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May 18

Dry skin itch

Why, precisely, does dry skin cause itching? What physical or chemical factors are involved? 69.224.113.202 (talk) 02:22, 18 May 2009 (UTC)[reply]

OR: What causes itching is people asking about itching on the Ref Desk - damn you! --Tango (talk) 02:26, 18 May 2009 (UTC)[reply]

Desquamation. I would say that the itching is an allergic reaction to dead epidermal cells, but researchers put it like this. Cuddlyable3 (talk) 11:45, 18 May 2009 (UTC)[reply]

Evidence of a civilization a billion years ago

Just a thought exercise: Assume that a civilization existed a billion years ago (and continued for about the same length of time as our own achieving roughly the same level of technology that we now have). Then, for some mysterious reason, this civilization disappeared. My question is: would there be any evidence that this imaginary civilization ever existed? —Preceding unsigned comment added by 209.161.213.227 (talk) 04:18, 18 May 2009 (UTC)[reply]

Well, sure. There'd undoubtedly be some physical artifacts left, even though much of it would be destroyed by the ravages of time. They'd show up in the fossil record, one way or another. More importantly, the ecological impact would probably be detectable. (A similar civilization couldn't have existed back then, though, because at the time there was really no ecosystem that civilization could've had an impact on, so the parameters would just have to be different.) I don't know if it would be anywhere near enough for anyone to actually be able to tell what the civilization was like, but I think there'd be evidence of its existence if one knew where to look and how to interpret what they were seeing. Actually, come to think of it, perhaps the most obvious piece of evidence would be that life on Earth would be completely different from what it is now: whatever bacteria and whatnot those people would leave behind would be the basis for our own life. Our primordial soup could literally be someone's leftover soup. -- Captain Disdain (talk) 09:42, 18 May 2009 (UTC)[reply]

Anything 'they' might have had or made back then would not likely be visible on the surface of the planet. Everything would be way underground buried deep in rock, as would any fossils of the creatures which had this civilisation. Life was around at that time, and had been around for at least 1.5 billion years before that, so it is certainly not impossible that some creature or another developed some sort of civilisation, which then got wiped out in some way or another. Changing the time frame to just 500 million years ago, we could say the same. It would, however, be very hard to detect, unless you knew what you were looking for. On a related note, you may want to look at this and see what you make of it. --KageTora - (영호 (影虎)) (talk) 11:13, 18 May 2009 (UTC)[reply]

Suppose the ancient civilisation having technology similar to ours did the same things (why not?). Expect aliens to arrive about now, attracted by the Arecibo message and asking if we want our probes back, like this. Cuddlyable3 (talk) 11:21, 18 May 2009 (UTC)[reply]

As the oldest fossils are "several billion years old", it seems likely that some evidence of human life would certainly remain. As many man-made objects are less susceptible to decay than living organisms, this would present evidence for our technological achievments (even simple mass-produced trash is dependant on a huge infrastructure). This, along with the planet-wide climate effects would make deduction of a widely-spread technological society pretty conclusive, imo.

Off course, none of this necessarily applies to any past civilisation, as they could easily have been entirely arable/biodegradable. Civilisation does not require an industrial revolution.Yob Mod 11:28, 18 May 2009 (UTC)[reply]

Having said that, there are a number of "ooparts" which bear consideration. I'm not saying I am a believer, but it would be interesting if we could find an explanation for them which fits our current 'knowledge' of history, or maybe we would have to rethink what we 'know'.--KageTora - (영호 (影虎)) (talk) 11:41, 18 May 2009 (UTC) I fixed the oopart link. Cheers.Cuddlyable3 (talk) 12:32, 18 May 2009 (UTC)[reply]

Cheers. I wondered why it wasn't working.--KageTora - (영호 (影虎)) (talk) 12:36, 18 May 2009 (UTC)[reply]

The book The World Without Us and the Discovery channel documentary that comes from it: "Earth Without People" follows the unlikely scenario where all humans suddenly vanish. That's not a very likely one - but if there were some kind of mass extinction due to (say) a massive meteor strike - then the story it tells of what remains of us after thousands of years is applicable - and it's pretty chilling. Aside from artifacts left on the moon, mars and out in deep space - there doesn't seem to be much that would survive. If that's true - then it might very well be that a sufficiently ancient civilisation would be very hard for us to detect. The fact that we find fossilised bones from the time of the dinosaurs (for example) leaves us with the impression that we'd also find fossilised Internet Routers and other indicators of civilisation. However, in reality, fossils are extremely rare - if a civilisation like ours were only to last (say) 100,000 years - then the probability of finding something within that window would be pretty tiny...the fossil record simply isn't that reliable. We have actually discussed this question before - and my opinion remains that the best place to look for such evidence would be out in space at places called Lagrange points - which would be a natural place to leave things that you wanted people to find a hell of a long time into the future. 13:16, 18 May 2009 (UTC)

Are even L-points stable on scales of a billion years? --Tango (talk) 14:54, 18 May 2009 (UTC)[reply]

I'm reading a book that depicts a similar scenario where aliens come to visit Earth 200 million years from now...and the only evidence they find of a past civilization is on the Moon (but it is difficult to tell if the artifacts were left there from a civilization on Earth or from another intergalactic traveller). Also, it really isn't a mystery why past ancient civilizations collapsed, leaving behind their ruins as if they disappeared suddenly: the biggest factors are war, famine, environmental depletion (think Easter Island), and climate change. The Arecibo message was aimed at the globular cluster M13, but it would take about 50,000 years for any reply message to reach Earth, assuming that the message actually arrives at its target, which it probably won't, because the motion of the cluster around the galaxy would likely have carried it away from the path of the message by then. ~A H 1^(TCU) 16:06, 18 May 2009 (UTC)[reply]

