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Does light make any sounds as it passes through different mediums (e.g. air, water, etc.)? If so, how is the "sound" measured? —Preceding unsigned comment added by 99.250.117.26 (talk) 00:59, 23 October 2010 (UTC)[reply]
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]
Did you mean to reply to me with that comment? TenOfAllTrades(talk) 18:04, 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]
There is always a finite chance that vibrational modes in the solid will be excited by the photon. This is described by the Debye–Waller factor. Count Iblis (talk) 18:41, 23 October 2010 (UTC)[reply]
And if the frequency of the light matches the frequency of a phonon mode in the solid... John Riemann Soong (talk) 23:25, 27 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]
This seems to resemble bakanae, except caused by storms rather than fungi. However we do not provide professional horticultural advice. ~AH1(TCU) 18:04, 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]
An almost identical question was asked a year ago. See HERE. Dolphin(t) 07:39, 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). Dbfirs 07:41, 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). Dbfirs 10:38, 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]
Why is the nuclear submarine here http://www.bbc.co.uk/news/uk-scotland-highlands-islands-11610817 giving off steam or smoke? On looking at it closely, it looks like it is issuing from a nozzel under pressure, so I presume it is steam. Although real steam hot gas would be clear, not white. 92.24.178.5 (talk) 09:36, 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]
Video of the sub shown today on CNN shows both the steam plume and water gushing out of a side vent. Cuddlyable3 (talk) 19:13, 24 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.)
Or since those engine designs only make fuel approximately 1.5x more efficient (instead of the 3x or more that we hear about in legend), the oil companies may have decided to let them slide. --Let UsUpdateWikipedia: Dusty Articles 10:33, 23 October 2010 (UTC)[reply]
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 had heard that the snake-headed aliens that run the oil companies staged the moon landings so they could hide the H2O-powered-car documents somewhere nobody would check ... underneath the moon's surface. But NASA had to get all uppity, publishing about the secret super-valuable hidden water in the moon rocks. The proverbial cat is out of the bag now! Nimur (talk) 18:28, 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]
Man, I wish some big oil company would bribe me to keep quiet... --Mr.98 (talk) 16:34, 25 October 2010 (UTC)[reply]
Mine already do, and they call them stock dividends.
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 ☻ Talk 21:34, 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 ☻ Talk 21:38, 23 October 2010 (UTC)[reply]
Fourth, if her battery dies, she can get a new phone. Nil Einne (talk) 01:10, 24 October 2010 (UTC)[reply]
Thanks, all - excellent - especially the Finlay rave - sets me off in fun directions .... Adambrowne666 (talk) 12:18, 27 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]
UDFy-38135539 is a galaxy and not one star but otherwize is the big picture correct. --Gr8xoz (talk) 14:55, 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]
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 ☻ Talk 13:29, 23 October 2010 (UTC)[reply]
Are there any human bodies floating around in space? Like if an astronaut got separated from their ship or whatever —Preceding unsigned comment added by 96.226.69.239 (talk) 12:57, 23 October 2010 (UTC)[reply]
No. Space accidents and incidents lists all the spaceflight fatalities, and all the bodies are accounted for. There are samples of a hundred or so people "buried" in space - see space burial. Laika spent 162 days in space, almost all of that time dead, on Sputnik 2. -- Finlay McWalter ☻ Talk 13:32, 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]
The "sugar cube" may be a stack of memory chips, increasing the capacity. --Chemicalinterest (talk) 16:32, 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]
Do you have a source for 2.3%? I would have thought it was lower myself, but I do not know how habitable is being defined here. Googlemeister (talk) 15:45, 26 October 2010 (UTC)[reply]
If you release a gas into a vacuum, it will expand. Will it also cool? I would think yes, intuitively, but where does the kinetic energy go? 76.68.247.201 (talk) 19:23, 23 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]
(ec) I very vaguely recall something from ancient school days about PV=nRT, no? PЄTЄRS J VЄСRUМВА ►TALK 19:49, 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]
Sorry, why does the gas do positive work? 76.68.247.201 (talk) 20:46, 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.
I will stop repeating myself now. Feel free to continue to believe whatever you believe. –Henning Makholm (talk) 02:11, 24 October 2010 (UTC)[reply]
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]
what r toothbrushes made out of —Preceding unsigned comment added by Kj650 (talk • contribs) 19:25, 23 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]
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]
Couldn't you just take the other 3 quadrants, calculate the position of their COMs then treat the problem as 3 discrete masses? Theresa Knott | Hasten to trek 19:46, 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 trek 22:02, 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. Dbfirs 07:52, 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]
Yes, or simply, from symmetry, when you find one co-ord, the other is the same. Dbfirs 12:35, 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]
It's not even Joules if you consider it as moles; molar masses are in g/mol, but J are kg m2s-2 ... --81.153.109.200 (talk) 17:46, 24 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]
Could it be the Creole Cattle of Guadeloupe? (Around 60% of the island's bovine livestock, 45,000 heads of cattle, 20,000 cows, according to that article which is unfortunately in French). Wikipedia only has a stublet on the Argentine Criollo. ---Sluzzelintalk 21:54, 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]
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]
If a penny is kept away from water, it will turn more black than green, indicating the formation of a copper(II) oxide. --Chemicalinterest (talk) 22:43, 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]
Don't know why they do it, but there is good reasons why the term pecking order is a long established part of the English language. HiLo48 (talk) 23:51, 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]
Do all chickens do it, or just the Republicans? Wnt (talk) 01:10, 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]
I'll bet that if you find loci linked to this behavior, at least one of them shows up more in Republicans than Democrats. See [2][3][4]. Wnt (talk) 09:23, 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? Dbfirs 07:50, 24 October 2010 (UTC)[reply]
I don't know about "inquisitive", but they do peck at red things for some reason. Red contact lenses can prevent this. Vimescarrot (talk) 16:11, 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]
![{\displaystyle dU=C_{V}dT+\left[T\left({\frac {\partial p}{\partial T}}\right)_{V}-p\right]dV\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e9fcc2a4d11ae7d9f91c9d979556624221d9061)
