The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.
No. (Except in ridiculously contrived situations involving deliberately modulated impossibly intense lasers or the like. But even so, the light that passes through a medium is completely silent; any sound would be due to light absorbed by the medium). –Henning Makholm (talk) 01:19, 23 October 2010 (UTC)[reply]
Well, not necessarily 'impossibly' intense (where for 'impossibly', I read 'sufficiently intense for nonlinear optical phenomena to occur'). Photoacoustic imaging discusses a technique relying on conventional optical absorption of pulsed light to generate transient sounds through local thermal expansion of the object under study. (See also photoacoustic spectroscopy, for another application. ) I agree with Henning, however, that sound travelling in a straight line through a transparent medium doesn't (or shouldn't) make a sound. TenOfAllTrades(talk) 01:36, 23 October 2010 (UTC)[reply]
Here you possibly get into misunderstandings based on the wave-v-particle nature of light. A photon that doesn't hit anything can't make a sound, right? But also it should travel at exactly c. However light in matter does not in fact travel at c.
You can describe this, if you like, as a bunch of photons that always travel at c, but are at intervals absorbed and re-emitted, and presumably if you average over all the relevant Feynman diagrams you come up with the right answer, but this is kind of unsatisfying because we want to know which Feynman diagram described what actually happened, and there isn't any answer to that. Also the way they all average out to give the right refractive and diffractive effects and so on seems kind of mysterious.
So it's much cleaner in that sort of context to use a wave description, with electric and magnetic fields.
And with a wave description, it's not implausible that in extreme cases the medium could respond to the electric and magnetic fields in a way that produces sound. --Trovatore (talk) 18:02, 23 October 2010 (UTC)[reply]
You and Henning, yes. Specifically to your final sentence about the transparent medium. Though admittedly I hadn't thought much about just what you meant by "transparent". --Trovatore (talk) 18:11, 23 October 2010 (UTC)[reply]
[I asked this a while back at Talk:Slow light, but without result] What happens when the speed of light in a medium is precisely equal to the speed of sound in the same medium? Can the light be transformed to sound in some kind of "sonic boom"? Can the sound be transformed to light in some kind of sonoluminescence? Is there an equilibration of energy between the two transmission modes? I wonder if there's something to use in solid state refrigeration. (hmmm, that last should not be a red link). See also [1] - pity I can't access it and don't know Mandelstam-Brillouin scattering or what an optical-acoustic soliton is... Wnt (talk) 01:32, 23 October 2010 (UTC)[reply]
Similarly, people often report hearing a sharp cracking sound from lightning, but this is usually the first audible expansion of air molecules caused by the static electrical discharge and not caused by the light itself. ~AH1(TCU)17:55, 23 October 2010 (UTC)[reply]
..dear friends..
my paddy crop drooped because of heavy rains and winds.Is there any remedy for that with agricultural scientists.Main problem is that grains were not yet grown properly. —Preceding unsigned comment added by Sameerdubey.sbp (talk • contribs) 02:23, 23 October 2010 (UTC)[reply]
A uniform 2m rope is resting on a table, with linear mass density λ. One end is slowly raised until half of the rope is left on the table. What is the height of the center of mass?
My attempt: If the end is being raised slowly, then a force with a vertical component of λlg needs to be applied, with l being the amount of rope off the table. I wanted to use this to find the work done, and then equate that with the total potential energy 2λghCM. But I need to know the max height of the rope, which is not given. I guess I could solve for it, knowing that the rope off the table assumes a catenary, but there has to be an easier way. 76.68.247.201 (talk) 02:43, 23 October 2010 (UTC)[reply]
You're making this problem harder than it is. In order for there to be a well-defined center of mass with the information given, the problem must be making the assumption that the portion of the rope that isn't on the table is purely vertical. (Consider the extreme cases to see how the center of mass isn’t otherwise well-defined.) Given the above assumption, it's clear what height the top of the rope is at from the specifications that the rope is 2m long, and that half of it is on the table. You don't need to mess around with force or potential energy for this problem; you can easily determine the center of mass directly from the definition of center of mass. Red Act (talk) 03:46, 23 October 2010 (UTC)[reply]
I'm sorry, but why wouldn't the center of mass otherwise be well-defined? If the portion of the rope that isn't on the table were magically a straight line at 45 degrees to the table, the height of the center of mass would still be calculable (h = 1/(4sqrt(2))). 76.68.247.201 (talk) 06:01, 23 October 2010 (UTC)[reply]
(ec) :::Yes, you can calculate the center of mass for the rope if the free portion is at any angle to the vertical (just multiply by the sine of the angle), or even if it lies along any algebraically defined curve (and possibly some others), but the point that Red Act was making is that this is a statics problem, you don't need to consider dynamics at all to get an answer (though it is not wrong to use your method). More interesting problems are created if λ is not uniform, or in calculating the motion of the rope when it subsequently slips off the table (assuming no friction). Dbfirs07:41, 23 October 2010 (UTC)[reply]
It seems to me that the problem is not sufficiently well defined: in other words, the answer is "There is insufficient information to answer the question". As noted above, the rope might be lifted in such a way that half of it is vertical, but it might also assume a catenary shape. We aren't told whether the end is being lifted straight up or what. Clearly different final positions are possible depending on how the rope is lifted. --Anonymous, 08:15 UTC, October 23, 2010.
Yes, agreed. If the rope is lifted straight up, then the answer is just 0.25 m of course, and if it forms a catenary then the center of mass is lower, possibly much lower, depending on friction and coiling). Dbfirs10:38, 23 October 2010 (UTC)[reply]
I've looked all over the internet for information regarding induction of flowering in Ocimum, but I have been unable to find any. My specific question is this: What causes Ocimum to be induced into the flowering stage? Is it photoperiod, age, temperature, a combination of these factors, or none of these? Thank you, 76.78.185.248 (talk) 08:03, 23 October 2010 (UTC)[reply]
Currently in physics, we are studying moment of inertia. I have tried to understand it but am having a very hard time understanding it conceptually as well as mathematical calculations involving it (stemming from my lack of concept). What are a few of the ways to conceptually understand moment of inertia? TIA, Ζρς ιβ'¡hábleme!08:11, 23 October 2010 (UTC)[reply]
I found it helpful to say in linear motion the concept of mass is very important. In angular (rotational) motion, moment of inertia plays the same role as mass in linear motion.Dolphin(t)08:16, 23 October 2010 (UTC)[reply]
Dolphin's is a very good starting point; in the same way that mass tells you how hard it is to accelerate/decelerate something given a certain force (remember that it is inertial mass, when all's said and done), moment of inertia tells you how hard it is to change something's rate of rotation with a certain torque. It might help to think about a whole set of correspondences (these are all in the sense of linear --> rotational):
Force --> Torque/Moment (depending on which word you go in for)
Velocity --> Angular velocity
Acceleration --> Angular acceleration
Momentum --> Angular momentum
Mass --> Moment of Inertia
And so on. If you look at more complicated derived quantities you still see some fairly direct correspondences -- for example, KE is given by while rotational energy is given by .
The thing that does make it all a bit harder than linear motion is that you have to start doing cross products between vectors to do it properly, but for a lot of the things you're trying to work out that's often hidden under the hood. --81.153.109.200 (talk) 10:48, 23 October 2010 (UTC)[reply]
The submarine reactor is cooled by sea water. I guess they boiled off some water to lighten the vessel to help get it off the ground at high tide. Cuddlyable3 (talk) 12:52, 23 October 2010 (UTC)[reply]
No, they aren't boiling off water just in order to lighten the sub. If there was water in the sub that they wanted to get rid of in order to lighten the sub, it'd be vastly more efficient to just pump it out instead of boiling it out. What I presume is going on is just a normal part of being powered by a nuclear reactor. Nuclear reactors generally work by using heat from the nuclear reaction to power steam turbines. Red Act (talk) 16:06, 23 October 2010 (UTC)[reply]
It should be noted that (according to our article) the Astute class submarine is equipped with a pair of 600 kW diesel generators in addition to its nuclear reactor. Cleanly-burnt diesel exhaust is mostly carbon dioxide and water vapour, and could produce the plume seen. I don't know where the exhaust snorkel is located on the Astute, but I certainly wouldn't be surprised if we were seeing the sub running on its diesels. (After running aground, they could have chosen to shut down their nuclear reactor as a precautionary measure.) It's almost certainly not exhaust from the Astute's steam turbines; venting bubbles of hot gas underwater is an excellent way to attract the attention of enemy sonar operators, and the turbines would be designed to chill and condense their exhaust with seawater. (Indeed, I would expect the steam to stay in a closed-loop system; the condensed fresh water would be recycled and reheated, not discarded.) TenOfAllTrades(talk) 18:19, 23 October 2010 (UTC)[reply]
Almost certainly diesel exhaust and (diesel) cooling water - the reactor would likely have been shut down to keep its cooling water loops from ingesting seafloor muck and debris, so the auxiliary diesel would be operating. Acroterion(talk)19:25, 24 October 2010 (UTC)[reply]
The steam generated by the kettle is used in the propulsion system. Nuc propulsion essentially is just a big water heater and the drive is steam.
The smoke discussed is from one of her diesel gennies.
Snopes says that the story of miraclly-efficient cars getting 200 MPG, but getting reclaimed by factories, disappearing, or their patents being bought up by oil companies, are false.
200 miles per gallon! It has been said that if it seems too good to be true, most likely it isn't true. These miracles of efficiency, and engines that run on water, are in the same class as perpetual motion machines - if they happened they would disprove the laws of thermodynamics.
I just get the feeling that the oil companies (or whatever organizations/entities where higher fuel consumption would be in their best interest) would buy up designs for highly-efficient engines. (And that the hybrid systems like the Prius and Chevrolet Volt has, weren't bought up probably thanks to recent government intervention.)
This is the Science Reference Desk. In science, we give no credit to feelings. Not even gut feelings. Ever since Galileo Galilei (died 1642) we have focussed exclusively on evidence and reasoning. Until you have some objective information your ideas don't really qualify for the Science Reference Desk. Prior to Galileo the whole world relied on intuition and superstition; and look where it didn't get them. Dolphin(t)10:51, 23 October 2010 (UTC)[reply]
This question feels a bit parochial to me, somehow. Note that in other parts of the world there are designs in common use which are far more efficient than those used in the majority US cars. Not at the "miracle" level necessarily, but still Bad News for the oil companies under your conspiracy theory scenario. Any explanation you come up with needs to account for why these aren't being bought up and quitely sidelined by the big bad shadowy conspiracy people, rather than being put into mass production. --81.153.109.200 (talk) 10:56, 23 October 2010 (UTC)[reply]
When 81.153.109.200 refers to other parts of the world that have engine designs more efficient than those in the US, I assume he is referring to parts of the world that don't have environmental legislation related to reducing pollution from motor vehicles. It is true that anti-pollution equipment reduces the fuel efficiency of the internal combustion engine. Dolphin(t)11:03, 23 October 2010 (UTC)[reply]
Probably talking about parts of the world that tax gasoline, increasing the economic incentive for (more expensive) efficient engines. The "anti-pollution = lower fuel efficiency" argument sounds like another of the urban myths that flies around about the subject. Physchim62(talk)11:14, 23 October 2010 (UTC)[reply]
No, not an urban myth. Unfortunately, in internal combustion engines the air-fuel ratio that yields maximum energy output is not the same air-fuel ratio that yields maximum fuel efficiency, and neither of them are the air-fuel ratio that yields minimum harmful emissions. So, like most engineering endeavours, some compromises are required. The end result is that the design engineers can optimise fuel efficiency but at the expense of emissions, or they can optimise emissions at the expense of fuel efficiency. And the best of those compromises will probably be found in the most expensive engine. Someone once said something like every successful engineering achievement is a thousand human judgements; none of them perfect and none of them foolish.Dolphin(t)11:33, 23 October 2010 (UTC)[reply]
I'm fairly sure that it is an urban myth. Both from an environmental and from a thermodynamic point of view you want complete combustion. Yes, it may be easier to achieve this with a leaner mixture, but the effect is minimal, especially with modern catalytic converters. And you can buy nearly the same fuel efficient cars in the US than e.g. in Europe (and vice versa). The reason why the average US car is relatively inefficient has little to do with engineering trade-offs, and more with prices and cultural preferences. In the US, fuel is cheap, and space is (mostly) plenty. As a result, comfort and convenience are valued higher than fuel efficiency and easy parking. To a certain degree, you can have both, but it will cost you. Modern BMWs, e.g. have automatic engine shut-down when the engine idles, and restart automatically when needed (I think that is illegal in some US states for unclear reasons). They also have cooling air inlets that open and close both to maintain an optimal engine temperature and to improve the aerodynamic profile of the car. All this engineering has a price, and when you pay US$35 for a tank of gas, you may opt to rather have extra cup-holders and a trunk that can fit 5 corpses and a sack of cement is big enough for that rare big load. If you pay US$70 per tank, things may look different.--Stephan Schulz (talk) 13:01, 23 October 2010 (UTC)[reply]
I also wonder what these countries are with their highly efficient engines due to lax regulations anyway. AFAIK, European countries tend to have decent regulations, particularly Western Europe. The countries with lax regulations tend to be developing ones. Price of the vehicles tend to be important there so small not too fancy vehicles sometimes from joint ventures and stuff. Second hand imports are sometimes common. In other words the tech level doesn't tend to be very high so I doubt that they have these ultra efficient engines even if it's possible. And many of these countries have subsidised fuel so relative to the price of the car, I don't think fuel cost tends to be quite so important. As StS said, cars in the US are known for being big and what many in other countries would consider overpowered so vehicles there tend to have low efficiency relatively and one of the reasons is the relatively low cost of fuel compared to much of the rest of the developed world so efficiency doesn't tend to be so important there. So the cars do tend to be relatively inefficient but it's not an inherent limitation in the engines per se as far as I know. Nil Einne (talk) 13:31, 23 October 2010 (UTC)[reply]
Patents are public documents. If the oil companies have bought up the patents to 200 mpg cars, what are the numbers of those patents? We could all go to the USPTO website (or one of many similar sites worldwide) and look them up... except we can't, because the people spreading the conspiracy theories don't give the patent numbers... because the patents don't exist... because the whole story is just an urban myth. Physchim62(talk)11:14, 23 October 2010 (UTC)[reply]
What's obviously happened is that the oil companies have just bribed the USPTO to remove those patents from their records. And yes, the earliest people to know about those leaked patents presumably made copies of the patents, or at least wrote down the patent numbers, but the oil companies just bribed any people like that to keep quiet, such that that information is now lost. You'd think some fraction of that large number of people the oil companies needed to bribe to keep this information quiet would have refused to be bribed out of principle, but I'm sure the oil companies just assassinated any of the people who couldn't be bribed. There aren't any records of people like that who have gone missing, but that's clearly because the oil companies have bribed the FBI to keep that information quiet. In fact, the more I think about it, I think the oil companies must have bribed pretty much everybody to keep this quiet, except for me and you – and I'm not too sure about you. ;) Red Act (talk) 17:13, 23 October 2010 (UTC)[reply]
Red Act, you have uncovered an inconvenient truth. I know too much about all this stuff and when the Rich Oil Companies offered me squillions of dollars to remain quiet I refused on principle. They killed me! To make matters even worse, they didn't bury me. I had to do that myself. 203.129.53.46 (talk) 22:52, 23 October 2010 (UTC)[reply]
By posting this, you are merely covering for yourself--what better way to maintain the flow of Big Oil hush-money than by publicly denying you receive it! I bet they even paid extra for that last sentence falsely accusing me of being in on the scheme. I don't need their dirty money (I'm already on the take as part of the Lumber Cartel) so I'm not afraid to speak the truth about the situation, even including your own involvement. DMacks (talk) 17:40, 23 October 2010 (UTC)[reply]
I just realised you are all right, particularly the OP and big oil is covering up all these wonderful inventions. I'm gonna go spread the world in a few days! Probably. Unless for some reason I decide not to. But I don't think one reason would be enough. I think closer to 100 million reasons would be ideal. BTW in case anyone is interested, I can be reached by e-mail via by user page. Nil Einne (talk) 23:02, 23 October 2010 (UTC)[reply]
One of the things that distinguishes snopes.com is that they always give references. I suppose their reference sites might also have been bribed by the oil companies ... --ColinFine (talk) 00:27, 24 October 2010 (UTC)[reply]
Mine already do, and they call them stock dividends.