A billion years ago all life was microscopic life, with stromatolites about the only macroscopic evidence of life. So, essentially any macroscopic fossil would stand out as evidence of the unexpected. If you are positing a civilization of people that spring up out of nothing (e.g. aliens landing on the Earth) and lasts for only 10000 years, then the duration of their existence and their spread might well be very hard to find in the fossil record. On the other hand, if you are imagining a civilization that arises organically, then it is hard to imagine doing that without developing a wider ecology of macroscopic life. For example there are far more trees on Earth than people. A naturally developed macroscopic ecology would be expected to have taken many millions of years. If you include all of those associated forms of life and their long history, then it would be shocking not to have some evidence of that in the fossil record. If we assume macroscopic life is a prerequisite for civilization, then I think we can rule out the development of civilization a billion years ago. Dragons flight (talk) 17:28, 18 May 2009 (UTC)[reply]

coefficient of volume elasticity

is there anything called coefficient of volume elasticity?? —Preceding unsigned comment added by 122.50.137.137 (talk) 05:10, 18 May 2009 (UTC)[reply]

I don't think there is anything called volume elasticity and I am sure there is nothing called coefficient of volume elasticity - DSachan (talk) 11:16, 18 May 2009 (UTC)[reply]

Possibly you are thinking of the elasticity tensor or one of the elastic moduli such as the bulk modulus. Gandalf61 (talk) 12:51, 18 May 2009 (UTC)[reply]

Here is a definition of the coefficient of volume elasticity. The term is also used in biophysics, as in this abstract. Isn't Google great? Looie496 (talk) 18:44, 18 May 2009 (UTC)[reply]

dimensional analysis

my teacher set me this question:

using dimensional analysis, check whether the relation is correct.

S_nth = u + a/2(2n - 1)

where the symbols have their usual meanings

i know that u stands for initial velocity and a for acceleration, but what about n and S_nth??? —Preceding unsigned comment added by 122.50.137.137 (talk) 05:16, 18 May 2009 (UTC)[reply]

This is a formula in Kinematics ( study of spped, distance, tima and acceleration ) for the distance traveled in the nth second. Say you are dropping a ball from a cliff. If it has fallen 5m after 1s, 20m after 2s, and 45m after the third second, the distance travelled in the first second is 5m, the distance travelled in the second second (ignore the pun) is 20 - 5 = 15m, and the distance travelled in the third second is 25m. So your Snth has the same dimension as S, distance. n is the number of seconds, so you should be able to tell its dimensions. Also don't forget to sign your posts by typing 4 '~'s in the end...Rkr1991 (talk) 05:29, 18 May 2009 (UTC)[reply]

I don't like this business of "where the symbols have their usual meanings" because 'n' isn't often used as a symbol for 'time' and S_nth isn't a "usual" symbol at all (although 'S' is conventionally used for a distance travelled). That makes this question very hard to answer definitively.

Does the constant '1' have a dimension (like '1 second' or '1 hour')? No - because if 'n' is a time - then (2n-1) is only dimensionally correct if '1' is also a time. As Rkr1991 explains it, the equation is only valid if the units happen to be seconds. If you were to provide 'n' in hours - then it would tell you the the distance travelled in the n^th hour...but only if that magic '1' constant is now assumed to be in hours.

But as Rkr1991 explains it, there is a 'get out' clause here. If the equation is truly stated to describe: the distance travelled "in the n^th second" then that leaves open the possibility that 'n' is "the number of seconds elapsed" - not "the elapsed time" (a subtle distinction that leaves n dimensionless and makes (2n-1) work). Using 'n' as a dimensionless number is certainly reasonable here because 't' would be the more usual symbol to use for a 'time'...but then the dimensions of 'u' and 'a/2(2n-1)' don't match anymore because u is a velocity (distance.time^-1) and a is an accelleration (distance.time^-2) so (2n-1) MUST be in units of time or else you can't add u to a/2(2n-1) legally. So that says that the units of 'n' MUST be time - and that constant '1' has to be in the same units as 'n' - which is unspecified in the equation. Furthermore if S_nth has units of distance - then the equation fails again because 'u' is a velocity and 'S' is a distance. If 'a' were zero then you'd have 'S_nth = u' - which doesn't work if S has units of distance. But if we believe Rkr1991's description of what the equation means then "S_nth is the distance travelled over a second" - which is a velocity - not a distance.

I think the equation is dimensionally incorrect - although the way Rkkr1991 explains it, there is some wiggle room. It would produce a meaningful answer providing that 'n' is defined as 'the time in seconds' and not 'the time' - and providing that '1' is really '1 second' and that S_nth can truly be construed to be a velocity. I think the question could have been better phrased - but the answer should be "No". SteveBaker (talk) 13:00, 18 May 2009 (UTC)[reply]

I agree with you, but to play devils advocate - the 2 could have units of time, and the (2n-1) not. The presence of constants in things like this always creates room for confusion. --Tango (talk) 15:00, 18 May 2009 (UTC)[reply]

Is this equation meant to be a discrete time version of kinematics? I really really don't like this "symbols have their normal meanings." That's a very meaningless statement. In the papers I've been reading lately, s is "slowness" (reciprocal velocity). And a is the anisotropy vector. And n is the Normal Move Out operator, or it can be almost anything else. You should verify the meaning of all of these variables, or you're really lost. "Conventional" variable names are only relevant in very specific contexts. Nimur (talk) 15:55, 18 May 2009 (UTC)[reply]

