m/s, and they are approaching at 120 degree angles so that at any instant, the atoms lie at the corners of an equilateral triangle. Find the speeds of the H2 molecule and of the single hydrogen atom that remains after the collision. The binding energy of H2 is 7.23*10^-19 J and the mass of the hydrogen atom is 1.67*10^-27 kg

Any help would be appreciated. —Preceding unsigned comment added by 164.67.66.113 (talk) 01:16, 24 February 2011 (UTC)

Consider the simplest possible case for (a), where the final product has no net velocity. To conserve momentum, each atom must approach with the same speed. Now calculate the kinetic energy before and after; and account for the binding energy, and determine if the reaction is possible.

For the second case (b), I think the problem is actually more straightforward, because they give you all initial conditions, and most of the final conditions; (so you don't need to "intuit" anything - just calculate). Nimur (talk) 01:31, 24 February 2011 (UTC)

Is there an assumption that hydrogen can not store excess energy as anything except kinetic energy? Can't it store the binding or collision energy as excited electrons (assuming the energy levels precisely match)? Can't diatomic hydrogen "bounce", i.e. stay bound but oscillate? Or slightly miss each other and rotate at high speed? Ariel. (talk) 05:08, 24 February 2011 (UTC)

That was the implicit assumption; energy is either kinetic or the fixed-value of the "binding energy." The validity of this assumption is at least partially predicated on the "perfect" alignment of inbound particles. (They can't start rotating if they had no rotational momentum in the initial condition). And electron energy might be negligible; you can actually calculate whether the kinetic energy is orders of magnitude larger than the hydrogen ionization potential, 13.6 eV; while the supplied binding energy is about 4 eV; so neglecting energy in the electron-orbitals might be a bit of a tenuous assumption. You can similarly compare order-of-magnitude to determine if diatomic oscillatory motion is "negligible." This is, after all, a hypothetical, simplified treatment. In practice, we should apply a nuclear scattering treatment and solve quantum-mechanical wavefunctions, and represent the collision as a quantum mechanical operation; but from the wording, I suspect that material is outside the scope of the physics class. Nimur (talk) 05:21, 24 February 2011 (UTC)

I suppose I feel the same as Ariel. The fact is, it is possible for H₂ to form in a collision of two hydrogen atoms, with the excess energy being shed somehow (probably as photons, but there are other ways). I guess it's unlikely, but it doesn't violate any conservation laws. Many physicists would describe this as "a collision that involves only two hydrogen atoms". I think there's no worse question you can pose to a student than "show X" where X is not really true. Aside from being misleading, it penalizes the students who understand the material best, since they are more likely to notice the hole in the obvious (intended) proof and waste time searching for a better one. -- BenRG (talk) 10:08, 24 February 2011 (UTC)

Exactly why can't momentum and energy be simultaneously conserved in a collision? Energy is always conserved. Kinetic Energy isn't conserved. And why can't you create H₂? If you send them towards each other at a high enough speed, assuming nuclear fusion does not occur, then they should collide inelastically, and you can calculate the required relative velocity with: ${\frac {1}{2}}\mu v_{(}rel)^{2}=\Delta$ (where μ is the reduced mass of the system). ManishEarth^{Talk • Stalk} 12:30, 24 February 2011 (UTC)

What is this compound?

Resolved

I'm actually doing a course in organic chemistry right now and yet I still can't figure out the name of this. Any help? --✶♏✶ 04:23, 24 February 2011 (UTC)

Looks like 5-hydroxytryptamine, or serotonin. Here's the standard numbering system for tryptamines [25], which is a little confusing to the uninitiated... and needs periodic rechecking by the initiate as well, I'm afraid. Wnt (talk) 04:53, 24 February 2011 (UTC)

Thanks! Thinking of it, that should have been one of my first guesses for "organic compound someone would tattoo on their body". --✶♏✶ 07:09, 24 February 2011 (UTC)

I don't see the appeal. Sounds painful... Wnt (talk) 16:18, 24 February 2011 (UTC)

Consequences of (admittedly magically) preventing every cell in your body from ever dividing again?

I understand that many parts of the body rely upon rapid cell division to maintain state (at least what we perceive at the macro scale, it's a Ship of Theseus thing). What would happen if you could snap your fingers and prevent every cell in your body from ever dividing again - in effect "freezing" the current state? Cells that die would not be replaced, clearly. I'm curious what would start to breakdown first from a medical perspective. I imagine death would not be long in coming, but I'm unsure as to how one would die? The Masked Booby (talk) 05:13, 24 February 2011 (UTC)

This should be pretty much like radiation poisoning - hematopoiesis (production of red and white blood cells) fails, the intestinal lining no longer regenerates, skin and hair can suffer, etc. I'm not sure offhand why headaches are a symptom of radiation poisoning and I'm not sure if the magical mechanism would replicate them. Wnt (talk) 05:29, 24 February 2011 (UTC)

See also Labile cell and Hayflick limit. Ariel. (talk) 05:35, 24 February 2011 (UTC)

I think your stomach juice would make short work of you. Vespine (talk) 07:26, 24 February 2011 (UTC)

Well, red blood cells have a lifetime of about 100 days, so in a couple of months you would die of anemia if nothing else killed you sooner. Looie496 (talk) 07:31, 24 February 2011 (UTC)