How to find an astronaut
The astronaut in question, lost on an alien world - anywhere, perhaps thousands of kilometres away - is wearing a spacesuit with biotelemetry - but the telemetry doesn't have much range - say a hundred kilometres. What equipment would you need to save them? A satellite? A radio receiver? What if you didn't have a satellite? Would you just barrel around in your rover scanning the landscape?
I'm open to all manner of discussion and speculation here.
It would seem odd to send one astronaut alone to another world, as the Apollo program always sent three astronauts at a time to land on the Moon. Also, as the Moon is the only solid object always within 1 million km of Earth, you'd have trouble applying this question to any object other than the Moon. For future planet exploration missions such as a hypothetical manned mission to Mars, it's possible that a transmitter would be planted on the original spacecraft, so that another astronaut could report his/her lost colleague to the transmitter and the information could be sent to Earth. ~AH1(TCU)17:17, 23 October 2010 (UTC)[reply]
It is depending on how you got there. By a stargate you would have to rely on short wave communication a small relay balloon or a plane, because the limited transmission distance is (I assume) due to the curvation of the planet. If you go there with a space ship you should leave a satellite in orbit doing the communication. The MER rovers have direct to earth and contact to the satelite and they are less than 200kg over all so it should be no problem to have a iridium telephone with you on an distant planet. You will only have a short communication window every pass of the satellite but this is enough.--Stone (talk) 20:47, 23 October 2010 (UTC)[reply]
The idea that "the telemetry doesn't have much range - say a hundred kilometres" is an arbitrary one. The range depends on the receiver. We can just build a bigger, more powerful one. Of course, he may be dead by the time we find him. HiLo48 (talk) 20:57, 23 October 2010 (UTC)[reply]
It seems downright negligent to embark on the manned exploration of a planet without first erecting an effective network of satellites for positioning (like GPS), satellite communication (like Iridium), radar telemetry (like SRTM), and optical and IR land observation (like Landsat). Even with current technology, our astronaut could carry an iPhone and an Iridium phone in her pockets; when she gets lost she looks up her location on Google Planet-X Maps and phones her location into her chums back at the base. If she's going off gallivanting by herself, we'll surely make her take a spare phone and a couple of extra batteries, and make her file some kind of route. In the event that she goes off the reservation, chucks her phones, sabotages the transceiver we put in her rover, and drives off into the sunset, we can have the satellites look for her. What sensors are best for that depend on the planet. If it's an airless dead place like the moon, she and the rover have IR and radar signatures that stick out like sore thumbs. If the planet has an atmosphere we can send aeroplanes, helicopters, and dirigibles after her (which is technologically much the same task as drone aircraft do now over Afghanistan). There's a limited amount we can do if the planet is so densely vegetated (a la Pandora) that she can't readily be distinguished from the environment, but I suppose there's always napalm... And all of this envisages going there with technology that's essentially what we have now; surely by the time we've figured out how to get someone even to Mars, satellites are cheaper and have longer endurance and their sensors more accurate and able to sweep more quickly. One thing military planners are thinking about is building tiny sensor bomblets that you spray across a target area - say we know she's somewhere in sector 4 of the Pandora forest, so we drop 100,000 of these 50¢ sensors over the whole sector - they operate for a week and have IR/UV/optical and maybe radar proximity sensors, and phone home any movement they see. The supercomputer back at base sorts out the clutter (wind moving tree branches) from the interesting stuff (blue native guy carrying off our plucky astronaut) and we can send the freedom copters in to "liberate" her. -- Finlay McWalter ☻ Talk21:34, 23 October 2010 (UTC)[reply]
"What's that iPhone for" you may ask. Firstly she uses it when back at base (where there's a cell tower), so she's not clogging up the Iridium with chitchat. Secondly it's got the GPS, so she knows where she is. And thirdly she can play Angry Birds to pass the time until we come to get her. -- Finlay McWalter ☻ Talk21:38, 23 October 2010 (UTC)[reply]
I am confused by the statement in the recent did you know article UDFy-38135539: "...has been calculated to have a light travel time of 13 billion years[1] with a present distance of around 30 billion light-years..." I thought if a star's light took X years to get here it was ipso facto X number of years distant. I've been thinking about it and I may have divined the answer but I'm not sure I'm right at all. If the star is moving away from us and we from it then I guess during the light's travel time we and and the star could have moved apart 17 billion light years during the intervening 13 billion years the light was traveling resulting in a total distance of 30 billion light years. Am I on the right track?--141.155.156.196 (talk) 12:04, 23 October 2010 (UTC)[reply]
I may as well trot out my diagram again. This is for an object that's 28 billion light years distant, so it's pretty close to the case you're talking about. The brown line on the left is Earth, the yellow line on the right is the distant object, the diagonal red line is the light (always at 45 degrees to the grid lines), and the orange line across the top is the "present-day distance" that people usually quote. The yellow line covers about 13 grid squares (13 billion years) between emitting the light (red) and the "present day" (orange), and the orange line crosses about 28 grid squares (28 billion light years). -- BenRG (talk) 02:25, 24 October 2010 (UTC)[reply]
DNA
I often hear about America and the UK having "DNA databases" of criminals. So when a crime is committed and the police find some DNA, how do they check it against the DNA database? Is it all on computers or do they have to compare samples manually? —Preceding unsigned comment added by 96.226.69.239 (talk) 12:56, 23 October 2010 (UTC)[reply]
DNA profiling explains. Essentially they boil a DNA sample down to a small-ish array of numbers, and the database is a collection of records (storing for each a similar array of numbers). The search examines the database for arrays that (mostly) match the sought array. Matches are then compared by a human expert. Much the same process is done for normal fingerprint searches too. -- Finlay McWalter ☻ Talk13:29, 23 October 2010 (UTC)[reply]
Laika was often mentioned in children's books about space travel when I was growing up in the 1960s. No one ever mentioned that she was dead! Alansplodge (talk) 23:08, 25 October 2010 (UTC)[reply]
What's the smallest sized computer chip you could fit the entire Library of Congress onto? I read an article from 1986 that said someday we could fit it on a chip the size of a sugar cube, so I'm curious how far we've gotten so far.
Our Library of Congress article states that the popular estimate is 20 petabytes of books and similar publications. Memory chips haven't really ever been "thick" (at least since back in the days of core memory), so the sugar-cube as a reference size is weird. But Random-access memory says that there are developmental/prototype RAM chips that can easily hold that much data. Even if LOC is actually an order of magnitude larger (from non-textual sources), those chips are still on that same rough capacity. DMacks (talk) 16:07, 23 October 2010 (UTC)[reply]
Latest-and-greatest technology in 2010 permits stacking maybe two (or even three) silicon dies in a SiP or stacked Multi-Chip Module. These are still pretty "flat" shaped; the total height of a silicon die is ~ 1/4 to 1/2 mm. (So stacking three high still gives you about a 1mm thick form-factor). To my knowledge, there is no way to create a "cubic" form-factor by stacking more chips vertically. The practical engineering details (like alignment, reliable wire-bond, potting thermal hazards, and so on) haven't been worked out yet for such a tall stack. I can think of no serious theoretical limitation which would prevent an arbitrarily-tall stack of silicon, but economic forces would need to push the technology development in that direction (which is unlikely). (Maybe power efficiency would suffer if the stack got very tall - a lot of power would have to flow vertically through metal interconnect instead of using copper PCB traces in a flat form-factor). Nimur (talk) 18:36, 23 October 2010 (UTC)[reply]
Our Library of Congress article is wrong and should be corrected, our article talks about the data in plain text while the source talks about scanned images. The calculation in our article 20 million books 1 MB/book gives 20 TB not PB.
It is very hard to give the size of a library in bytes whiteout specifying the storage format, a uncompressed scanned high resolution image of a page can be 3GB, the plain text can be about 3 KB and with state of the art compression it can be stored in 300 bytes, an difference by a factor 10 000 000. In addition to this most libraries contains music, movies and so on how are they counted, they are often already digital and requires much data storage.
If 20 PB of data is stored on microSD cards they can fill a cube 0.5 m*0.5 m*0.5 m, I think this is the most compact storage in any consumer device. If the data were to be stored without a mechanism for accessing it, it could be done in a much smaller volume by a atomic force microscope manipulating atoms. --Gr8xoz (talk) 18:44, 23 October 2010 (UTC)[reply]
I haven't built a circuit with memory chips lately, but in early ttl chips, you could easily tell logic from memory chips because the memory chips were warm. So I wonder if the latest memory chips have problems with failure due to high temperature if they are tightly stacked without provision for cooling? Edison (talk) 19:55, 23 October 2010 (UTC)[reply]
Flash chips can be turned off when not reading or writing that particular chip so that should be no problem if the data access patern are not to demanding. --82.209.130.40 (talk) 20:09, 23 October 2010 (UTC)[reply]
TTL chips draw current all the time hence the warmth of MSI chips with many gates. Modern CMOS chips at rest draw much less current and therefore remain cool unless being read or written at a high rate which causes them to heat. The heating is due to currents charging and discharging the internal and external capacitances. Cuddlyable3 (talk) 18:56, 24 October 2010 (UTC)[reply]
My guess is the sugar cube thing is partly to do with expectations, since the concept of data cubes in science fiction is not uncommon. Also particularly in the days before things like floppy disks let alone CDs, flash memory cards etc were something people were familiar with, a sugar cube may be easier to imagine and seem smaller to someone then say, a half credit card size card 5 mm thick even if the later is more likely. Chewing gum stick may be an alternative of course. Nil Einne (talk) 00:56, 24 October 2010 (UTC)[reply]
Hi. Many Hot Jupiters have been found orbiting certain stars, some of them very close to their parent star. We also know that some of these star systems have gas giant planets orbiting near the star, and terrestrial planets further out. Since Jupiter's magnetosphere in our own solar system is very large and stretched out, sometimes to the orbit of Saturn, would a similar effect take place in extrasolar systems? Suppose that this "Hot Jupiter" orbits close to the star, and either has a rapid rotation rate or is tidally locked to its star, in which case the rotation rate would still be quite rapid. The planet would likely have a powerful magnetic field, and the stellar wind would be concentrated around the outside of that field. Would the strength of the radiation be considerable even near the end of this magnetosphere? Now, if a terrestrial planet in the habitable zone were to exist in an orbit reached by the "tail" of the gas giant's magnetosphere, this magnetosphere would often overlap the habitable planet, perhaps giving it a dose of stellar wind each time the edge of the field moves across the planet. Being in the larger planet's magnetosphere would of course shield the smaller planet from most incoming radiation, but would the intense bursts of solar wind be enough to limit the said terrestrial planet's ability to support life? Would the presence or absence, or strength of the terrestrial planet's magnetosphere change this greatly, and would possessing a magnetosphere for the small planet shield it from the gas giant's channeling of stellar radiation, or would the larger magnetosphere simply disrupt the smaller one? Have any theoretical modelling studies been done on this topic? Thanks. ~AH1(TCU)16:56, 23 October 2010 (UTC)[reply]
I do not know the answer but I think it is important to note that the reason we see so many Hot Jupiters are that they are easy to find, not that there are many of them.--Gr8xoz (talk) 18:56, 23 October 2010 (UTC)[reply]
About 2.3% of stars are likely to have habitable planets. The expected error of that measurement is a topic of interest among those who are interested in Fermi paradox-friendly colonization regimes. Ginger Conspiracy (talk) 08:28, 25 October 2010 (UTC)[reply]
Our article on the Joule–Thomson effect goes into the technical details, but not always very clearly. For an ideal gas, the temperature wouldn't change, but real gases are not completely ideal. If we stick to the normal situation (there are three exceptions at room temperature, gases which warm up when they expand), there is a slight net attraction between the molecules of the gas due to what are known as van der Waals forces. As the gas expands, the van der Waals forces get weaker: for a gas such as nitrogen or oxygen, the strength of the van der Waals forces is proportional to 1⁄d6 (where d is the distance between two molecules). Because van der Waals forces are an attraction between molecules, there is potential energy stored in that attraction. So some kinetic energy has to be lost to overcome the potential energy, or to "pull the molecules apart" if you prefer to look at it that way. So the average kinetic energy of the molecules drops, which is equivalent to a drop in temperature. Physchim62(talk)19:47, 23 October 2010 (UTC)[reply]
If you work through the thermodynamics using pV = nRT, you find that the temperature doesn't change on expansion: this is the ideal gas situation. A better approximation for a real gas is the van der Waals equation:
where a is a measure of the attraction between the molecules and b is a measure of the volume the individual molecules take up (not the gas as a whole, which is mostly vacuum between the molecules). Using the van der Waals equation, you will find a rise drop in temperature when the gas expands, for the reasons I described above. Physchim62(talk)20:03, 23 October 2010 (UTC)[reply]
PV=nRT says that temperature might decrease and/or pressure might increase if "volume increases" (as I understand your question setup). Both don't have to happen, and I assume the exact combination of those two results depends on the nature of the exact experiment. The ideal gas law page has a nice table of the specific cases for what "other variable(s)" are vs aren't controlled and the specific results/relationships. DMacks (talk) 20:09, 23 October 2010 (UTC)[reply]
(edit conflict) Note, though, that the change in temperature due to this effect will be quite small, in general. If you release a pressurized gas (even an ideal gas) into the atmosphere, on the other hand, it will cool quite appreciably, because positive work is done by the gas. Buddy431 (talk) 20:19, 23 October 2010 (UTC)[reply]
Imagine opening the valve of the cylinder of pressurized gas into the general atmosphere. The gas escapes, of course, but, in order to escape, it has to push the air out of the way to make room for itself. This is the "force" of the escaping gas, in the general sense of the word force. The amount of work done by the escaping gas is p∆V, where p is the atmospheric pressure and ∆V is the change in volume of the compressed gas. Physchim62(talk)21:49, 23 October 2010 (UTC)[reply]
It's a bit subtler than that, I think. If you have a pressurized container floating in the vacuum of space and poke a hole in its wall, the gas that escapes will pick up a macroscopic velocity away from the container, and the kinetic energy for this has to come from the heat energy of the original gas, so it must cool down as it escapes. However, if the vacuum the gas escapes into has walls and a finite size, the escaping stream of gas will eventually hit walls, be reflected towards other parts of the stream, and eventually all the kinetic energy will be lost to turbulence, at which point the equilibrium temperature will be the same as for the gas we started with, for reasons of energy conservation. –Henning Makholm (talk)
Not so. If we stick with the simplest case of non-turbulent flow, the molecules of gas that escape from the cylinder don't "pick up" a macroscopic velocity away from the cylinder. They simply keep the same velocity they had inside the cylinder (as per Newton's first law): the only thing that changes is there's no longer a cylinder wall to force them back inside. Physchim62(talk)21:49, 23 October 2010 (UTC)[reply]
I will not mince words about whether "pick up" is the right term to use. The fact is that the escaping gas has a macroscopic velocity, and the kinetic energy corresponding to this no longer counts as thermal once the molecules are no longer mixed with ones that go in the opposite direction. –Henning Makholm (talk) 23:33, 23 October 2010 (UTC)[reply]
I disagree that the kinetic energy can no longer be counted as thermal once the molecules escape from the cylinder: you seem to be choosing a particular definition of temperature in search of a paradox. The escaping gas molecules imply an impulse on the [cylinder + remaining gas] system, from conservation of momentum, but I don't think anyone is suggesting otherwise. The escaping molecules will still have a Boltzmann distribution of kinetic energies, even if all the initial velocities must have a component in the direction "out of the cylinder". If you don't wish to define temperature in terms of the Boltzmann distribution of (scalar) kinetic energy, how do you wish to define it? Any thermometer placed in the gas jet would only register a temperature change if gas molecules collide with it, immediately solving the apparent paradox of "unidirectional thermal motion". Physchim62(talk)00:02, 24 October 2010 (UTC)[reply]
There is no paradox. Temperature is the the denominator in the exponent of the Boltzmann distribution, which in non-quantum situations is proportional to the average kinetic energy of a molecule measured in the rest frame of the surrounding gas (if every observer were to choose his own frame, very fast-moving observers would think that almost all molecules had very high kinetic energies, in violation of the Boltzmann distribution). The energy of a closed system is constant. At the start of the experiment, the only energy of the system is in thermal motion. At some later time, there is also nonzero kinetic energy from the macroscopic movement of the gas. Since energy is conserved, at that point in time there must be less energy in thermal motion. None of this is paradoxical. No laws of thermodynamics are broken. Which "apparent paradox" do you think needs solving? –Henning Makholm (talk) 00:56, 24 October 2010 (UTC)[reply]
If you accept that temperature is the denominator in the exponent of the Boltzmann equation, then you should see that the "non-zero kinetic energy from the macroscopic movement of the gas" is no different from the "thermal energy" that was there before opening the cylinder. Hence the temperature cannot change for that reason alone (the ideal gas case). The temperature does drop for almost all real gases at room temperature, but not by any mechanism of the type you describe. In an isolated system containing a real gas (other than hydrogen, helium or neon) at 298 K expanding into an equally isolated vacuum, the final equilibrium temperature will be less than than the initial 298 K. Physchim62(talk)01:31, 24 October 2010 (UTC)[reply]
The macroscopic energy rises from 0 to something; therefore the microscopic energy that shows up as heat must fall similarly. Otherwise the First Law is violated. In the ideal gas case, and if the gas escapes into a bounded vacuum, the eventual equilibrium temperature will be the same as the initial one. At some points in time in before final equilibrium is reached, the temperature of some parts of the gas will be lower. (This can also be seen backwards: As the macroscopic energy of the gas is eventually lost to turbulence, the gas will be heated. Since it ends at its initial temperature, it must have been lower before the macroscopic flow ended). If the gas expands into an unbounded vacuum, total equilibrium will never be reached, because there will always be more space left to expand into.