Well let me tell you that this equation is perfectly correct, both dimensionally as well as physically. S_nth is the distance traversed in the nth second, not the average velocity. Note the distinction. So it has dimension L. The only trick in the question is the first term on the right hand side. It is not velocity. It is Velocity seconds. Velocity multiplied by one second. This will have been given in the question, because it is not the usual meaning. n is a number, dimensionless. So now the rest should be obvious. Though it is not mentioned here, i'm quite confident the teacher would have mentioned that u is velocity seconds in class. This is a standard equation for distance traveled in the nth second, so it is obviously right. Rkr1991 (talk) 05:27, 19 May 2009 (UTC)[reply]

The notation is far from standard, so I'm guessing "usual meaning" means usual meaning in that class, so we can't really tell if the question is right because we don't know the notation. Sure, we can come up with interpretations that make it correct, but that doesn't mean those interpretations are the "usual" ones. --Tango (talk) 15:52, 19 May 2009 (UTC)[reply]

Well, Rkr1991 can hypothesise that our OP has miscommunicated the question to us - but for absolute 100% certain, 'u' is not "velocity seconds" in it's "usual meaning" in the context of dynamics/ballistics questions. The usual meaning is found in equations like:

S = ut + at²/2

...where the meaning of 'u' is "the initial velocity"...and our OP even tells us that explicitly. If the meanings of these terms are not "the usual ones" then all bets are off. 'u' could be the phase of the moon, 'a' the dissolution rate of resublimated Thiotimoline, 'n' the airspeed an unladen swallow and 'S' the amount of wood a woodchuck could chuck if a woodchuck would chuck wood. So either our OP has the question wrong - or the question was written incorrectly - or the answer is "No". Pick any one - but the answer isn't "Yes". SteveBaker (talk) 20:24, 19 May 2009 (UTC)[reply]

The reason i say the equation is dimensionally right because of two reasons : (i) Any equation which is right is also right dimensionally. You can derive the given equation yourself, writing the formula S = ut + at²/2 for t = n seconds and t = n+1 seconds and then subtracting. Note that the n seconds part cancels out in the first term, which is why u here represents velocity seconds, not velocity. This equation is pretty much standard, and it would be against my senses to call it wrong just because the teacher has told usual meanings. (ii) The OP has said usual meanings. For this particular formula, this is the usual meaning. Now, since the dimensions of each quantity has not been explicitly stated, there is a very high chance that the teacher has actually mentioned this in class. Ya, you guys are right, for the exact question the OP has asked, the answer is NO, but i am pretty confident that u here refers to velocity secons and NOT velocity. So if this question was asked to me in an exam, i would right YES as the answer. Having known a formula, my instincts bristle at having to write it as not correct. But i think this question really doesn't require so much debate. You got to ask the teacher what she meant by u. The other confusions, regarding n and 1, should be quite obvious by now. n is a number. So is 1. Rkr1991 (talk) 04:52, 20 May 2009 (UTC)[reply]

I disagree. If you derive the equation the way you suggest then 'n' has the dimension of time and the '1' in(2n+1) is really 1 second - I don't see a unit against that constant in that equation - which means that it's dimensionally incorrect...not because the equation is incorrect per-se but because the person who wrote it down omitted the 'time' dimensionality for the constant '1'. You certainly can't argue that the "usual meaning" of the symbol '1' is "1 second" - so the equation is quite simply wrong. If I take it as written and choose 'hour' as the unit of time for 'n' and whatever contorted units you then arrive at for 'S', 'u' and 'a' then it produces the wrong answer. It's a fundamental precept of dimensional analysis that you can't add a dimensionless constant to something that is not itself a scalar. This equation (as written) is therefore incorrect...and I'm quite sure that's what the teacher expects you to point out. SteveBaker (talk) 11:58, 20 May 2009 (UTC)[reply]

Can you please explain how n will refer to time here ? Remember the formula is for distance in the nth second. n here is clearly a number. When you substitute t in the equation, do not make the common mistake that t is n. t is n seconds. Catch the subtle difference. But to accept whether the equation is right dimensionally or not lies with each individual, but certainly it should be clear that n and 1 are numbers. Rkr1991 (talk) 12:35, 20 May 2009 (UTC)[reply]

'n' has the dimension of time because of the way you derived this equation. Look above - you said it yourself: "You can derive the given equation yourself, writing the formula S = ut + at²/2 for t = n seconds and t = n+1 seconds and then subtracting. - you can ONLY substitute t=n and t=(n+1) if 'n' (and '1') is in the same units as 't'. Hence if 't' is time - then so is 'n' - and so must be '1'. So your derivation of the equation is wrong unless both 'n' and '1' have time as their dimensions. You may be about to claim that you aren't setting t=n but t=nseconds - thereby making 'n' into a scalar - but that's just introduced a non-scalar constant so that 't=n.k' where k=1second (it has to be like that in order to make your 't=n' substitution be dimensionally correct) - but then the problem is that when you do the subsitution, you get "S = u.n.k + a(n.k)²/2" - which would leave the equation our OP gave us littered with k's which would be carrying these '1 second' constants and making it dimensionally correct...but it most certainly doesn't have those...which is why it's incorrect. SteveBaker (talk) 18:16, 20 May 2009 (UTC)[reply]