Yes, it's more the immune effects of lethal irradiation that tend to be a problem - though completely disabling gastrointestinal regeneration would probably work the quickest end here, barring extraordinary interventions. I suspect that lethal irradiation is relatively less damaging to the slowest-growing stem cells of the intestine than a magic stop to all cell division would be. Though there's also a certain uncertain ability of hematopoietic stem cells to seed the intestine when need be.^[26] Wnt (talk) 16:17, 24 February 2011 (UTC)

A human being's body experiences about 10,000 trillion cell divisions in a lifetime, see Cell division. The OP's magic snap, if its effect could be focussed, would incapacitate Cancer cells. Cuddlyable3 (talk) 10:35, 24 February 2011 (UTC)

Battery rack earthing

Whether battery rack has to connected to the safety earth pit or not? If it has to be connected what is the reason behind this? If not then why it should not? — Preceding unsigned comment added by Magesh1581 (talk • contribs) 12:01, 24 February 2011 (UTC)

You might want to tell us which sort of battery rack you are talking about, but in short, the answer is probably yes. The reason is that earthing the rack prevents harm should an inadvertent connection be made between the rack and the positive terminals of the batteries. Charge will be routed to earth, rather than to the next poor soul who touches the rack. --Tagishsimon (talk) 12:31, 24 February 2011 (UTC)

Electric Field strength at a distance x from an infinite charged sheet

I was solving a problem based on Electric Field, and I noticed, that, in my derived formula for electric field at a distance x from a charged disc of radius R and charge density σ, $E={\frac {x\sigma }{2\epsilon _{0}}}({\frac {1}{x}}-{\frac {1}{\sqrt {R^{2}+x^{2}}}})$
When R tends to infinity, the surface becomes an infinite charged plane, and E is: ${\frac {{}\sigma }{2\epsilon _{0}}}$

Note that it's no longer dependent on x, i.e. the magnitude of the field is the same at all points in space (Except those on the plate)
I rederived the same formula for the infinite case with Gauss' theorem (the surface is a cylinder with its height perpendicular to the plane), and by summing up infinite strings.
My question is, why is E no longer dependant on x? Thanks, ManishEarth^{Talk • Stalk} 13:49, 24 February 2011 (UTC)

I fixed your formula for you. Your answer is correct. The field doesn't depend on x. Just use Gauss' law in a cylinder of length 2x with the charged plane placed symmetrically at the center of the cylinder to get the same result. Dauto (talk) 15:47, 24 February 2011 (UTC)

Semen omelet

Unpalatable question, but curiosity is king: Could you make an omelet or a quiche or something out of semen? For the sake of decency, let's say it comes from a free-range bull fed on alfalfa and organic blueberries. Perhaps I'm wrong to suppose that the substance would have the culinary properties of egg. If not, what would it be like when cooked? (Bonus question: Have the Japanese already done this?) LANTZY^TALK 14:05, 24 February 2011 (UTC)

Not a direct answer, but are you aware of Rocky_Mountain_oysters? SemanticMantis (talk) 15:30, 24 February 2011 (UTC)

Focusing camera mounted on biological gel imager

On the top of our gel imaging equipment is a camera, allowing us to capture digital images of fluorescent bands of (usually DNA or RNA). The camera has a ring that we can adjust for zoom. It also has another ring which appears to alter exposure, which I surmise is the aperture. If I adjust the zoom, must I also adjust the aperture for a properly focussed image? Or will the image be equally focussed at all zoom level?

Changing the zoom on a lens is pretty fundamental, and you'll probably have to change the focus by a lot. Changing the exposure (time) or aperture (diaphragm) depends more on how desperate you are to get that faint little band you see to show up on the picture. As a rule, running the gel is enough of an expenditure of time and effort, even money, that you shouldn't hesitate to fiddle with every control on the camera trying to make it better (especially when it's digital!) ... just remember what it was before you started and put it back again so that people don't start murmuring. Wnt (talk) 16:05, 24 February 2011 (UTC)

Knowing what camera would help to give a better answer. Aperture affects the depth of field and the best (and same) f stop should satisfy all images in this application. This leaves only the speed to consider. So in other words: use the recommended f stop and don't fiddle with it!--Aspro (talk) 16:34, 24 February 2011 (UTC)

There are conceivable reasons to fiddle with an f stop - for example, depending on the thickness of your gel and thus the thickness of your bands, or if the surface of the gel is showing up too much for some reason and you want to try to blur that out a bit - but admittedly most of the time this isn't that important to fool with. Wnt (talk) 17:17, 24 February 2011 (UTC)

Note that if you zoom in more, the image will get darker. Some fixes for this are:

1) A brighter light. But too much might cook your sample.

2) A wider lens or lens opening (aperture/f-stop).

3) A longer exposure. Could cause motion blur if the subject moves.

4) Post-processing to brighten the picture. There's a limit to how much can be done this way, though. StuRat (talk) 17:11, 24 February 2011 (UTC)

1) When is flash going to cook a southern blot

2) What's wrong with keeping Aperture#Optimal_aperture

3) Just how fast do you think a southern blot needs to gallop along to cause blur?
--Aspro (talk) 17:28, 24 February 2011 (UTC)