Consider the isolated and rigid system as the cylinder and the bounded vacuum. In the case of an ideal gas, there can be no change in internal energy on releasing the gas: where would such a change come from? You can't say that the "macroscopic energy" changes, because the system is isolated, any energy change would remain within the system. No net work (because the system is isolated and rigid), no heat change ⇒ no change in temperature.
In the case of a real gas, there is a change in internal energy (loss of heat), because of the change in interactions between the molecules (absent in the case of an ideal gas). There is still no net work (because the system is isolated and rigid), but some kinetic energy is used to counteract the potential energy of the van der Waals forces (total energy conserved) ⇒ drop in temperature. Physchim62(talk)02:54, 24 October 2010 (UTC)[reply]
I agree, but we have to note that when the gas escapes and has a macroscopic momentum in one direction, it isn't in thermal equilibrium. You could assign it a lower temperature based on the motion of the molecules in the center of mass frame, but that can be a bit ambiguous. The gas certainly does have a lower entropy (which can be defined for general non-equilbrium states), compared to the final equilibrium state. It should be obvious that you can let the gas perform work when it shoots out of the cylinder. The entropy in fact stays exactly the same before thermalization happens. E.g. considering a thought experiment where you reflect all the gas molecules back into the hole where they escaped from. Clearly this is a reversible process as long as the molecules haven't thermalized. Count Iblis (talk) 02:51, 24 October 2010 (UTC)[reply]
You can compute the temperature change for the van der Waals gas quite easily. Internal energy stays constant in free expansion. The change in internal energy as a function of temperature and volume changes is given by
For a van der Waals gas, you can write the pressure as a function of T and V as:
P = N k T/(V - N b) - a N^2/V^2
Plugging this in the above equation gives:
dU = Cv dT +a N^2/V^2 dV
Putting dU = 0 and integrating will give you the temperature as a function of volume. If we assume that the heat capacity Cv stays constant, you find that the temperature increases from V to r V is given by:
Delta T = -a N^2/(V Cv) (1 - 1/r)
What you can also see from the equation
dU = Cv dT +a N^2/V^2 dV
is that Cv only depends on T, and that therefore the error in the approximation made by assuming constant Cv will be small, as the temperature doesn't change much. This is because Cv, being the coefficient of dT in dU in terms of dT and dV, is the partial derivative of U w.r.t. T at constant V. Similarly, the factor a N^2/V^2 in front of dV, is the partial derivative of w.r.t. V at constant T. The second derivative of U w.r.t. T and V is thus the derivative of Cv w.r.t. V at constant T, but this is also the derivative of N^2/V^2 w.r.t. T at constant V, which is obviously zero. Count Iblis (talk) 02:51, 24 October 2010 (UTC)[reply]
Usually plastic, now days. I'm not sure what type specifically. Incidentally, the first commercial use of nylon was in Toothbrush bristles (sold under a different trade name, I'm drawing a blank right now, and neither Toothbrush nor Nylon has that information). I assume that Nylon's still used for bristles, but I'm not positive. Before Nylon was invented, toothbrush bristles were pig hair (all together now: ewww). Buddy431 (talk) 19:53, 23 October 2010 (UTC)[reply]
center of mass problem
I have a uniform square of side a with one quadrant missing. I'm to find the center of mass.
Taking my origin at the center, I know that the x-coordinate of the COM should be (1/M)Sxdm (i'm using capital s for my integral sign)and similarly for the y-coordinate (1/M)Sydm. I'm just having some trouble seeing where to proceed from here. Thanks199.94.68.201 (talk) 19:35, 23 October 2010 (UTC)[reply]
the COMs of each quadrant would be in the exact center of each (-1/2 a, 1/2 a), (1/2 a, 1/2 a), (-1/2 a, -1/2 a) assuming the fourth quadrant is removed--how do I use that to find the COM of the whole?199.94.68.201 (talk) 21:21, 23 October 2010 (UTC)[reply]
In the original formulation, you need to work out an expression for the mass of a strip of width located at x-coordinate , and plug the result into the integral; same process for . Since this is clearly a homework problem, I'm not going to give you the details. Looie496 (talk) 21:53, 23 October 2010 (UTC)[reply]
It's easy now, you have three point masses, you must have learned how to calculate the CofM of a system of point masses before you moved on to having to integrate for a uniform density object. If you find three hard to do then chose to do two first {I'd go with the ones at (1/2 a, 1/2 a), (-1/2 a, -1/2 a)} Calculate the mass and the position of the CofM for that, then calculate the CoM for that one with the last point. Theresa Knott | Hasten to trek22:02, 23 October 2010 (UTC)[reply]
You can also get the same answer by subtracting the moment of the missing piece from the moment of the complete square (about an axis along one side). There's no need to integrate unless you are asked to work from first principles. Dbfirs07:52, 24 October 2010 (UTC)[reply]
And the calculation is even simpler if you remember that if a uniform object has a line of reflective symmetry then its centre of mass must lie somewhere on that line. So now you just have to work out one co-ordinate i.e. how far along that line the CofM is. Gandalf61 (talk) 08:02, 24 October 2010 (UTC)[reply]
a 32-u oxygen molecule moving in the +x direction at 580 m/s collides with an oxygen atom (mass 16-u) moving at 870 m/s at 27 degrees to the x-axis. The particles stick together to form an ozone molecule. Find the kinetic energy lost in the collision.
I calculated that the initial kinetic energy of the system should be (1/2)(32u)(580m/s)^2 +(1/2)(16u)(870m/s)^2
The final kinetic energy of the system I calculated to be (1/2)(48u)(658m/s)^2 so the kinetic energy lost is 1.046x10^6 J. My question is, in using the mass of oxygen molecules as u--do I need to convert them to kg first? Or is my answer correct as is?199.94.68.201 (talk) 20:22, 23 October 2010 (UTC)[reply]
You certainly need to take into account the mass of the particles, and you had better decide the units. Your assumption that the answer was in Joules is not correct as you did not convert. You have calculated the energy if a Mole of ozone formed. So yes you do need to multiply by the conversion factor. Energy per molecule is going to be fairly small, you should have spotted that 10^6 J is large. Graeme Bartlett (talk) 20:37, 23 October 2010 (UTC)[reply]
I was in Guadeloupe some months ago and noticed the depicted cattle breed several places. Being a Dane the breed reminded me a lot of Danish Red cattle, but can that really be true? I guess there are quite some red cattle breeds of which some are hard to distinguish unless you are an expert (which I am certainly not). I took a look at the ten red cattle breeds of which there are images on Commons, and I could rule several of those out, I think such as Belmont Red, Harzer Rotvieh, and several others. But still, there are some candidates left, where I cannot see which one it is, or do not know what to look for. Any cattle breed experts here, who knows, or anyone knowledgable of common catlle breeds in Guadelopupe, who could help me identify it? Thanks, --Slaunger (talk) 21:37, 23 October 2010 (UTC)[reply]
Sluzzelin, very interesting hint. That could indeed be the case. I am glad for autotranslation, as that article was actually very interesting. Moreover that article has an emaillink to a person, who is apparently an expert within the field. It also gives me a pointer for some further googling. Thanks, --Slaunger (talk) 22:05, 23 October 2010 (UTC)[reply]
As a followup I found this article on the French Wikipedia: Créole. The descriptions seems to fit well. Mostly used in small family farms, placed at road sides and slopes (that is where I observed it). Resilient to heat and parasites. --Slaunger (talk) 22:14, 23 October 2010 (UTC)[reply]
Green pennies
So I have always assumed and heard that green pennies have oxidized copper on them, you know copper oxide (I always just assumed). I just randomly decided to lookup copper oxide, and I noticed that nether copper I or copper II oxide were green, but that copper II acetate was blueish green. Is that what is on pennies? —Preceding unsigned comment added by 76.22.133.121 (talk) 22:38, 23 October 2010 (UTC)[reply]
Not quite. The copper(II) oxide made (black penny) reacts with water and carbon dioxide in the air to make a patina, which is a mixture of green copper(II) carbonate and green copper(II) hydroxide. Hope this helps. --Chemicalinterest (talk) 22:42, 23 October 2010 (UTC)[reply]
Why do chickens get violent with other chickens in their flock if the "target" chicken is bleeding? I've seen this with my own flock and my wife has told me that she has read that it's common. If a chicken gets hurt and starts to bleed, the other chickens will start to attack it. What's the reason for this? What is it called? Do we have an article about it?
Note: This isn't due to any overcrowding issue or other cause for stress in the chickens. We once had to trim the beak of one chicken and accidentally took too much off and a small drop of blood appeared. When we tried to put it back with the rest of the flock, it got attacked. We were able to keep it alone for a little while until the bleeding stopped and then re-introduce it to the flock just fine. Dismas|(talk)23:46, 23 October 2010 (UTC)[reply]
Amongst gulls, it has been known for the flock to gang up on and swiftly kill an injured or sickly individual. One theory I've heard that this is because an individual behaving differently (say, wandering around with a broken wing, or lying on the ground obviously having some sort of seizure) amongst the stoic uniformity of the rest of the flock will stick out like a sore thumb and draw the attention of predators, potentially leading to nearby birds, or their young becoming 'collateral damage'. I suppose that it could be a similar situation with chickens. --Kurt Shaped Box (talk) 00:11, 24 October 2010 (UTC)[reply]
A dead member of the species is one less competitor for food and sex. Of course, killing your competitors has its risks, particularly that you might get killed yourself... but if the competitor is weakened, the risks are obviously lower. Physchim62(talk)00:33, 24 October 2010 (UTC)[reply]
Chickens do not have detectable political opinions. In the wild, individuals who are genetically disposed towards too enthusiastic killing of conspecifics risk going extinct either because they've eradicated so much of their community that their offspring are left to inbreed, or their offspring will spend so much energy fighting each other that they will lose out to a more peaceful neighboring subpopulation. In the end, evolution will find some equilibrium point between the various effects (or the species will not survive, and be replaced by something that can find an equilibrium). –Henning Makholm (talk) 01:20, 24 October 2010 (UTC)[reply]
Could some of the more excessive aspects of this behaviour be a function of poorly-raised, maladjusted captive birds? I've seen it mentioned on here before that wild/feral roosters raised by their parents in the natural way are not nearly as violent and intolerant with each other as they are on the farm (i.e. you can have more than one of them around without them fighting to the death at the first opportunity). --Kurt Shaped Box (talk) 01:28, 24 October 2010 (UTC)[reply]
I'm not an expert on the psychology of chickens, but are they not just inquisitive and peck at anything bright red? Dbfirs07:50, 24 October 2010 (UTC)[reply]
Wow. Just wow. Live and learn. Personally, I'd have thought that it would be almost impossible to make a bird wear contact lenses - and keep them in... --Kurt Shaped Box (talk) 23:21, 24 October 2010 (UTC)[reply]
October 24
UDFy-38135539 both exists AND may have been seen by space aliens, or something
The new article on UDFy-38135539 states that the galaxy is "the most distant object in the universe known to have existed and which has been observed from Earth." As opposed to objects that have been observed but are NOT known to have existed? I assume the qualifier is meant to exclude hypothetical objects, but it's redundant for something that has been scientifically observed to be further qualified as "known to have existed." Comments? Clarifications? (Also, why "observed from earth"? Isn't "observed," by convention, a human-centric concept? A bit pedantic to worry that saying "observed" instead of "observed from earth" might leave out "observed by aliens." And clearly "from earth" isn't meant as opposed to "from a space telescope.") So why not: "the most distant object in the universe that has been observed."? 63.17.82.132 (talk) 01:03, 24 October 2010 (UTC)[reply]
Scientists--real scientists--prefer precision. Given that we do not know if other species exist out there, it would be silly to say simply 'observed,' because that presupposes we are the only observers. Actually by the same token, 'observed from earth' is a bit of a presupposition. → ROUX₪01:34, 24 October 2010 (UTC)[reply]
OK, "real scientist." I suppose when a particle is said to be the smallest particle yet observed, it's necessry to add "by human beings" and "on earth"? Seriously, this is the sort of thing that gives pendantry a bad name. Anybody else? 63.17.82.132 (talk) 02:01, 24 October 2010 (UTC)[reply]
Well, one might consider it perhaps a bit "pedantic" just to raise the issue you are raising yourself, though (but I am sure you are doing so in good faith, and it is an awkward sentence that ought to be improved, so I am glad you mentioned it and would encourage you, again, to go ahead and make the change yourself and/or raise the issue on the talk page there). Let's not get into a "real scientist" contest here, though; let's just focus on making this improvement to this article. WikiDao ☯ (talk)02:15, 24 October 2010 (UTC)[reply]
To clarify, by 'real scientist' I meant actual researchers, not the professional sound-bite providers to news organisations that elide details. → ROUX₪02:23, 24 October 2010 (UTC)[reply]
The "known to have existed" thing could be referring to the fact that, just because we've detected the light from this object, doesn't necessarily mean it still exists anymore. All we know for sure is that it was there 13 billion years ago, so it is known to have existed. But a lot of water can flow under the cosmic bridge in that time. Heck, we never even see the Sun as it is right now, but as it was 9 minutes ago. -- Jack of Oz ... speak! ... 02:36, 24 October 2010 (UTC)[reply]
That's funny! There should be a category of "most distant objects that DO exist," and we could quote Descartes. Not a very great distance ....63.17.51.20 (talk) 03:51, 24 October 2010 (UTC)[reply]
Stay with me, buddy; we're gonna make it through this
On television and in movies, wounded people are often encouraged to stay concious. Is this necessary to improve chances of survival? Why not just let them be unconscious and continue rescue efforts while they sleep? --90.215.213.167 (talk) 01:51, 24 October 2010 (UTC)[reply]
It's more dramatic to see the Wounded Hero struggle. However, some injuries--e.g. concussion--carry a danger if the patient loses consciousness. → ROUX₪01:54, 24 October 2010 (UTC)[reply]
WP:OR here: I suspect that, because an injured person who is verifiably conscious, is not dead, this is reassuring to the others around. Fighting to stay conscious also demonstrates that the injured person has not "given up". Bielle (talk) 05:27, 24 October 2010 (UTC)[reply]
With a (suspected) concussion loosing consciousness or sleeping is not dangerous. The problem is that going to sleep can mask serious injury. That's why the recommendation is not to let the person go to sleep have a head injury - you want to make sure they are not seriously injured (which you would notice by altered consciousness states, pupils of different sizes, etc.). From this people extrapolate a little too far.