Well, i'm sorry, but there appears to be a lot of confusion over nothing. Lets forget for now whether the equation is dimensionally right or wrong. Lets just analyze n and 1. First question. What is this formula for ? Ans : It gives the displacement covered is the nth second of motion. That's it, I'm through. Just think about it, how can n possibly be in seconds ? The whole formula wouldn't make any sense. Second, think physically. As i had explained in my very first post, you substitute t as n seconds. Now you tell me this formula would be littered with ks. Unfortunately, while writing equations, we don't have the practice of writing units along with the numbers. You just write the numbers. So in the formula you just substitute k = 1, assuming it is in seconds. You might argue now that see it is not the usual convention. But now you're coming to analysis of the whole equation. Personally, i think this is a standard enough formula to be asked as a trick question, but its up to the teacher, there's really not much point in arguing about that. But remember, the whole physical meaning of the formula is lost if you put t = n. When the question is how many meters has the ball traveled in the 5th second, then you must substitute n as 5. There is no meaning in substituting n as 5 seconds. All the ks are left out because of conventional convenience. Another earlier confusion was whether Snth is distance or velocity. Its distance, as can be clearly seen from the physical meaning of the formulaRkr1991 (talk) 03:39, 21 May 2009 (UTC)[reply]
Let me just Give an example. Say there are a set of 5 vectors. I want to know the magnitude of the 5th vector. Is 5 a vector here ? Or is it just a number ? Think about it in relation to this question. Rkr1991 (talk) 04:59, 22 May 2009 (UTC)[reply]

Brain health

Is there any explanation for a condition that brings a headache when a person is studying or trying to focus on a taskBotosh (talk) 05:48, 18 May 2009 (UTC)[reply]

Eyestrain causes headaches, and can be caused by "tedious visual tasks". -Running On Brains 07:01, 18 May 2009 (UTC)[reply]

See Tension headache.71.236.24.129 (talk) 09:40, 22 May 2009 (UTC)[reply]

Houseplant identification please...

I've looked on the internet and on W, and the best I can do is narrow these down to Succulents. I understand that's not much help, but I'm an amateur ornithologist, not a botanist.

Just bought this pair today in China and they were completely unlabeled...

Thank you!

succulents maybe?

--61.189.63.170 (talk) 07:04, 18 May 2009 (UTC)[reply]

try Kalaonchoe (spelling?) Sorry, gotta run. 67.193.179.241 (talk) 07:51, 18 May 2009 (UTC)Rana sylvatica Spelling Kalanchoe[reply]

Aloe aristata? --Dr Dima (talk) 08:25, 18 May 2009 (UTC)[reply]

More probably a Haworthia species, pretty. To the right is an Agave, I think. --Ayacop (talk) 17:02, 18 May 2009 (UTC)[reply]

Earth Rotation Effects

I asked this question few weeks ago but couldn't get an answer, so I will rephrase my question in another way:

If a rocket is initially launched (or an object thrown at very high initial speed) normal to the Earth (Vertical) at the equator for example then what would be the path it flows with respect to its origin on Earth (recall that earth is spinning and the rocket was initially having the same spin)? I'd appreciate it if the governing equations are included.--Email4mobile (talk) 07:21, 18 May 2009 (UTC)[reply]

Sorry, doctor said no equations this late at night for me :-D

In all seriousness, a rocket which appears stationary to an observer in the rotating frame has an initial tangential velocity v=[omega]xr (I never did learn how to make wikiequations). Thus if you add thrust normal to the surface of the sphere, its velocity will be v=at^2+v_0, with the a being in the, say, x-direction and v_0 being that initial tangential velocity in the y-direction. A few considerations when considering Earth: add an atmosphere and you have drag accelerations as well, and don't forget that "normal to the surface" isn't always necessarily what you think of as "up": Depending on latitude, the sum of the gravitational force and the centripetal "force" will not point to the center of the earth, but rather a small, but significant deflection away from the center.-Running On Brains 07:46, 18 May 2009 (UTC)[reply]

Thanks -RunningOnBrains, but I'm only curious about this stage when the rocket or any trajectory is being thrown 90o normal to the Earth surface and assuming ideal conditions where no air effects or any other kind of friction. This is because my original question was how the Earth rotation or spin would affect the released bodies from its surface (i.e. how it would be different from doing the same experiment at the Arctic or Antarctica). Indeed this question arose when a friend of mine was reading an entertainments physics book and asked us the following question: "Assume you're ideally jumping up and down, continuously at the Equator, can you slip from the Earth surface due to its spin? " His answer according to that book was No, never! But to me it wasn't convincing because I analyzed it starting with the initial conditions (same spin velocity and same angular speed) and so I realized their must be a slight slip in every jump due to the increase in the new radius (Earth's radius + height). When I used simple equations and approximations I concluded a full circle slip every 58000 years if the jump is around 0.5 m height (I eventually reached this formula, :: $T_{r}=3T{\frac {R_{e}}{2h_{max}}}\,$ , where Tr is the required time for full circle slip, T is the normal 1 day time, Re is the standard Earth's radius and hmax is the max height of the jump. For this reason I wanted some one to help me justify this answer. --Email4mobile (talk) 08:25, 18 May 2009 (UTC)[reply]

OK - so you have an equation - what's the question? Do you not trust your equation? If your math is right - then the amount of 'slip' is proportional to the radius of the earth divided by the height the object reaches - which certainly fits with the idea that the effect is essentially negligable for things like people jumping and rocket launches. When you add in the atmospheric drag effects, then with a perfectly calm atmosphere, it would act such as to reduce the effect you are describing - when you add in wind speed, then the randomness of the result will completely dominate the results. 12:28, 18 May 2009 (UTC)

I'm sure the OP is aware about those caveats. He wants to confirm his calculations. I got an extra factor of 1/2.

T={\frac {3}{4}}{\frac {R}{h}}T_{Earth}

.

But I didn't double check it, so I may be the one that made a mistake. Dauto (talk) 14:55, 18 May 2009 (UTC)[reply]

Conserve angular momentum. Roughly speaking, the stationary rocket begins on the ground, "orbiting" with a circular orbit equal to the radius of the Earth. When you add an impulse to kick it to a higher orbit ("flying vertically from the surface of the earth"), you have not changed its angular momentum, but you have moved it to a new orbital radius. You will find that the rocket is now traveling in an elliptical orbit which intersects the surface of the earth (crash!). Eventually, the rocket will return to the Earth and crash, unless you give it sufficient velocity to escape orbit. If you wanted to get into a useful orbit, you would need to add some angular momentum, which can only be done with a non-vertical launch trajectory. (Also, you should account for atmospheric drag, which will have an important effect on the orbit anywhere near Earth). Nimur (talk) 16:04, 18 May 2009 (UTC)[reply]

These equations are way beyond me, but I believe the question being asked is: Will the rocket return to earth at the point of origin, or further along the equator?Drew Smith What I've done 04:51, 25 May 2009 (UTC)[reply]

name that snake

Near San Francisco Bay, my cat killed a little snake, about a foot long, brown back, light beige belly. Should we worry? —Tamfang (talk) 07:49, 18 May 2009 (UTC)[reply]

This site and this site can help you a great deal in the identification of the snakes of California. Hope this helps. I'm actually curious what species it is. --Dr Dima (talk) 08:32, 18 May 2009 (UTC)[reply]

I wouldn't worry. I'm sure the snake didn't feel a thing :) --KageTora - (영호 (影虎)) (talk) 11:00, 18 May 2009 (UTC)[reply]

Alas that I didn't photograph it. The leading candidate is rubber boa. —Tamfang (talk) 17:10, 18 May 2009 (UTC)[reply]

Are you worried about the snake (species) or the cat or your family/self? Nil Einne (talk) 12:56, 18 May 2009 (UTC)[reply]

Yes, maybe it's just a phase the cat's going through. She'll grow out of it. Just stop playing GTA IV in front of her.--KageTora - (영호 (影虎)) (talk) 13:39, 18 May 2009 (UTC)[reply]

He is four years old and I think this is the first prey he has brought home other than leaves. —Tamfang (talk) 17:10, 18 May 2009 (UTC)[reply]

Mostly about the cat. It's futile to worry about the safety of vermin in the city. —Tamfang (talk) 17:10, 18 May 2009 (UTC)[reply]

Well I would question whether a snake in what is potentially its native habitat and which from the discussion appears to be non-posoinous is unlikely to, for example, get into your house and chew your food etc and is not know as a carrier of disease can be called a vermin especially since the snake may help to kill rats and mice which may get into your house and chew your food and are known as carriers of disease. And definitely it seems to me it's worth worrying about endangered snake species (I don't know if this one is). As for the cat, well to be honest if you're asking on the RD by the time you get an answer your cat is likely to be either dead or out of danger. If your worried your cat may make a habit of catching snakes, well to be honest as with birds, there's not so much you can do. You could try putting a bell on your cat but I don't expect it will help much Nil Einne (talk) 05:54, 19 May 2009 (UTC)[reply]

vermin was the first word that came to mind for 'wild vertebrates in the city', likely because its cognate worm (whose cognates in some languages still mean 'snake') was on my mind. —Tamfang (talk) 03:32, 25 May 2009 (UTC)[reply]

There are only a few venomous snakes in California and they are all pretty unpleasant (rattlesnakes—relatively large, mean-looking, lots of warning signals, don't generally live near populated areas) —I doubt your cat would try to mess with them. The smaller snakes cannot do any damage to your cat. Psychologically, your cat is both having fun and trying to please you. There's no reasoning with it about such things. --98.217.14.211 (talk) 18:39, 18 May 2009 (UTC)[reply]

Don't rule out a house cat going after a rattler. For example check this out [1] (Otherwise I can only offer OR.) Males seem to be generally less inclined to do so, but females, particularly the unfixed ones, will tackle snakes. As with other prey there seems to be a size limit, but that can get thrown to the wind if a litter of kittens gets threatened. The good news is that by the time kitty brings one home it is very unlikely to still be alive. (Other than more harmless trophies like mice or rabbits.) Look at the bright side, you'd probably rather have your cat deal with snakes than meeting them yourself while they are alive. BTW. I've never heard of a house cat that got killed by a rattlesnake, so they seem to know what they are doing or stay away if they don't. 71.236.24.129 (talk) 23:05, 18 May 2009 (UTC)[reply]

I did lose a cat to unknown causes but a snake bite was a potential candidate. It died frothing at the mouth. This was in Malaysia however so no rattlers, it could have been a python or something. This ref [2] suggests rattlesnake bites are definitely not unheard of and often fatal but this site seems to suggest rattlesnake bites don't usually kill cats [3] and this one is in between [4] Nil Einne (talk) 05:54, 19 May 2009 (UTC)[reply]

Pythons do not have venom, they are constrictors. 65.121.141.34 (talk) 13:09, 20 May 2009 (UTC)[reply]

8D type batteries

Explain about 8D type Lead acid batteries —Preceding unsigned comment added by Kumar3214 (talk • contribs) 09:30, 18 May 2009 (UTC)[reply]