The only other thing I can think of is that a conscious person can help in their recovery, but for example pinpointing injuries (pain), and giving a medical history. But besides that I can not think of any reason not to let the person become unconscious - in fact one of the first treatments for serious head wounds is a medically induced coma. Physically forcing (shaking) a person to stay awake can probably injure them more! (However all that said, I would be reassured if a doctor could confirm/deny what I wrote.) Ariel. (talk) 08:02, 24 October 2010 (UTC)[reply]
There can be a number of reasons to keep the casualty awake. For example a study discussed here into pre-hospital deaths from road accidents "showed that at least 39% and up to 85% of preventable pre-hospital deaths may be due to airway obstruction”. Much easier to monitor and remedy airway problems in a conscious person. Another serious risk with accident victims or people suffering from blood loss is hypothermia. At least two reasons you want to keep someone awake who's at risk of this, firstly again problems are easier to monitor in a conscious patient, and secondly the body temperature naturally decreases during sleep, so thereby heightening risks of hypothermia. Of course once a person has been transferred to professional care and a proper medical facility the equation changes completely. --jjron (talk) 11:55, 24 October 2010 (UTC)[reply]
The most significant mortal risk in loss of consciousness is loss of airway. An unconscious person cannot protect their airway, so no gag reflex. That means you can inhale stuff (blood, snot, vomit, bone, brain, whatever) into your lungs (aspiration pneumonia) and this can progress to ARDS. Then you have a really good chance of dying. In hospital we have plenty of technology at our disposal to artificially maintain airways. We just need to be aware of the problem! But in the field, the best means of protecting an airway is to keep the victim conscious. Failing that there's airway manoeveres like jaw thrust. If there 's a seriously life-threatening head injury, then unconsciousness may be as preventable as nightfall, but deterioration of milder injuries may be prevented when the victim stays awake. Mattopaedia Say G'Day!05:38, 25 October 2010 (UTC)[reply]
Large rigid structures on the ocean
I've heard that some of the large ocean rigs/cruise ships nowadays are so huge that you can barely notice any rocking, even when the waves are quite large. Is this true, and if so, how much so? I looked for some description of the stability of these structures, and I found mega-float, which sat in Tokyo bay for a while, and it claims on that page that pilots didn't notice anything different from a landing on a land-airport. I'm wondering if the same would apply to someone standing on the platform, and if not, if the structure was a few kilometers in size, is it feasible to think that it could be build so that there was no noticeable "seismic activity", at least for a human? This is assuming a rigid (within reason) structure! Thanks! 173.180.219.65 (talk) 04:45, 24 October 2010 (UTC)[reply]
Yes, the wider the structure the more waves and swells it spans and the less the influence of each wave/swell. Even the heaving (up and down) motion would become insignificant if the dimensions of the structure were large enough. Unfortunately I don't have any measurements. Dolphin(t)05:09, 24 October 2010 (UTC)[reply]
Defend the statement that 100%of the electrical energy that goes into lighting a lamp is converted to thermal energy.Are the first and the second laws of thermodynamics violated? —Preceding unsigned comment added by 99.28.147.106 (talk) 05:48, 24 October 2010 (UTC)[reply]
I don't wish to defend the statement (or to do your homework for you), but I think a very tiny amount goes to produce sound. I suppose that this eventually dissipates as thermal energy, as does the light output. Dbfirs07:43, 24 October 2010 (UTC)[reply]
I'm not so sure that it goes into sound...the filaments in light bulbs are in a vacuum, as far as I can recall, and thus there would be no medium within which sound could travel (inside the light bulb, at least). Ks0stm(T•C•G)02:27, 25 October 2010 (UTC)[reply]
What about those photons which make it out the window, through the atmosphere, and into space? How long will it take for their energy to be converted to thermal energy? But isn't the light emitted by the filament itself black body thermal radiation? So why should that original escaped photon be considered any different from those photons which are emitted by the roof shingles which were, rather indirectly, heated by light from the bulb? -- 124.157.218.132 (talk) 02:08, 25 October 2010 (UTC)[reply]
There is no air left in a wreck, so it won't be crushed by the pressure as, say, a submarine would be if it exceeded its crush depth. Any gaps in the wood will be filled by water, and the individual fibres of the wood will not be affected by the external pressure. A wooden wreck is very fragile if it is recovered and allowed to dry out (see Mary Rose#Conservation), but if left deep underwater, where the cold and dark will slow down decay, then I imagine it is quite tough. Gandalf61 (talk) 07:53, 24 October 2010 (UTC)[reply]
I only watched the first video here: http://forum.grasscity.com/science-nature/686839-we-being-watched.html
out of that big crowd of people, news cameras, etc, for hours and hours, did anyone film that object with a telescopic lens? How about just renting a helicopter and flying up closer and taking a closeup? Is that video real? What is it? Why do we just get such a small, far-away version when obviously a few hundred dollars of equipment will give you a clear closeup? (Or you can use a $20 pair of binoculars and put your iphone camera to them...) 84.153.221.42 (talk) 10:42, 24 October 2010 (UTC)[reply]
I think the answer is obvious, if anyone did actually take a closer look they would have seen exactly what they were. My money is on something completely innocuous such as some balloons. But then there wouldn't be any news story and you wouldn't here about it. Theresa Knott | Hasten to trek10:54, 24 October 2010 (UTC)[reply]
Of course are it real, the question is what it was. It was per definition an UFO. I saw a interview with some on school in the area that had released a number of party balloons by mistake, it was probably what they observed. --Gr8xoz (talk) 14:34, 24 October 2010 (UTC)[reply]
Don't forget that your equation will give you the number of full cycles of the pendulum: the full cycle includes both the "forward" swing and the "backwards" swing. Physchim62(talk)15:23, 24 October 2010 (UTC)[reply]
measuring id, ego, etc.
I was just reading a recent response in the Ref desk re: the use of evidence and reasoning rather than "feelings" or "gut feelings" to "prove" something. This led me to wondering about some things that have been "accepted" as truth. For example, many psychologists (as well as the public in general) accept the "truth" or "validity" of such thikngs as id, ego, superego and other such psychological concepts. However, I have spoken to many psychologists (I work with many of them)and they are not able to provide support for any evidence for these terms/concepts yet they still generally accept them. So, I am wondering: is there actual way of measuring these terms/concepts or have they just been referenced so much in common vernacular that they are simply accepted as fact? What about the concepts put forward by other theorists (such as Piaget, Jung, et al)?
We also accept 'love' 'anger' and 'that pleasurable feeling of comfort you get when you flip over your pillow to reveal the cool underside' as truth, and they are just as immeasurable. The basis for accepting those names--ego, superego, id--is that they serve to describe concepts well enough that we may as well treat them as real. Other paradigms in psychotherapy use different terminology. Freud just happened to publish first, so got to be the dominant nomenclaturist in the public eye. → ROUX₪16:41, 24 October 2010 (UTC)[reply]
I agree with Roux here: in a sense, the necessary-and-sufficent "evidence" for these things is simply that they are there and that whatever they are (in themselves) can be described or defined to some extent. It is a good question, though, and also true that the naming and defining of "intangibles" can cause them to take on a reality that is different from what they are "in themselves", which can be a source of systemic error in some cases. The formal "defining" of mental illnesses is another example. WikiDao ☯ (talk)16:54, 24 October 2010 (UTC)[reply]
That's just tautological reasoning, though. It's not scientific. Saying, "we know it's there, that's the evidence" is fairly silly with regards to things like the id or the ego. Do we know they are there because we "feel" them, or because we've been told they are there? People think auras and psychic activity and superstition and ghosts "feel" real too, but it doesn't make them so. "Anger" and "love" both correlate with distinct emotional states that can be quantified and measured (adrenaline, facial temperature, activity in the Limbic system). They are much stronger empirical grounds than are Freud's concepts, which have essentially no empirical backing, and there is no reason other than "as part of a therapeutic regimen it has a good track record" to believe in them, and even there you're going to have a hard time convincing a skeptical outsider that they do in fact have a good track record. The validity of psychoanalytical concepts (and their inability to be falsified) has been a major point of contention in the mind sciences for a long time and I don't think you can just waive them away by saying we "know" they are true. We know no such thing. --Mr.98 (talk) 17:44, 25 October 2010 (UTC)[reply]
But, that's not what I am saying.
I am not a Freudian myself, btw, 98, and certainly have no bone to pick with all the very reasonable criticism of Freudianism. But I think a lot of criticism of Freud's basic concepts and theories would benefit from greater familiarity with what is being criticized.
I am also not a clinical psychologist or a psychiatrist, much less a psychoanalyst, but the OP implies that they are still teaching the concepts of id, ego, and superego to students in those fields. If so, there is obviously some reason to do so – ie., presumably those concepts are still useful for some purposes. And it is utility and not "scientific validation" that is what's important in such general conceptual frameworks as this. It is important to note the distinction, though, and avoid making misleading assumptions about the framework under the misunderstanding that it has been "scientifically validated" in ways that it has not been. WikiDao ☯ (talk)20:23, 25 October 2010 (UTC)[reply]
The experimental method is the gold standard of finding out if something is true or not. If a conceptual model has been verified by experiment, then I'm willing to accept it as probably true. A lot of people such as Freudian psychologists do not require experimental verification and just accept things on faith or the assertion of others. Its not too difficult to create a model which is intenally consistent but still untrue: eg. Phlogiston theory or Aether theories. People without scientific training will accept all sorts of twaddle as being true if it seems superficially coherent and it is supported by others. People may have invested years of training and experience in Freudian psychology and get a good income and prestige from it, so they are not going to give it up. I understand that experiments to confirm the truth of Freudian psychology have not verified it. 92.15.31.47 (talk) 17:15, 24 October 2010 (UTC)[reply]
There is no experimental verification that 'love' actually exists. It is the name given to a complex set of social norms coupled with some verifiable brain activity plus feelings, which likewise cannot be independently verified by experiment in the way you propose. Do you say that love does not exist? → ROUX₪17:28, 24 October 2010 (UTC)[reply]
That's a glib saying: are you sure there have been no experiments regarding love? I think I can vaguely recall at least one, to do with scanning the brain. I expect if you search the scientific literature then there have been many experiments regarding love. You might as well claim that pain or pleasure do not exist. 92.15.31.47 (talk) 17:32, 24 October 2010 (UTC)[reply]
Yes. But there are many "conceptual models" that have not been "verified by experiment" (at least, not in the sense of "experimentation" as used in the physical sciences) but are nevertheless not only "probably true" but quite useful as well!
One really must try to be as "scientific" as possible in the face of material that by its very nature evades "scientific investigation" – which is what Freud (who wanted to be a scientist but became a medical doctor for the $$$), for one, always tried to do. WikiDao ☯ (talk)17:35, 24 October 2010 (UTC)[reply]
lol - I always get a kick out of this misconception. 92.15, experimentation is (literally) the antithesis of a truth claim. Non-scientifists make claims about truth or falsehood; scientists create theoretical structures and create evidence-based arguments that those structures should be accepted by others. No scientist would ever claim that the (massive amount of) experimental verification of the Theory of Gravity makes the Theory of Gravity true. All such experimental verification does is make the ToG believable.
If you think this is ever about truth, you misunderstand science drastically. The entire rasion d'etre of science is to replace unfounded truth-claims with founded beliefs, where the foundation of the new beliefs lies in empirical evidence.
Back on the original topic: Freud's inner structures (id, ego, superego, cathexes, neuroses, etc...) are mostly important for the insight that there are inner structures. Prior to Freud, people had a very superficial understanding of the workings of the mind (assuming that every outward action was either the result of a conscious choice or the manifestation of a physical defect). Freud created the understanding that there was an internal mental life that was neither strictly conscious nor strictly biological. There are certainly problems with Freud's theory, which even Freud recognized (he was constantly revising his conception over his lifetime), but the basic idea of id, ego, and superego (corresponding to to the tension between purely egoistic interests and purely social constraints, moderated by conscious rational processes) is appealing and intuitive, and a helpful tool for understanding both functional and dysfunctional behavior. No doubt that particular terminology will disappear over time, but the concept behind it is too useful to disappear. --Ludwigs218:33, 24 October 2010 (UTC)[reply]
If you are already aware of this, then please don't say things like "The experimental method is the gold standard of finding out if something is true or not"[6], which would tend to imply the opposite. clarity in communication is important. --Ludwigs219:54, 24 October 2010 (UTC)[reply]
how useful can solumedrol be in patients suffering from viral fever ( malaria /dengue fever )especially when it is associated with thrombocytopenia ?
Does it play a role in such condition ? —Preceding unsigned comment added by Pravinbathe (talk • contribs) 19:09, 24 October 2010 (UTC)[reply]
In Antarctica, occasionally a rock will be found on top of the ice in a place where a rock has no business being. Some Antarctic ice-sheets are kilometers thick - so how would a rock float its way up through 1000 m of ice? In such cases, it is typically assumed to be a meteorite. This assumption can be supported by geochemical analysis of the specimen. Other ways that a rock can "surface" to the top of an ice sheet include being carried there by animals and humans, being blown by wind across the surface (and uphill), or even "convecting" through the ice. Glaciers behave as "fluids" over very long time-scales, and the ice moves and flows very dynamically, carrying debris with it. If the ice is the highest geographical point for many miles around, the number of ways rock can get on top of it are very limited. But bear in mind, Antarctica is not the same as Patagonia - but we do have an article on the Southern Patagonian Ice Field. You might find the article Grey Glacier instructive, especially photos of the gritty surface below a mountainous landscape. We also have an article called colluvium; you might find this term helpful. Nimur (talk) 23:16, 24 October 2010 (UTC)[reply]
There are several volcanoes in the Southern Patagonian Ice Field, some of which have been active within the last century. Lautaro, for example, last erupted in 1979. Gandalf61 (talk) 12:02, 25 October 2010 (UTC)[reply]
Define "potable". The easy answer is that Humans used pre-modern methods to obtain potable water. See spring, well, and aqueduct. Moreover, humans can get used to all but the worst biological contaminations. Most running water in low-population density areas of the world is pretty safe to drink. Indeed, I spend some time on the East coast of Australia, and they put up signs at the one river where it was not safe to drink the water. --Stephan Schulz (talk) 23:54, 24 October 2010 (UTC)[reply]
I'm not sure about how the numbers work out for this, but it's possible that pre-modern societies in some regions can obtain a large percentage of their water from the foods they eat, such as fruits. I know that in pre-Industrial times it was common in Europe, at least, for people to drink beer or wine instead of water because it kept fresh for much longer. --Laryaghat (talk) 00:00, 25 October 2010 (UTC)[reply]
And then there's the simple fact that in many places around the world (China, India, etc.) we've made water far, far more poisoned in the last 100 years than it ever would have been normally. The Masked Booby (talk) 02:20, 25 October 2010 (UTC)[reply]
Interestingly, it was the invention of alcohol that really let humans congregate in high population density areas [7]. When the water source has cows pooping in it upstream, drinking beer can be a lifesaver. Buddy431 (talk) 02:46, 25 October 2010 (UTC)[reply]
No, but the processing of (for example) beer involves boiling the water: a major danger was the watering down of beer, which added contaminated water to the boiled-water beer. Similarly, drinking tea is considered to have had health benefits for the English working and middle classes, as the water was (again) boiled before drinking. 86.163.212.182 (talk) 18:39, 26 October 2010 (UTC)[reply]
Possibly our ancestors, as well as "primitive tribesmen," were better able to just drink water from whatever creek or puddle they encountered, than the delicate hothouse flower people of today, who would suffer from the contamination in such water sources. Edison (talk) 04:21, 25 October 2010 (UTC)[reply]
Yes, the human gut adapts to minor bacterial contaminants in drinking-water (and the acidity of the stomach kills off most of the bacteria). Many years ago, a farmer was prevented from selling milk because the local water supply was not of a sufficiently high standard for the cows, despite the fact that he and his mother (who lived to 92) had been drinking it all their lives. I regularly drink from streams when walking on the local hills, but I am selective in my choice! Personally, I wouldn't drink from rivers under normal circumstances, but some people drink water from the Ganges! Dbfirs08:01, 25 October 2010 (UTC)[reply]
Light speed and giant sticks
This is a self-evidently stupid question, because I know what the answer is, but I'd like to know exactly *why* the answer is the one it is from someone who understands this stuff. Here it is:
Say we have two people, Alice and Bob, facing each other but standing a whole light year apart. If Alice were to try to radio signal Bob, it would take a year for that signal to reach Bob. However, let's say that Alice has a giant stick or rod, that is exactly a light year long and is somehow magically incapable of snapping or bending. Assume also that Alice has the unnatural strength that would be required to lift this rod. Now let's say Alice thrusts this rod forward. How long before Bob is poked by the other end of the rod?
My common sense tells me the answer is "immediately", but I know that the answer has to be "a year or more" - I'm just not sure why this is the answer. Can anyone explain this to me? Is the entire premise of the idea so infeasible that the question is meaningless?
Alice isn't a light year long so she would only be able to lift one end of the rod. The rod would flex slightly and a signal would travel down that rod towards the other end telling it to "lift". Yhe speed that this signal travels at would be ( I think) the speed of sound in the material of which the rod is made. Much slower than lightspeed. Theresa Knott | Hasten to trek00:02, 25 October 2010 (UTC)[reply]
This is a standard question, and I'm sure we've discussed this before. If you allow magic, all bets are off. On the other hand, if you have a real rod made of matter, then that is held together by electromagnetic forces. That is, if you push one end, the rod does not move as a whole, but the push creates a longitudinal wave that runs down the rod at at most the speed of light (as the electromagnetic force is mediated by photons which travel at the speed of light). --Stephan Schulz (talk) 00:06, 25 October 2010 (UTC)[reply]
Ah, of course! Electromagnetic energy is what's keeping it together in the first place, and that propagates at light speed. Thanks, I think I understand it now. --Laryaghat (talk) 00:11, 25 October 2010 (UTC)[reply]
As Theresa notes, the response of solid objects to being pushed or pulled is generally limited to the speed of sound in the material. For typical solids this is about 3 km / second, which is much less than the 300,000 km / second that light travels at. Dragons flight (talk) 00:17, 25 October 2010 (UTC)[reply]
If "perfectly rigid" means instantaneous communication down the length of an object, then there is no perfectly rigid object. If "perfectly rigid" means transmission down the length of an object, then how does a small segment in the middle know which end has been pushed with greater force? Also see Ehrenfest paradox. Wnt (talk) 15:35, 25 October 2010 (UTC)[reply]
October 25
Dark spots/bruises on old people's skin
What are the dark spots and bruises that old people (60+ years old)? Most of them look like dried blood under the skin or a bruise. What are they? Why are they there?
76.169.33.234 (talk)Dave —Preceding undated comment added 03:10, 25 October 2010 (UTC).[reply]
I looked at pictures and I don't think thats it. They are purple or black spots, some are small, some are large. I've seen them mostly on old people's arms, but I think they many be all over, not sure. —Preceding unsigned comment added by 76.169.33.234 (talk) 04:48, 25 October 2010 (UTC)[reply]
Aging skin is susceptible to a number of conditions that present themselves as dark spots. The lentiginous conditions are are a range of pigment depositions which may be benign, pre-malignant or malignant. Older skin may also be more susceptible to injury and bruising, for a number of reasons, including medication effects, bleeding and clotting disorders, and age-related thinning of the epidermis and dermis. I would encourage anyone with an undiagnosed pigmented or dark spot on their skin to seek a doctor's opinion in real life, not on wikipedia. Mattopaedia Say G'Day!05:53, 25 October 2010 (UTC)[reply]
Many elderly people take blood thinners or anticoagulants such as Plavix (to counter the risk of stroke and heart attack) which can result in much easier and sometimes quite severe bruising. -- 117.47.204.34 (talk) 11:57, 25 October 2010 (UTC)[reply]
Yes, as I said, this is one of a number of possible causes. Again, see a doctor if you don't know what the cause or nature of the "dark spot" is. Not all the causes are innocuous. Mattopaedia Say G'Day!14:26, 25 October 2010 (UTC)[reply]
Thanks for the answers. I don't have it, I just see it on a lot of my older customers at the pharmacy I work at. I didn't get a chance to ask the pharmacist so I thought I would ask here. 76.169.33.234 (talk)Dave —Preceding undated comment added 05:05, 26 October 2010 (UTC).[reply]
Calorimetry
6.60g of were dissolved in 100 mL of water. The temperature dropped 0.895°C. What is the molare heat of dissolving? . This is not accurate because a lot of energy goes into the vessel, the glass rod/thermometer etc. If the 100 mL of water in the calorimeter fell by 0.700°C when 4.00A was passed through at a potential difference of 3.00V for 35 s.
I've been told the calibration factor is , which yields = 600 J°C-1. How do I use this to find the molar heat of dissolving? Do I just multiply this by so ?--MrMahn (talk) 10:09, 25 October 2010 (UTC)[reply]
Atom pic
I need a pic of an atom that may be used to illustrate that much of the volume of anything consisting of atoms is actually empty since the nucleus is small (compared to the atom) and the electron cloud is very sparse. I looked at the atom article, but the illustrations there aren't very helpful. -- kainaw™15:57, 25 October 2010 (UTC)[reply]
In ballpark figures, the atomic radius is around 10,000 times larger than the nuclear radius. So in any realistically scaled diagram that scales the whole atom to the width of your screen, the nucleus will still be too small to see. Maybe some animation zooming in from atom to nucleus would illustrate their relative sizes better than a static diagram. Gandalf61 (talk) 16:14, 25 October 2010 (UTC)[reply]
An atom isn't mostly empty space. Okay, electrons are pointlike in a certain sense. But quarks and gluons are pointlike in the same sense, so in that sense an atom is entirely empty space, as is everything else. To the extent that there is such a thing as occupying space, atoms do it; they're the quintessential example of something that does. It's true that most of the mass is concentrated in a small volume, but there's no threshold mass density at which space becomes full. -- BenRG (talk) 21:10, 25 October 2010 (UTC)[reply]
I agree. The atom isn't mostly empty space. What is true is that an atom has a very small collisional cross-section, as compared to the atomic spacing. Probably, if you want to be accurate, you want a plot of scattering-angle probabilities from a Rutherford scattering experiment (surprisingly, we do not have this plot in our articles, but I can look one up in a textbook and create it later tonight if it's necessary). Or, if you prefer to use a classical approach, consider one of the diagrams in the Bohr model article. "If drawn to scale", the atomic "solar-system" diagram will look like you expect - mostly sparse - but it isn't exactly a correct picture of the atom. Nimur (talk) 21:25, 25 October 2010 (UTC)[reply]
The point-like nature of the electron is an abstraction. I mean, if you want to measure "how big the electron is", you get something based on its wavelength, which is on the scale of an atom. There's no way to measure how big the "point-like spot" is. When a white dwarf compacts a star into an object the size of the Earth, it's a million times denser than water, because it's made up of electron-degenerate matter where electrons no longer form neat little atoms in contact with one another but are all smooshed together, but it's still nowhere near as dense as a neutron star. So even there the electrons aren't "points", and they can't be made to get out of the equation without actually joining up with protons to form neutron-degenerate matter.
I'd say, as an opinion, that the most sensible way to think of the size of an atom is just to take a piece of ordinary matter, divide by the number of atoms, and there you are. There really aren't many spots of vacuum at the atomic level - molecules of water, air, etc. work their way into any significant crevice, and things crystallize in a way to make the atoms happy. They're elastic - they'll expand a bit if you atomize them (so to speak) into space, and they'll compress depending on the amount of force - but they're not empty space, except in the physics-philosophical sense that everything is empty space when you ignore the fields and phenomena that you use to describe the existence of matter. Wnt (talk) 22:25, 25 October 2010 (UTC)[reply]
That may be tough to do, as I would suspect (but not guarantee) that under normal conditions, the copper (I) chloride would disproportionate into metallic copper and copper (II) chloride, or possibly oxidizde in air just to make copper (II) chloride. That's why the only pic we have in our article is the green-tinted powder. As it is dried, its oxidizing away. It may be possible to do it in a completely inert (nitrogen or argon) atmosphere, assuming it doesn't disproportionate. --Jayron3200:50, 26 October 2010 (UTC)[reply]
The picture on the right claims to be copper(I) chloride. It's a different green tint from the blue-green copper(II) chloride, but I can't vouch of its authenticity. Maybe a picture of a flame-test like the ones on [8] could be an alternative way to represent the compounds? EverGreg (talk) 08:57, 26 October 2010 (UTC)[reply]
It's two or three years now that we removed that image from the copper(I) chloride article after complaints. I remember one professor saying that, if he received CuCl that looked like that from a chemical company, he would send it straight back! The problem is that the sample in the image is so oxidized as to be pretty much useless. Pure CuCl is white; the stuff you find in the bottle at the back of the shelf in the storeroom is often grey (because of oxidation). It's tough to avoid oxidation when you're preparing it if you don't have access to an inert atmosphere: I'd say it's more hassle than it's worth (although I'm sure Chemicalinterest wants to try!) because it's a fairly common lab chemical, and someone will take a photo at some stage. Physchim62(talk)09:50, 26 October 2010 (UTC)[reply]
The main problem is that when people hear "plasma weapon" they think "Star Wars, cool." But in reality they would have pretty limited application, and as crude as it is, the plain old bullet (accelerating a piece of metal to high speeds) works better in most situations than a plasma (or laser) would. There are a few situations where the possibly high accuracy and speed of a laser weapon would be better (e.g. shooting down a missile) but in most cases, the tried-and-true method seems pretty clearly superior. --Mr.98 (talk) 17:37, 25 October 2010 (UTC)[reply]
The reference desk does not answer requests for opinions or predictions about future events. See information at the top of this Project Page. Dolphin(t)00:52, 26 October 2010 (UTC)[reply]
Most likely they would not be as useful as you think on earth because the atmosphere will quickly kill the power of your plasma. If you had a plasma gun in space, perhaps that would be more useful. As far as practicality goes, I don't think you will ever see widespread plasma weapons on earth, though some research group might make one just to see if it can be done. Googlemeister (talk) 15:31, 26 October 2010 (UTC)[reply]
(1) I don't believe the third formula is supposed to be derived from the previous two. But the previous two can each be derived from the third formula, by setting the appropriate variable to zero. So at least they're consistent.
(2) I didn't have any trouble with them on my Linux machine running Firefox. My bet is that any user of Netscape 7 is going to have those same problems. Netscape Navigator 7.02 is about seven years old, and Firefox (its spiritual successor) has gotten a lot more stable over the years. I haven't seen a page that crashes Firefox in a long time. Unfortunately, it looks like Firefox isn't available for OS 9. I searched around a bit and found iCab and Classilla (which I'd never heard of before); I bet one of those would work a lot better for a lot of pages, because support for browsers as old as that is declining rapidly among web developers. Paul (Stansifer)17:29, 25 October 2010 (UTC)[reply]
Deriving the third equation from the first two is quite straightforward. It's useful to define additional variables, to keep and from being overused. Define as the frequency of the wave as of a moving source, as the frequency of the wave as of a moving receiver, and as the frequency of the wave as of a point in between the source and the receiver, which is stationary with respect to the wave's medium. The first formula in the article section gives the frequency as of a stationary point given the frequency as of a moving source, which using the variables we defined is
The second formula in the article section gives the frequency of a wave as of a moving receiver given the frequency as of a point that's stationary with respect to the wave's medium, which using the variables we defined is
Substituting the expression for given by the first equation above into the second equation above gives
P.S. If you're seeming to remember that the equation for the ratio of frequencies in the Doppler effect involves a square root, perhaps that's because you're remembering the equation for the ratio of frequencies in the relativistic Doppler effect, which does have a square root. Red Act (talk) 19:19, 25 October 2010 (UTC)[reply]
Response: 1) Thanks very much for the clarificatons. However I feel even more, that a rewrite of the original article is needed. Your 'adjustments' of the denotations are more than adornments. They correct misleading denotation, especially in light of the last line of the section Analysis:"... but hold for the relativistic Doppler effect as well. Furthermore the use of the expressions "the wave" and "stationary with respect to the wave's medium" is obscure.