Explain what? Perhaps you would care to read our article on lead-acid batteries? -- Captain Disdain (talk) 09:47, 18 May 2009 (UTC)[reply]

8D is a standard size of lead-acid battery for marine use [5] [6] that fits within a space 11-1/2 x 21-1/4 x 12-3/4 (height including cables) inches[7]. Do not confuse 8D with 8 D-cell batteries. Cuddlyable3 (talk) 11:04, 18 May 2009 (UTC)[reply]

I don't think D-cell batteries come in lead-acid form. Hence our OP is certainly talking about the 8D marine batteries. These appear to be pretty serious bits of equipment. Prices range from $250 to $700 and those seem to assume you're returning an old battery for recycling - otherwise they stick you with another $70 charge. They look a bit like longer car batteries - but according to the various web sites that deal with them, they have all sorts of specialisations such as vibration protection. Google "8D marine battery"...there's a ton of information out there. SteveBaker (talk) 12:17, 18 May 2009 (UTC)[reply]

Name that fluffy white lint bug!

WikiProject Reference Desk Article Collaboration

This question inspired an article to be created or enhanced:

Eriosomatinae

Please consider contributing

I'm looking for an ID on this lint-like bug. I'm not the photographer by the way. Thanks. 152.16.223.48 (talk) 18:21, 18 May 2009 (UTC)[reply]

A location would be great if you have one.Popcorn II (talk) 19:37, 18 May 2009 (UTC)[reply]

The photograph was in southeastern Ohio, and I've seen them in North Carolina's Appalachians. 152.16.223.48 (talk) 21:01, 18 May 2009 (UTC)[reply]

It is an Eriosomatid, a kind of aphid. We have article on Eriosomatinae. --Dr Dima (talk) 20:37, 18 May 2009 (UTC)[reply]

An after-thought: Eriosomatidae family is better known as Pemphigidae. I'll update the Wiki pages for Eriosomatinae and Eriosomatidae if no objections land on my talk-page today. --Dr Dima (talk) 21:08, 18 May 2009 (UTC). Done. --Dr Dima (talk) 20:26, 19 May 2009 (UTC)[reply]

I'm not so sure. It doesn't look much like an aphid in my opinion. The antennae make it look more like a moth to me - no idea whatsoever what type though I'm afraid! Smartse (talk) 23:54, 18 May 2009 (UTC)[reply]

No, Smartse, it is definitely an Eriosomatid (a "woolly aphid"). The picture on the Eriosomatinae page is not so good though, that's probably what's confusing you. I wish we could use the Original Poster's picture... In the meanwhile, you can look at the pictures of Eriosoma sp. at this site. --Dr Dima (talk) 00:12, 19 May 2009 (UTC)[reply]

Here is the original image. I suppose someone could ask the user for wikipedia appropriate rights. Sifaka ^talk 01:44, 20 May 2009 (UTC)[reply]

Radios in tunnels

Why does my car radio continue to work all the way through the Rotherhithe Tunnel? It usually soon dies to static in much smaller motorway tunnels. Do they have a leaky feeder or something down there? Why do you need to know what I was listening to to answer the question? Radio 4 on FM if you must know. SpinningSpark 19:22, 18 May 2009 (UTC)[reply]

"Drivers will be able to listen to normal radio programmes including BBC and commercial radio programmes on FM, AM and LW and also on the increasingly popular DAB (Digital Audio Broadcasting) channels while travelling through the [Rotherhithe] tunnel." -- Press release. Cuddlyable3 (talk) 21:46, 18 May 2009 (UTC)[reply]

I bet there are few of the drivers using that tunnel in rush hour who realise how much technology went into that, if they even noticed anything at all. SpinningSpark 22:52, 18 May 2009 (UTC)[reply]

That is a sign of good technology - people not noticing that it exists. --Tango (talk) 00:23, 19 May 2009 (UTC)[reply]

fuel stability

In reference to storing fuel, Our article on gasoline states that it will go stale in time frames of 3-24 months. How long can you store Avgas before it will go stale? 65.121.141.34 (talk) 19:43, 18 May 2009 (UTC)[reply]

According to this page, the answer is 1–2 years, but that doesn't look like a very reliable source. Bovlb (talk) 22:02, 18 May 2009 (UTC)[reply]

The reason regular gasoline 'goes stale' is because the lighter fractions evaporate to a greater degree than the heavier stuff - but it's the lighter fractions that produce much of the engine power. Avgas has more of the lighter fractions to start with - but if you're going to use it to fly a plane, then you need that. So it doesn't really surprise me that the times are comparable. SteveBaker (talk) 01:22, 19 May 2009 (UTC)[reply]

It would depend on how big a container it is in and how well it is sealed and whether there is oxygen or just pure nitrogen filling the rest of the space. Well anyway I don't believe if it is well sealed that it will go off it is just that it is very volatile so you need a very good container or a huge amount to keep it a long time. I dfon't know what the times quoted are for but I'd guess it's for those small containers people sometimes carry petrol around in rather than tankers. Dmcq (talk) 10:10, 19 May 2009 (UTC)[reply]

Does space feel cold?