2) I do now not believe my troubles at obtaining a text from the links I mentioned originate with my using Netscape 7.02 for Mac . Why would I otherwise succeed in acquiring the text I required, when I simply used an alternative parallel source from the same http://www.archive.org/search.php?query=theory%20of%20sound ? It has only been in cases of programming deficiencies, that Netscape 7.02 for Mac has failed me. —Preceding unsigned comment added by 83.226.97.246 (talk) 23:20, 25 October 2010 (UTC)[reply]
Sticking the Landing
If a person were strapped inside the man-sized hollow chamber of a large ball of foam rubber and dropped from high enough to reach terminal velocity upon impact with the ground, how large would the ball of foam rubber need to be to not injure the person inside? TheFutureAwaits (talk) 18:19, 25 October 2010 (UTC)[reply]
You're asking how much foam rubber padding would be required for a person to survive a 200 mph impact unharmed? Considering the amount of crumpling metal required to protect someone in an auto accident at 65 mph, my first instinct is to say "lots", but I'm sure someone can come up with a more sophisticate response. the real problem (honestly) is more a matter of thickness then absorbency - a person traveling at 200 mph can only decelerate so quickly without suffering catastrophic injuries, and the rate of deceleration is primarily dependent on the distance over which deceleration occurs - I'm too lazy to do the math, but try calculating the amount of force needed to change velocity of a 160 lb person by 200 mph over the space of (say) one meter, then ask yourself if that would leave you feeling happy afterward. --Ludwigs219:10, 25 October 2010 (UTC)[reply]
Don't forget he said "terminal velocity". If you make the ball large enough it'll be a parachute of sorts. You can make the ball large, with thin rubber. Or smaller with thicker rubber. There are many many kinds of foam, each with different levels of impact protection. It's not really possible to calculate this, at best maybe someone knows a terminal velocity formula for a ball. Ariel. (talk) 19:34, 25 October 2010 (UTC)[reply]
I just assumed that a ball was relatively aerodynamic, so it would have an equivalent terminal velocity to a plummeting human. I could obviously be wrong. plus, you could fill the ball with helium to cut down terminal velocity further, though you'd talk funny when you got out. --Ludwigs220:01, 25 October 2010 (UTC)[reply]
There is also a great dependency on the orientation of the person at the time of impact. If the impact occurs with the person's head lowest and his feet highest the ball would have to be of very large diameter indeed to protect the person's head and spine. If the impact occurs with the person's feet lowest and his head highest that would be a little better than the previous case. The best configuration would be if the ball had a parachute or tail that ensured it remained orientated so the person was always lying flat looking upwards - the human body can withstand much greater acceleration or deceleration in that position than in any other. During the launch of astronauts and cosmonauts, when very high accelerations are employed for a long period of time, the astronauts and cosmonauts are lying on their backs. Dolphin(t)21:37, 25 October 2010 (UTC)[reply]
Assuming the foam compresses nice and evenly to bring the test subject from an initial speed v to rest over a distance d at a constant deceleration ng, then
Taking v as 200 mph or approx 90 ms-1 and n as 2, I get d to be about 200 metres. Taking n to be 10, this reduces to 40 metres. Still quite a large air bag ! Gandalf61 (talk) 13:07, 26 October 2010 (UTC)[reply]
You might even be able to get away with 15 or 20g's if the human is in the back down landing configuration, but obviously the risk would be even higher. that might get you as low as 20 meters. Googlemeister (talk) 15:28, 26 October 2010 (UTC)[reply]
how would we even know
I figured this would segue off into a discussion of Christianity. Archiving as off-topic --Ludwigs2
The following discussion is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.
This question may be inappropriate for the Reference Desk, and may precipitate non-productive arguments or disputes. Please restrict responses to neutral, factual, sourced statements.
We wouldn't. there's no way to tell from fossil remains the particular activities of the animal (except in very general ways). if you personally rode a horse, could you tell that you had ridden it from examining its bones? You might be able to determine if the horse had been ridden extensively or left to run in the fields (I assume that over time riding may make some detectable changes in the bone structure of the horse), but there would be no way to determine which or how many people had ever ridden the horse. --Ludwigs219:14, 25 October 2010 (UTC)[reply]
It also depends on your definition of Jesus. Depending on your particular views, Jesus was a specific individual who lived approximately 1 A.D. until approximately 33 A.D. (though even these dates are considered dubious, and most scholars (scientific and theological) believe Jesus was born about 4 years "before Christ." (Rather, it is the Gregorian calendar that incorrectly calculates the years since Jesus' birth). If you subscribe to this point of view, you would be able to use radiocarbon dating to determine the age of the dinosaur bones: if the fossil age does not overlap with the period of time that Jesus was alive, there is no way that he could have ridden on the animal. Most scientists will say with a resounding degree of confidence that radiocarbon dating is a well-established and accurate method for dating fossils that are around 2,000 years old; they will note a clear lack of any species of dinosaur that was alive at that time; and will conclude that it was impossible for Jesus to ride a dinosaur. (Indeed, most dinosaur bones are so old that their carbon signature is too weak to accurately use for dating; other forms of radiometric dating exist for older types of fossil, including dinosaur bones. If your views of theological Jesus are different from those of established scientific and historical evidence, you will have to explain them in a clear and empirical way before we can evaluate their scientific merits (if they have any). Nimur (talk) 20:12, 25 October 2010 (UTC)[reply]
I suppose it's not worth pointing out that even according to a very literal reading of the Bible, there should not be any Jesus-dinosaur overlap. --Mr.98 (talk) 20:32, 25 October 2010 (UTC)[reply]
Incorrect. Some flavors of Christianity, in particular the Orthodox sects, view Jesus as fully man and fully god who has always existed; in other words, "Jesus" is both son and father, in the Holy Trinity. Other flavors of Christian thought prefer to view Jesus as distinct (the "son" of God), and while of "one substance", not the same as God. Still other flavors of Christianity, particularly various Protestant factions, completely discount the notion of "trinity" and consider Jesus as totally distinct from God. See Religious perspectives on Jesus. It's futile to describe any of these as the literal interpretation of the Bible. They are all considered "literal interpretation" of the Bible. "The Bible" is self-contradictory, and contains passages to support all of these points of view. Furthermore, there is great debate about which translation is the authoritative version of the Bible; these quibbles usually boil down to exactly one of these types of interpretative schisms. A thorough survey of the different sects of Christianity will unveil that (in addition to political reasons), the various factions split from each other because of fights about these subtle distinctions. For example, a particular Syriac Orthodox Christian faction might very well contend that literal interpretation means that Jesus has always existed, and always will exist, and Jesus was one and the same with the Creator God of Genesis. Thus, Jesus always existed and could have ridden dinosaurs many millions of years before Christ. An American Baptist Christian would deny this wholeheartedly, and insist that literal interpretation of the Bible clearly indicates that Jesus is distinctly and only the son of the God Creator of Genesis, and did not exist until four years before Christ, when he was summoned into existence to die for the sins of mankind. A different American Methodist faction might contend that Jesus existed in spirit but was not on Earth until 4 before Christ. A Catholic would insist that Jesus has always existed but did not take the form of flesh until four years before Christ. All of these are well-founded in literal readings of Bibles of various languages and translations. Scientists do not believe that any humans co-existed with dinosaurs. Nimur (talk) 20:44, 25 October 2010 (UTC)[reply]
All birds are dinosaurs and therefore are dinosaurs not extinct so your calculation are clearly wrong. Our dinosaur article states "Consequently, in modern classification systems, birds are considered a type of dinosaur — the only group of which that survives until the present day." I do not think Jesus have ridden any birds since they are generally to small. --Gr8xoz (talk) 20:42, 25 October 2010 (UTC)[reply]
(EC) 'generally' but not always - [9][10]. Also during Jesus's time the Moa was still around. Of course Jesus wasn't near anywhere ostriches are found AFAWK, let alone moas who AFAWK weren't known to any humans at the time. Nil Einne (talk) 20:56, 25 October 2010 (UTC)[reply]
And let us not forget the elephant bird of Madagascar which I think is the size of a young tyrannosaurus rex and could be ridden if it didn't kill you first.
The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.
Probability amplitudes
Why does ? I understand that it has to do with making , which is certainly convenient, but it can't all be just convention. implies that the probability for a state φ to be in the state ψ is equal to the probability for a state ψ to be in the state φ, which seems very non-trivial. How can this assumption be justified? 76.68.247.201 (talk) 20:49, 25 October 2010 (UTC)[reply]
I believe that equivalence is only true if the operators are Hermitian adjoint. As such, it is a matter of definition, not an assumption. See our article on invertible matrix, specifically:
The rules of Hermitian conjugation apply equally well to states as to operators. I should say, I have never been a fan of Dirac notation, because this ambiguity does not help simplify the problem (it just makes it use fewer characters to write). Nimur (talk) 22:15, 25 October 2010 (UTC)[reply]
I think there's something I'm still missing. If you say that a bra is the hermitian conjugate of a ket, then isn't that the same as saying ? Certainly you can't just define the bra to be the hermitian conjugate of a ket, no? 76.68.247.201 (talk) 23:05, 25 October 2010 (UTC)[reply]
I think this is the quantum mechanical version of the time-reversibility of physical laws, and as such it's reasonable to call it a physical assumption. It's justified in the same way as any other assumption, by consistency with experiment. That's not to say that you could drop it and get a quantum mechanics without time reversibility. It's too fundamental to the whole mathematical structure of the theory. -- BenRG (talk) 22:47, 25 October 2010 (UTC)[reply]
You define because it is Hermitian adjoint. Then, you have, by distributive property:
I think that does it, no? I have a more elaborate "rigorous" scheme, that involves pre-multiplying and post-multiplying and canceling out a bunch of terms that exactly equal 1, but that's a lot of TeX to type for essentially the same result as the above. Nimur (talk) 23:13, 25 October 2010 (UTC)[reply]
I think things have been defined differently for me than they have for you, so I'm having some trouble following your argument. For me, the "dagger" operation † can be defined such that |Ψ>† = <Ψ|. I imagine my "dagger" is your *, but then you apply * to the amplitude <φ|ψ>, in which case * should be meant as the complex conjugate...so my guess is that * applied to an amplitude means something different than * applied to a ket. But then you say that <φ|ψ>* = (<φ|)†(|ψ>)†...isn't that tantamount to saying <φ|ψ>* = <ψ|φ>? Unless I'm missing something obvious (and I have a horrible feeling I am), then what you said seems circular. 76.68.247.201 (talk) 23:47, 25 October 2010 (UTC)[reply]
You can also think of the ket vectors being the elements of a Hilbert space and the bra vectors as linear functionals that map elements from the Hilbert space to the set of complex numbers. Then the set of all these functionals form a Hilbert space too (the so-called dual of the original Hilbert space). You then want to define a mapping between the Hilbert space and its dual. We can choose this mapping such that any ket vector |psi> is always mapped to that bra vector <V| which, when applied to |psi>, maps it to its squared norm. It is convenient to denote that particular bra vector <V| that |psi> is mapped to by <psi|. It is easy to see that this mapping from the Hilbert space to its dual can be chosen to be anti-linear.
That works IF we force <ψ|ψ> to be real (and it seems that <ψ|ψ> = 1, as noted above), which you're obviously doing by making it the norm-squared of |ψ>. My question is, why is it justified that <ψ|ψ> = 1 or real or whatever? It has to be more than definition, because setting <ψ|ψ> = 1 has a physical significance beyond the mathematical formalism, namely that the probability for a state φ to be in the state ψ is equal to the probability for a state ψ to be in the state φ. 76.68.247.201 (talk) 01:08, 26 October 2010 (UTC)[reply]
That these inner products give the probabilities is postulated by QM. We postulate that the Hilbert space structure applies to Nature. There is no proof that this has to be true. You can prove the Born rule postulate, i.e. that |<a|b>|^2 gives the probability that a system in state |a> is found in state |b>, from weaker assumptions, see e.g. Zurek's proof. Count Iblis (talk) 02:01, 26 October 2010 (UTC)[reply]
But the only requirement is that |<a|b>|2 is real. But what's stopping <a|a> from being complex? You could argue that, by definition of an inner product, <a|a> must be a real number greater than zero, but that just shifts our burden because now we have to show that every property of the inner product matches with observation. I'm okay with saying that |<a|b>|2 = probability, because that's just the definition of the amplitude. I'm okay with saying that because that follows from 1) the adding of amplitudes when an observation isn't made (justified by the observed interference effects of quantum states), and 2) the amplitude for the state |ψ> to be in the state |φ> by first going into the state |i> is just , which can be proved easily. can be rewritten as if we assume some sort of distributive property. This allows us to write , and so we enter the domain of the vector space. The only step missing from making this a Hilbert space is to show that <a|a> is real and positive...but this implies, which I can't for the life of me see why it would be true. I hope this gives you some idea about how I'm thinking about this problem. In effect, we don't need Hilbert spaces to do quantum mechanics. 76.68.247.201 (talk) 02:41, 26 October 2010 (UTC)[reply]
I concur with BenRG that this is a time-reversal thing. The is part of the definition of the inner product of a Hilbert space, which existed as a concept before it was ever used to model quantum mechanics. One of the reasons that representing quantum states as vectors in a Hilbert space works well is because the way the inner product of a Hilbert space is defined results in an automatic invariance of joint probability densities under time reversal. There's a time order dependence to the meaning of . That expression is the amplitude that the system is in the state as considered at some earlier time, and is in the state as considered at some later time. is then the time reversal of that evolution of the system state, in that it's the amplitude that the system is in the state as considered at some earlier time, and is in the state as considered at some later time. For time reversal symmetry to hold, we require that the probability that that combination of before-and-after states occurs is the same, whether looking at the system forward in time or backward in time, i.e., we require that , which happens automatically due to how the inner product is defined. The above explanation may make more sense if you think of the state as being the superposition of various state vectors, at both the earlier time and the later time.
There's actually a complication to the above in that a state function is actually complex conjugated under time reversal. I.e., the time-reversal transformation of is , where * means complex conjugation, not Hermitian conjugation. However, that doesn't affect the basic argument above because .
I don't know how weak interactions fit into the above.
As a caveat, I'm just basing the above on my own understanding; it isn't something I got out of a quantum mechanics textbook. The QM textbooks I have generally effectively just say "quantum states are modeled with vectors in a Hilbert space", without saying anything about why quantum states are modeled that way. Hopefully bigger physics heavyweights like BenRG will shred my above post if my reasoning is way off. Red Act (talk) 03:27, 26 October 2010 (UTC)[reply]
If we're running out of helium, why not plunk down a bunch of radioactive waste in (well-sealed) room full of air (or some gas). Unless I'm missing something, the beta and alpha particles it gives off should slow down and then combine into helium... that's how the helium we get out of natural gas pockets forms, right? The only problem I could see would be that the room would potentially need to be REALLY big, and a lot of the walls would get irradiated... but surely if helium becomes valuable enough these aren't insurmountable costs. 96.246.58.133 (talk) 20:53, 25 October 2010 (UTC)[reply]
No need for a big room, nuclear reactors do this every day. I'm not sure if they collect the helium or simply vent it, but they definitely do need to handle it. But, I don't think they make enough to replace our uses of it. Maybe someone has time to do some order of magnitude calculations? Ariel. (talk) 21:07, 25 October 2010 (UTC)[reply]
The total production in the Earth by all the radioactivity all the way in to the centre is less than 10 000 ton/year our consumption was 12 000 ton/year in 1996 CE. We can not collect all the radioactivity in the Earth in a room and not prevent most of the helium from escaping the Earth.
Derivation:
Total amount of geothermal power, 30 TW (Twice the total human energy consumption so nuclear power will produce no significant amount of He)
Hello! I'm currently transcribing my great-great-great aunt's memoirs, and she mentions a strange phenomenon that took place on a Good Friday, circa 1890. She describes it thus: "at Easter the moon is always at the full, and right across the moon was a beautiful cross in vivid gold [...] We hurried home to be in time to tell our parents & they came out to look, but by that time one part of it had faded. Thus † became [right half of a cross]." Does anyone have a scientific explanation for this event?--Midgrid(talk)20:54, 25 October 2010 (UTC)[reply]
One obvious possibility would be a cross-shaped cloud at sunset that happened to be in front of the moon. If a group of people who happen to be at the same place at the same time point it out to each other after one of them notices, then there would be multiple witnesses. And I totally could see one of those little news-starved small town newsletters printing something as minimally earth-shattering as that. Decades ago, I visited my girlfriend at the time and her parents overnight in the tiny town she grew up in, and that event was big enough news that it wound up in the local newspaper. Red Act (talk) 22:01, 25 October 2010 (UTC)[reply]
I assume people could recognize any ordinary cloud, but there are a long list of oddities under the general heading of "optical phenomenon" which are quite rare in many areas and very striking to observe, which are produced by a relatively transparent layer of ice crystals. For example a moon dog produces extra spots that can be part of a cross-like pattern, though the ring would seem jarring to someone expecting a +-shaped cross - still, some religious traditions use a cross with a broad ring, so that might not disqualify it. The article sun dog gives a broader variety of how such things could look. Wnt (talk) 22:11, 25 October 2010 (UTC)[reply]
From the same source as my above question, my relative mentions a bright comet visible in the sky, circa 1879. Can anyone identify a possible candidate for me?--Midgrid(talk)20:57, 25 October 2010 (UTC)[reply]
(ec) Many comets were visible from Earth in 1879, and if your "circa" is interpreted as "plus or minus a few years," the list grows even larger. Tempel 1 is one possibility; several small comets are known only as "Comet 1879a, 1879b, 1879c", and so on. A letter to Nature dated July 3, 1879, describes astronomical details for one comet observed in early summer. Nimur (talk) 21:08, 25 October 2010 (UTC)[reply]
Thanks, but it's almost certainly not that one, as from the letter I gather that one needed a telescope to see it, whereas the one my relative mentions was clearly visible to the naked eye.--Midgrid(talk)21:31, 25 October 2010 (UTC)[reply]
Brorsen sounds like a likely possibility, as it passed close the to the Earth and my relative writes that the comet she saw was very bright and had an extremely long tail. "Circa" is not a big variation in this case: she was born in 1870 (or possibly 1871) and says she was "about nine" when she saw the comet. Curiously, she goes on to say that she doesn't think a comet has been visible since (writing c. 1935), despite the fact that even Halley's Comet, the most famous of its kind, passed by in the intervening period!--Midgrid(talk)21:30, 25 October 2010 (UTC)[reply]
Not only that, but the Great January comet of 1910 was the brightest comet of the century. No offense, but you're talking about a person who said she saw a golden cross on the Moon on Easter. Her reliability with regard to astronomical observations, I would say, is low. Comet Tuttle (talk) 21:42, 25 October 2010 (UTC)[reply]
No offence taken. :) I did initially think that she might have confused the 1879 comet with Halley, but there was such a large time difference between them.--Midgrid(talk)21:49, 25 October 2010 (UTC)[reply]
To fall does not hurt but hitting a rock or the ground may be hurt but I think there are some probability that the brain is destroyed before you feel any pain, are here anyone that knows more about neurology than me that can answer that? --Gr8xoz (talk) 22:21, 25 October 2010 (UTC)[reply]
The only conclusive way to answer such a thing is to do an experiment, e.g. a fMRI scanner is rigged to a subject tossed out of a helicopter... I don't think it's been done, though I haven't gotten through to Wikileaks lately. Note that an action potential in long myelinated axons in vertebrates can get up to 110 m/s or greater, which I think is 250 mph, versus the figure given for skydivers at terminal velocity of 120 mph. Which means that the hapless jumper landing feet first after falling prone should have 2 m = ~ 1/30 s to appreciate the sensation, minus the time needed to send and receive the signal. Note however that subjective time behaves strangely even under the mild alterations imposed by psychoactive substances, and I would not want to make any assumptions about how long the last fraction of a second might last under those conditions. Please note that Wikipedia does not give medical advice and disclaims all responsibility for any conclusions you might draw, in this world and the next. Wnt (talk) 22:42, 25 October 2010 (UTC)[reply]
On a recent NGC channel documentary they showed how you can jump safely from a height of about 10 meters on a hard concrete surface. You have to make sure you have enough forward velocity and make yourself small and roll when you hit the ground. Count Iblis (talk) 01:09, 26 October 2010 (UTC)[reply]
But one might aim for the water not the ground. This is a discussion about the question What is the maximum height from which a person can make a free dive without incurring physical injury?. A useful factoid observed is that Feet-first impact can be lethal due to explosive rupture of the large colon. Presumably that is not a pleasant experience. Cuddlyable3 (talk) 10:51, 26 October 2010 (UTC)[reply]
Having finally watched that YouTube video, I have to say that the idyllicness was somewaht spoiled by all the human chatter in the background. Maybe that's what made our poster think of jumping. HiLo48 (talk) 08:34, 26 October 2010 (UTC)[reply]
I doubt that. Humans have the tendency to get quite excited and make a lot of noise when you start pushing people off very high cliffs Nil Einne (talk) 11:24, 26 October 2010 (UTC)[reply]
Only of those you successfully push. Under the conditions where there are plenty of people or when other people are likely to see what you do as seems likely in the above case, I stand by my comment. Nil Einne (talk) 23:25, 26 October 2010 (UTC)[reply]
My textbook derives the relationship [Jx, Jy] = ihJz (I can't make h bars) by defining the J's as the generator of their respective Rotation operators...but only for the Rotation operators in regular Euclidean space (ie ). How do we know/show that this holds for the more general rotation operator over a ket space? 76.68.247.201 (talk) 04:37, 26 October 2010 (UTC)[reply]
I've tried to pass IQ test, but seems like it always involves some tough math calculations, which I don't like (so I still don't know my IQ). Nonetheless I'm quite familiar with geometry. Is there some other way to measure the IQ? 85.222.86.190 (talk) 08:36, 26 October 2010 (UTC)[reply]
Wikipedia has an article Intelligence quotient about IQ testing. The picture at the top of the article shows an example test item which you will see is a geometrical test. Properly designed IQ tests are not based on math. Cuddlyable3 (talk) 08:43, 26 October 2010 (UTC)[reply]
I've taken multiple IQ tests and none of them have contained "math". They do contain numbers, sequences, pattern matching, spacial analysis, etc... All of that could be mistaken as math, but it isn't. It is the primary point of an IQ test. It isn't a test of how intelligent you are. It is a test of how well you learn new patterns and processes and apply those to problems. -- kainaw™13:06, 26 October 2010 (UTC)[reply]
I agree, I've taken a couple IQ tests through the public school system, and besides perhaps a little basic arithmetic of small numbers, there was no math. There were some problems that had numbers, but no advanced computation was required, just pattern recognition, problem solving, and such. APL (talk) 16:05, 26 October 2010 (UTC)[reply]
Any IQ test you take that is "tailored" to your likes and dislikes is as useful as a screen door on a submarine. Take the real IQ test and whatever you get as a score is your IQ. Go learn some math! —Preceding unsigned comment added by 165.212.189.187 (talk) 13:50, 26 October 2010 (UTC)[reply]
Apologies for the unhelpful, incorrect and snarky reply from 165. There is no `real' IQ test, and picking one that is `tailored' to minimize bias is probably helpful in getting reliable results. The usefulness / meaning of the IQ concept is also contentious, see IQ#Criticism_and_views for starters. The OP may like this IQ test [13], which uses no words or numbers, just visual geometric patterns. This makes it a good tool for assessing how good one is at pattern recognition; extrapolate additional meaning at your own risk. --SemanticMantis (talk) 14:53, 26 October 2010 (UTC)[reply]
Quantitative analysis and pattern recognition are related to mathematical thinking. Many IQ tests use algebra or other "learned" skills in order to rapidly assess quantitative skills. There are many different IQ tests, and many different definitions of IQ, but it is safe to say that most psychologists consider mathematical capability to be an intrinsic part of intelligence quotient. You should read our article to understand what IQ actually is and how it is used, as well as the preferred ways to measure it. On the one hand, test designers want to avoid cultural biases and test inherent ability as opposed to learned skills; on the other hand, they also want to tests mathematical aptitude, because they consider mathematical ability an important element of intelligence. Imagine the SAT without a math section! (Of course, the SAT is an "aptitude" test and not an "intelligence quotient" test, but that is a euphemism and a trademarking/marketing ploy. SAT is highly correlated to IQ; for practical purposes, they both indicate the same thing. Nimur (talk) 15:34, 26 October 2010 (UTC)[reply]
I think a lot of the criticism of IQ would go away if it were called something like "Logical Thinking Quotient" instead of a name that grandly implies that the test measures "intelligence" as a whole. APL (talk) 16:05, 26 October 2010 (UTC)[reply]
I agree; that's why prominent agencies use a "euphemism" for it - such as "scholastic aptitude." This is a good thing - it's not just "political correctness," it's an effort to use a more proper terminology. As our article notes, the validity and significance of IQ is highly debated. In fact, every scholarly analysis I have every seen about standardized intelligence metrics prefaces all discussion with a disclaimer to the effect that intelligence metrics only measure whatever it is that they measure. Nimur (talk) 18:26, 26 October 2010 (UTC)[reply]
Just to clarify, the SAT is no longer intended to be an aptitude test, and no longer correlates with IQ as well as it used to. For a while, SAT stood for "Scholastic Assessment Test", and now is not an acronym for anything (officially). The SAT was originally designed to pretty much be an IQ test, but as "IQ testing" got sort of a bad rap in the mid to late twentieth century, it was redesigned to not be an intelligence test. Mensa, for example, will not accept SAT scores from after the 1994 redesign as evidence of intelligence [14]. Buddy431 (talk) 19:17, 26 October 2010 (UTC)[reply]
My UK freeview digital reception through a TV aerial is very weak (as measured by the signal indicator on the digibox) but usually sufficient for trouble-free viewing. However, I've noticed that when I iron clothes the signal deteriorates significantly (tends to break up). There's some tendency to get even worse when I tilt the iron downwards and it generates steam.
They're plugged into the same power outlet. I don't observe other devices suffering a similar problem when ironing (no crackles from the hi-fi speakers, etc.) so I'm not convinced there's an electrical fault on the iron).
Is it possible that the power used by the iron means that the digibox is somehow less able to process the signal it receives? Is the iron generating a field that interferes with the box? --Frumpo (talk) 10:18, 26 October 2010 (UTC)[reply]
Are we referring to an internal aerial here? If so have you considered the possibility the steam for example is interfering with the reception? Nil Einne (talk) 10:40, 26 October 2010 (UTC)[reply]
When the USA was having its debate over digital TV standards, the U.S. manufacturers claimed the European standards were particularly susceptible to "impulse noise" -- essentially a sudden electrical spark or electrical arcing, which creates a short momentary radio wave pulse. Even though it would be too short (and maybe too low power) to hear on analogue hi-fi speakers, the little radio wave click would be enough to upset all of the digital signal bits being transmitted at that moment, which could create a major glitch in the picture. The European standard was said to be much better at dealing with "multipath" interference -- ie the effects of reflections causing multiple versions of the same signal to be received, that would show up as "ghosting" on an analogue TV; but a bit worse at dealing with impulse noise. Impulse noise particularly gets caused by electric motors, where you get little electric arcs at the brushes that keep electrical contact between the stationary and the moving parts -- so vacuum cleaners, kitchen equipment, washing machines were particularly identified as troublesome. But it seems conceivable that if you had a low-level dodgy connection inside the iron (or if the iron's thermostat keeps switching on and off) that this might produce something similar. Jheald (talk) 11:01, 26 October 2010 (UTC)[reply]
A iron should not generate any interference except when the termostat turns on or off. If it does it is probably faulty, probably bad contact somewhere. --Gr8xoz (talk) 14:25, 26 October 2010 (UTC)[reply]
You also get noise when you put the iron down on the metal resting place. When I was a teenager I sometimes listened to shortwave radio and whenever my mother was ironing, I would hear the interference. This precisely when she was putting the iron down to move or change the clothes she was ironing. Count Iblis (talk) 14:44, 26 October 2010 (UTC)[reply]
Thanks everyone. I'll try using a different socket but I suspect that Jheald has identified the problem. The iron is designed so that steam is only generated whilst it's in the horizontal position. So, when I'm ironing, I guess there is a switch that's constantly flipping as I move the iron between the horizontal and vertical positions (in addition to the thermostat maintaining the temperature). What Count Iblis says is also observable but maybe that's because the iron is returning to the vertical position at the same time as it's being put on the metal resting plate. Or, is there some additional interference that could be caused by placing the iron near the metal plate? --Frumpo (talk) 15:29, 26 October 2010 (UTC)[reply]
What rank is the highest that (in any domain) has only one known species? I found an order: Tubulidentata (aardvark), but there might be a higher rank: Family, Class, ... with only one species. --Eu-151 (talk) 13:54, 26 October 2010 (UTC)[reply]
I would like to know whether one could generate an electric current in a conductor that is moving between two magnets REPELLING each other on either side of the conductor?
In other words, must the magnets be attracting each other or can they repel also?
It requires a magnetic field. When the magnets are repelling each other, their magnetic fields cancel out. It won't necessarily be perfect. If one magnet is bigger or closer, there will still be a current, but you'd be better off just having one magnet. — DanielLC16:48, 26 October 2010 (UTC)[reply]
Yes, a permanent magnet generator can be constructed with magnets mounted in a repelling orientation. If the conductor moves directly between the two repelling magnets, not much will happen, as Daniel mentions above. However, repelling magnets still create a field that coils can take advantage of if shaped and located properly. Here's an example; [15] __TunqstenCarbide XXXII (talk) 23:35, 26 October 2010 (UTC)[reply]
It would be wasteful to use magnets in a repelling orientation in a generator. At the null point, there would be absolutely no electricity generated. Move the windings closer to one or the other poles, and some electricity would be generated, but less than if the other magnets had not been installed. Thus it would seem to be an idiotic and wasteful design. Edison (talk) 05:13, 27 October 2010 (UTC)[reply]
that's perhaps a strong statement. Are you sure you understood how the design I mentioned works - seems not? Look at the FEA, mid page. Most coreless PMAs have iron backing rings to complete the magnetic circuit - the example I give is an ironless design, so perhaps it's useful if one had a reason to eliminate weight. Also, Halbach arrays have a fair amount of repulsion in their design (if you've ever tried to assemble one you'd know this), and they definitely have uses, including the elimination of iron or other magnetic material to complete the magnetic circuit. __TunqstenCarbide XXXII (talk) 07:02, 27 October 2010 (UTC)[reply]
In advanced physics, we use theories about particles to explain certain measurements. A particle is a small "object" that has some property - usually, it has a position and a velocity, and most of the time also has a mass and a momentum and a few other numbers associated with it. In high-energy physics, we sometimes expect an extra particle to exist, even when we don't see it, in order to explain why our measurements are as they are. So, we sometimes "guess" that there is a particle that we haven't seen, but that is responsible for some measurement. Tachyons were one such "guess" - but we have never actually seen any reason that they should exist, or do exist. Experiments have almost ruled out the existence of tachyons, by requiring them to have physical properties that do not make sense (like having an imaginary number value for the mass). Nimur (talk) 15:40, 26 October 2010 (UTC)[reply]
As far as anyone knows tachyons do not exist. So why do we talk about them? They basically are a result of the math used to calculate speed. If you plug in a speed above the speed of light you get a weird result, and we call the result "tachyons". In physics if you don't have the math for it, it doesn't exist. Conversely if there is math for it there is a claim that it may exist. That doesn't mean it always does, but very very frequently it does. There are many particles that were found because someone noticed something in the math, and looked there. Ariel. (talk) 18:37, 26 October 2010 (UTC)[reply]
Tachyons are about one step up from unobtainium, in the sense that we can invent mathematical derivations of existing models which perhaps, maybe, could allow for the existance of tachyons. But they probably don't exist. --Jayron3205:41, 27 October 2010 (UTC)[reply]
in physics we read frictional force does not depend upon area of contact provided normal remains same. In chemistry under topic of viscosity we read frictional fore is directly proportional to area.