I recently got into a discussion with some friends about enduring the vacuum of space. I mentioned that this article states that, because of the lack of a surrounding environment, heat doesn't transfer out very quickly, so freezing to death isn't an "immediate risk". This makes sense, but what then does outer-space feel like, if there's little transfer of heat going on? -- MacAddct1984 ^{(talk  contribs)} 21:39, 18 May 2009 (UTC)[reply]

There is a significant pressure difference between the outside pressure your body is adjusted to and the vacuum of space. My guess would be that your body would use existing sensors for heat/cold and pain to signal that. Extreme pain is often felt as extreme chill. A 'burning" sensation is also possible, though. 71.236.24.129 (talk) 22:28, 18 May 2009 (UTC)[reply]

One of the few to have experienced this was Joseph Kittinger during Project Excelsior when his hand was exposed to low pressure. His main reported symptom was severe pain due to swelling. See also Vacuum#Effects on humans and animals. SpinningSpark 23:30, 18 May 2009 (UTC)[reply]

That's not really comparable; the rest of his body was pressurized. Essentially the pressure on the rest of his body was trying to squeeze his insides out through his hand. With full-body exposure to vacuum, obviously you have other problems, but you don't have that one. --Trovatore (talk) 23:47, 18 May 2009 (UTC)[reply]

Keep in mind, I'm asking specifically about the perceived temperature of space, ignoring, if it's even possible, the other effects, such as a the body's reaction to a vacuum. I'm thinking a true vacuum is temperatureless? -- MacAddct1984 ^{(talk  contribs)} 00:06, 19 May 2009 (UTC)[reply]

That's correct, the temperature of a true vacuum is undefined. Space, even without the few particles dotted around, is not a true vacuum though, since it is filled with radiation. It is possible to define a temperature of that radiation (for example, the cosmic microwave background radiation has a temperature of about 3K). --Tango (talk) 00:14, 19 May 2009 (UTC)[reply]

A quick google search turned up Phil Plait's article[8] which states:

Space doesn't feel like anything, because there is nothing to feel! Space is a vacuum (or near enough). It's a common question to ask how hot space is (or how cold), but space itself has no temperature. However, the Sun is hot, and radiates that heat in the form of light. You absorb that light and feel heat. Near the Earth, a person floating in space would actually not receive enough light to keep from freezing! You yourself would radiate away your heat, and that's why spacesuits have heaters in them.
-Phil Plait

--MacAddct1984 ^{(talk  contribs)} 15:53, 19 May 2009 (UTC)[reply]

That doesn't sound right. Of course you radiate heat, but you also produce heat metabolically. Black body#Radiation emitted by a human body says you radiate about 100 W (oops, see below), which is roughly what you produce (100 W × 1 day ≈ 2000 kcal). So I'd think you'd be reasonably comfortable temperaturewise. That's ignoring heat from the Sun. Insolation at 1 AU is around 1400 W/m². I don't know what the human cross-section is, but I guess you'd be getting about a kilowatt. I also don't know how much of that you'd absorb (it depends on the color of your clothing/skin), but at any rate it looks like overheating is your problem, not freezing. I guess sweating would help—it would evaporate instantly and at least carry away the heat that was in it at the time (like Tango, I'm unclear on whether it would draw any extra heat as it evaporated). I don't know if it would be enough. Note the Apollo suits were cooled, not heated, and they were white, which seems to support this analysis. (They were in fact evaporatively cooled—I'm not sure what that says about the sweating issue.) -- BenRG (talk) 18:54, 19 May 2009 (UTC)[reply]

Never mind, I just realized that the figure in the article is net radiated heat, not σT_body⁴, for which I get around 1 kW. So Phil Plait is probably right. -- BenRG (talk) 19:21, 19 May 2009 (UTC)[reply]

I thought space suits have refrigeration units in them, not heaters. Insolation from the Sun is, indeed, about 1400 W/m² and the side of the person facing the Sun is probably about 1 m², so that is 1400W from the sun. There is also about 100W from metabolism. By my calculations, completely exposed to vacuum one would radiate about 1000W. That's about 500W left over that needs to be dealt with by refrigeration. --Tango (talk) 19:37, 19 May 2009 (UTC)[reply]

The extreme vacuum would be sucking the moisture out of your skin. I'll bet you'd get frostbite as your sweat boiled away. APL (talk) 02:44, 19 May 2009 (UTC)[reply]

Would you get any evaporative cooling in a vacuum? The sweat wouldn't need any heat in order to evaporate since the boiling point in a true vacuum would be absolute zero, wouldn't it? --Tango (talk) 15:46, 19 May 2009 (UTC)[reply]

You have to supply the latent heat of vaporization in any case, so yes, there would be cooling.

Water doesn't really have a "boiling point" in a vacuum. At pressures below the triple point there is no liquid phase — it simply passes directly from solid to vapor (see sublimation (chemistry)). If this picture

is accurate, it appears that the solid phase continues up to almost 200 K even at zero pressure. I suppose every now and then you'd lose a molecule to space and it wouldn't come back, so it doesn't seem like a true equilibrium, but I gather that there isn't any tendency for this to happen quickly. --Trovatore (talk) 19:37, 19 May 2009 (UTC)[reply]

Excellent point, I completely forgot about the latent heat! Thank you for that. If the loss of pressure were gradual (a leaking spacecraft/spacesuit say), then the sweat would presumably boil before the pressure reached the triple point (if I'm reading that graph correctly), so the lack of a liquid phase isn't an issue. If the loss of pressure were more sudden, what would happen? What happens if you put a liquid into conditions in which there isn't meant to be any liquid? Presumably it would boil as well. Once the sweat that was produced prior to the loss of pressure is all gone, does anyone know what would happen? Are sweat glands capable of producing sweat in a vacuum? The pressure of the skin would allow liquids to exist inside the body, but how would pores behave? --Tango (talk) 19:51, 19 May 2009 (UTC)[reply]

I think if you put liquid water into a vacuum it would simultaneously boil and freeze. As it boils, ~~giving~~ taking up its latent heat, the temperature drops and the remainder freezes.