Also according to classic physics
" friction exists becoz of locking of two layers"
if it is so area should be directly proportional to the frictional force as more the area, more will be the locking means more force needed to overcome it ie more friction.
according to Modern physics
"friction is the force needed to break the bonds(cold welding)between the two surfaces "
[i.e. why if we make something extra smooth friction will increase tremendously]
if it is so then also area should be directly proportional to the frictional force as more area means more bonds means more friction.
This has been discussed many times here. (here's one such discussion.). Basically, the approximation you learn in highschool for calculating friction really only works well in certain situations with perfectly ridgid objects. (Like the steel spheres you might use for experimenting!) In real world calculating friction is surprisingly complex, and not a 100% solved problem. APL (talk) 16:38, 26 October 2010 (UTC)[reply]
Yes, friction is complex in reality, but there is a much simpler explanation for the apparent discrepancy. Friction is indeed proportional to area, but also to normal force. In a fluid, the normal force (per unit area) tends to be independent of area, whereas in the simplified school physics example, the normal force (per unit area) is inversely proportional to the area of contact, hence the total friction force for a solid block is approximately independent of the contact area. Dbfirs17:23, 26 October 2010 (UTC)[reply]
When there is less area, there is a stronger force per unit area pushing the two surfaces harder together. This makes the friction approximately the same. --Chemicalinterest (talk) 00:56, 27 October 2010 (UTC)[reply]
There are a lot of ways to derive the Lagrange points. Conventional derivations use Lagrangian mechanics (surprise!) Our article links to a few derivations; Here is a derivation by Prof. Cornish at Montana State University, and more mathematical elaboration. "The brute-force approach for finding the equilibrium points would be to set the magnitude of each force-component to zero, and solve the resulting set of coupled, fourteenth-order equations for x and y," but using symmetry, Prof. Cornish simplifies this significantly. Such simplification could be written algebraically as a 5th-order equation (surely, the five points come from the five roots where some equation defining the magnitude of the net force equal zero). But writing that out, and then solving it, might be horrible. As I recall, this is similar to the approach in Marion & Thornton's textbook on classical dynamics - though they don't provide a full solution. Nimur (talk) 18:52, 26 October 2010 (UTC)[reply]
How strong are adult male orangutans? I know that chimps have about five times a human's strength and gorillas have about 20 times as much strength, but does anyone have such a figure for orangutans? And how aggressive are they? Have there been any documented cases of attacks on humans (not counting sexual assaults)? --TheHighFinSpermWhale23:29, 26 October 2010 (UTC)[reply]
Hello. If the conditions for a double displacement reaction to proceed are one of:
precipitation,
evolution of gas, or
acid-base neutralization;
can CuO react with H2SO4 to yield CuSO4 and water? Since the products are soluble and not gaseous, is CuO a basic oxide? What determines an oxide salt to be a basic oxide? Thanks in advance. --Mayfare (talk) 23:51, 26 October 2010 (UTC)[reply]
Copper(II) sulfate is quite soluble in water, so you wouldn't so much form CuSO4, but rather Copper(II) ions and sulfate ions in solution. But yes, copper oxide is a basic oxide, and will thus dissolve in mineral acids like sulfuric acid (though "double replacement" is really a simplistic way of looking at it). As for basic versus acidic oxides, more electronegative elements form acidic oxides, while less electronegative elements form basic oxides (in general). Some oxides of elements that are sort of in the middle can go both ways: ZnO and Alumina, for example. Higher oxidation state metal oxides can also form acidic oxides, like Chromium trioxide. Buddy431 (talk) 01:47, 27 October 2010 (UTC)[reply]
"Double replacement reactions" is the sort of stuff we teach to first semester high school students to simplify models of chemical reactions into basic model reactions (synthesis, decomposition, single replacement and double replacement). It's not really used as a concept in "real" chemistry; rather it is a gross simplification used as a means build upon later in teaching more accurate models. This is a common pedagogical tool; you teach a very inaccurate, but very easy to understand model, and then you build more accurate models upon it, abandoning the old models. It's like teaching a kid to ride with training wheels, knowing eventually he will be riding a regular bicycle. See Lie-to-children. During the first few weeks of class, you start with the "double replacement" model, eventually you abandon it when you start to work with net ionic equations. --Jayron3205:36, 27 October 2010 (UTC)[reply]
October 27
Naturally-occurring promethium
I understand that promethium doesn't occur naturally in any place to which we have easy access, but if it did, would we always find it to be radioactive? I understand that all known bits of this metal have been radioactive, but they've always been produced by human-created nuclear reactions, and I know that nuclear reactions can also create radioactive isotopes of elements that aren't normally radioactive, such as carbon. In other words, if we found promethium that hadn't been created by human-started nuclear reactions, would it necessarily be radioactive? Nyttend (talk) 00:50, 27 October 2010 (UTC)[reply]
See Isotopes of promethium. The most stable is 145Pm with a half-life of 17.7 years. So all of them are radioactive. Note that the more stable the easier it is to find the isotope, so it's not likely there is another more stable isotope out there that we did not find, plus the table seems pretty complete. Ariel. (talk) 01:30, 27 October 2010 (UTC)[reply]
(edit conflict with Ariel) We have an article Isotopes of promethium, which asserts that there are no stable isotopes, but doesn't really give an explanation. However, if we look at Isotopes of technetium, the Liquid drop model, which is pretty good at predicting nucleus binding energies, rules out the possibility of any stable isotope of technetium. I assume that a similar reasoning applies for promethium, and if someone more knowledgable than me wants to, an explanation similar to the one in "Isotopes of technetium" article could be put in the "Isotopes of promethium" article. Buddy431 (talk) 01:34, 27 October 2010 (UTC)[reply]
Note that when a reactor creates radioactive isotopes, it's actually just creating isotopes that happen to be radioactive. To put that (and others' comments) another way, it's an intrinsic property to the isotope, not a result of how it was made. So you'd have to figure out a previously-unknown isotope. And models of the whole general idea here don't seem to find any. DMacks (talk) 01:48, 27 October 2010 (UTC)[reply]
One caveat (which doesn't in any way detract from the general truth of DMacks statements) is the existence of nuclear isomers. Most notable is technetium-99m which has a half life of 6 hours, versus "regular" technetium-99, with a half life of 211,000 years. The two have the same number of protons & neutrons (they are the same isotope, for certain definitions of isotope), but have different half lives. The half life of 99mTc (or plain 99Tc) is an intrinsic property of that state, and doesn't depend on how it's made, but whether you get 99Tc or regular 99mTc depends on how you make it. -- 174.31.221.70 (talk) 02:25, 27 October 2010 (UTC)[reply]
Hmm, DMacks' comment is quite illustrative. I'd never known this before; I'd assumed that these isotopes were radioactive because of the way they were made, not because of their intrinsic natures. For that reason, I had no interest in the isotopes article. So...does promethium not occur naturally in any place to which we have access because it's radioactive? Nyttend (talk) 02:34, 27 October 2010 (UTC)[reply]
Basically if it is only and always radioactive and has very short half lives (which it does), you won't find much of it around. You can still find some of it in very trace quantities because other radioactive elements will convert into it. But if it is very unstable, those won't stick along very long, and the fact that only a few reactions seem to lead to its creation make it pretty rare. If it happened to be created by a more common reaction (e.g. the alpha decade of uranium or thorium or whatever), you'd see more of it, even though it is radioactive. Radon, for example, has no stable isotopes either, but is present in far greater quantities because the reaction that creates it is pretty common by comparison. --Mr.98 (talk) 02:40, 27 October 2010 (UTC)[reply]
The promethium article notes that there is probably about 570 g that exist in the earth's crust at any one time. But given the ~18 year maximum half life, it disappears quickly. Even if the entire earth was made of solid promethium, in just over 100 half lives (2,000 years), you would have less than 1 mg of it left. The only promethium you'll find was made relatively recently. -- 174.31.221.70 (talk) 06:29, 27 October 2010 (UTC)[reply]
How do gold and other valuable (and relatively non-chemically-reactive) metals collect and concentrate into very localized deposits, veins, lodes, etc.? I did not get a very good sense of how that is thought to occur from the Gold article; Placer deposits make some sense, but are there other mechanisms? (Would separation-by-density, as in centrifuge enrichment but using gravity and a lot more time, be a possibility?) WikiDao ☯ (talk)03:09, 27 October 2010 (UTC)[reply]
some one will helpme that in which place i should ask my questions
i am a student of electronics engineering and want to ask question about my course and my technology related to be answared by wiki friends so plz some one show me the particuler place where this technology will under discussion thank you —Preceding unsigned comment added by Dawoodian 09 (talk • contribs) 08:01, 27 October 2010 (UTC)[reply]
You've come to the right place! Wikipedia's science reference desk (this web page) is an appropriate place to ask questions about electronics engineering, especially if you haven't been able to find the answer to a question within Wikipedia's articles. We won't do homework problems for you, but we can help you to do it yourself if you get stuck, and help you grasp the concepts that you are learning. Do you have a question currently, and if so, what is your question? Red Act (talk) 08:34, 27 October 2010 (UTC)[reply]
Hi, I recently asked a question about electrical current in a conductor placed between 2 repelling magnets. I got some interesting and sensible responses, thanks!
I have a few images of what I have in mind, but this site seems too cumbersome to upload images for communication purposes.
If someone is at all interested in entertaining an idea of mine, could they kindly beep me e-mail redacted.
I know too little about associated electromotive force or magnet technology (despite googling everything I could get my hands on), so any assistance would be very much appreciated!
...in response to my first question on magnets and electricity generation (let my try and use a few words to paint a thousand pics!):
I am picturing a magnetic bearing designed in a toroidal shape (equilateral triangular in section) which spins around the toroid's central axis.
It is supported by repelling magnets on the upper and lower triangular faces and on interfacing surfaces of a spherical holder.
A solenoid coil wraps around the toroid, but is attached only to the holder on the inside face, so that the toroid runs inside the solenoid.
The bearing therefor is the generator, one and the same. No air inside and no friction, hopefully = no resistence to motion/acceleration(??)
If this ain't gonna work, one could piggyback the generator on the magnetic bearing, but this is plan B...
What are your goal? Compact generator, efficient generator, Free energy, eye catching demonstration or what? It is hard to coment on a plan without knowing the goal. How are you going to drive the rotor if it is encapsulated in a toroidal coil? (I do not think it is a solenoid if it is bent to a toroidal shape.) --Gr8xoz (talk) 13:37, 27 October 2010 (UTC)[reply]
Tillage
Discuss the following tillage equipment under mouldboard plough and Disc plough.
-Functions and constructional detail sand components
-Principle of operation.
-Power requirement. —Preceding unsigned comment added by 80.87.90.71 (talk) 12:39, 27 October 2010 (UTC)[reply]
When people offer a chocolate bar to a caveman or an alien they almost always eat it and enjoy it, thus establishing a bond between the human and the creature. Would an alien or a caveman actually enjoy a chocolate bar in real life? —Preceding unsigned comment added by 82.31.238.189 (talk) 13:27, 27 October 2010 (UTC)[reply]
I don't think you're going to have to worry about it, but the "caveman" (i.e. a pre-sapiens hominid) would enjoy the chocolate quite a bit, as many/most primates have a well developed love of sweet things. As soldiers and travelling merchants have found throughout the ages, giving people sweet stuff (and booze) is a great way to make friends. If aliens exist, we have no way of knowing what their metabolism would be like. Maybe they'd like it, maybe they'd drop dead from theobromine poisoning. Matt Deres (talk) 13:36, 27 October 2010 (UTC)[reply]
I think it really depends on whether the alien is anthropomorphic. The caveman, we would know some caveman would like it, and others wouldn't, because they are human, and we are human, and they are human, and they have tastebuds, like a human. But with the aliens we can't be anthropocentric because we have not met aliens. Unless they are already living among us. If they are, then yes, they would enjoy chocolate. Unless they don't. AdbMonkey (talk) 13:59, 27 October 2010 (UTC)[reply]
I thought the most common theme was that of the alien offering the chocolate bar to the native, as with the G.I. giving his Hershey Bar to a local child. -- 124.157.254.112 (talk) 14:42, 27 October 2010 (UTC)[reply]
The Vitamin B12
This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis or prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the talk page discussion (if a link has been provided). --TenOfAllTrades(talk) 13:57, 27 October 2010 (UTC)[reply]
I don't see anything wrong in answering this question, so I will. If you disagree, take it up at my talk page, but as far as I'm concerned the OP has made a valid scientific question and has expressed no intent to harm him/herself. Taking vitamin B12 in large amounts has generally been considered safe, and any excess B12 is removed through the lower digestive tract as usual. That said, no study to my knowledge has established a safe upper limit for B12 consumption, so it's clearly best to stick the recommended daily amount unless otherwise instructed by a doctor. If you're considering taking more than the recommended daily allowance, you should discuss this with a physician as B12 can interact with certain drugs, and has been known to cause side effects in rare cases. Unfortunately I can't answer the other half of the question about how long it takes to act. Regards, --—Cyclonenim | Chat 15:22, 27 October 2010 (UTC)[reply]
mosquito eggs and larva
1.what would be happen to the human when swallow the mosquito eggs and larva?
2.what would be happen to the mosquito eggs and larva inside the human body?
If i was to put a carrot in a blender till it was a soup like consistency, would be considered physically to be a liquid? it would pour like one and fit the shape of the container. also in chemistry solids can't normally be pured or made to fit their container, but salt,sugar and sand can be poured and Play-doh can change its shape.--Lerdthenerd (talk) 14:20, 27 October 2010 (UTC)[reply]
Maybe? A puree is a Suspension (chemistry). If you left it alone for long enough (a week?), the tiny solid carrot particles would sink to the bottom. Salt, sugar, and sand are not liquids because each grain is actually a very small solid. In the case of liquids, each "grain" is a single molecule. The behavior of Play-doh is an example of Plasticity (physics). Anyway, I don't think "being able to fit their container" is a perfect rule for separating liquids from other states of matter, it's just a general rule told to schoolchildren. Indeterminate (talk) 14:52, 27 October 2010 (UTC)[reply]
This is a magnet motor design idea.
In response, the idea is to try and combine the function of a magnetic bearing with that of a generator (solenoid) -
so that you have one set of magnetic array, instead of 2 seperate ones. The coil is fixed to the outer shell on the inside, not fixed to the rotor.
The big question still remains: would the repelling flux with the wire between the magnets cause problems instead of electrical induction?
If it does, one needs 2 systems then and the whole thing gets more complicated.
Another consideration: without air or friction bearings, just how fast could this rotor spin (or keep accelerating) before it disintegrates to pieces - centrifugal forces there...?
![{\displaystyle dU=C_{V}dT+\left[T\left({\frac {\partial p}{\partial T}}\right)_{V}-p\right]dV\,}](/media/api/rest_v1/media/math/render/svg/8e9fcc2a4d11ae7d9f91c9d979556624221d9061)
Agree – and maybe also consider raising the issue for discussion on that article's talk page just as you have here – happy editing! :) WikiDao ☯ (talk) 01:58, 24 October 2010 (UTC)
--Ludwigs2 19:54, 24 October 2010 (UTC)
-chloride-sample.jpg?lang=en)