As for the human body, I don't really know. At some point I'm going to go challenge the information in the articles about "skin swelling to twice its normal size" or whatever it claims now. There isn't a cite for it (or wasn't last I checked), and I suspect that someone may have extrapolated from Kittenger's experience. I explained above why that's not a valid inference. --Trovatore (talk) 20:07, 19 May 2009 (UTC)[reply]

My understanding is that the elasticity of the skin is sufficient to keep internal pressure normal with only minimal swelling. Death is from suffocation (air gets "sucked" out of the lungs [assuming you don't try and hold your breath and make them explode], oxygen then defuses from blood to empty lungs very quickly and you lose conciousness within seconds). What happens to your corpse, I don't know. I would guess not a great deal at first and it eventually freezes or burns depending on whether or not you are in sunlight. --Tango (talk) 20:22, 19 May 2009 (UTC)[reply]

The pressure axis on that diagram is logarithmic. I suspect that only the gas phase exists in the limit of zero pressure, because of the irreversibility of casting a molecule out into the infinite void. Of course, at absolute zero the solid phase kinda has to prevail. --Tardis (talk) 22:11, 19 May 2009 (UTC)[reply]

Those molecules would only be cast out very slowly, though, there would be a solid for a long time. --Tango (talk) 22:30, 19 May 2009 (UTC)[reply]

I'm a latecomer to the party, I know, but remember that the temperature of human skin is much greater than 200 K, so a great deal of evaporation/sublimation would occur initially, until the skin cooled to well below what we'd consider "frozen solid". -RunningOnBrains^{(talk page)} 17:19, 20 May 2009 (UTC)[reply]

Plastics, Playdough and thermoelastisity

Several questions here.

What is the melting point of #2 plastic (used in milk jugs in usa)

What is the melting point of playdough

If I make an object with #2 plastic using a playdough mold, how much smaller would the final object be from the original thing pressed into the playdough after the plastic cools?

thankyou and have a nice day. —Preceding unsigned comment added by 173.25.242.33 (talk) 21:52, 18 May 2009 (UTC)[reply]

According to Plastic recycling, a #2 cycle code is for High-density polyethylene - and according to that article: "HDPE has little branching, giving it stronger intermolecular forces and tensile strength than lower-density polyethylene. It is also harder and more opaque and can withstand somewhat higher temperatures (120 / 248 °F for short periods, 110 °C /230 °F continuously)". According to playdoh, "Play-Doh's current manufacturer, Hasbro, reveals the compound contains water, salt, and wheat flour. While its 2004 US patent indicates its composed of water, a starch-based binder, a retrogradation inhibitor, salt, lubricant, surfactant, preservative, hardener, humectant, fragrance, and color.[7] A petroleum additive gives the compound a smooth feel, and borax prevents mold from developing.". This seems like a bad thing for your idea - the plastic won't melt until 110°C to 120°C - but the water in the playdoh may well be boiling by that point...that might not be a good thing. The RepRap Wiki says that: "HDPE Is one of the more finicky plastics...It does not stick together well, and...it shrinks a lot when it cools tends to distort readily."...unless you REALLY specifically want to do recycling, you'd be better off heading to an art supply store and getting some Polycaprolactone - which is sold under various trade names as stuff to make jewellery out of. It softens in warm water and melts at about 60°C - so you can even mold it with your fingers. SteveBaker (talk) 01:16, 19 May 2009 (UTC)[reply]

Maybe another modeling clay would be better, like the polymer based Fimo (can be heated to higher temps, and is harder).Yob Mod 07:16, 19 May 2009 (UTC)[reply]

One warning from the palycaprolactone article... "PCL has been known to become brittle, lose its tensile strength and fall apart after several months so is not suitable for permanent or critical applications." So I wouldn't make anything too important out of it. Low density polyethylene might be a better answer as it has a melting point of 95 C. 65.121.141.34 (talk) 13:06, 19 May 2009 (UTC)[reply]

Yeah - I noticed that. But I have stuff that I made of polycaprolactone (specifically the "Polymorph" brand) that's at least 16 years old that hasn't obviously degraded (I know it's that old because Polymorph is the British name for the stuff - so I must have made it while I still lived in the UK - which must be more than 16 years ago). I suspect that our article is referring to the plastics based on polycaprolactone that have starch mixed in with them specifically to make them biodegradable. The article mentions this - but I wonder whether the authors have correctly put two and two together. SteveBaker (talk) 19:58, 19 May 2009 (UTC)[reply]

Identifying some Australian animals from Sydney Wildlife World

Hi all. I'm looking for help identifying the following Australian animals that I photographed at Sydney Wildlife World in April 2007 so that I can upload them to Commons:

Some kind of skink
Some other kind of skink
Some kind of spider
Some nocturnal creature in its nest (maybe spotted-tail quoll, Spinifex Hopping Mouse, or bilby?)
This grasshopper
This large bird

Thanks for any help. :-) Dcoetzee 22:11, 18 May 2009 (UTC)[reply]

We don't have a page, but the Psudocheirus cinereus - Daintree Ringtail Possum [9] looks like the "nocturnal creature in its nest" I'm not into creepy crawlies so I'll leave those for others. 71.236.24.129 (talk) 23:58, 18 May 2009 (UTC)[reply]

o.k. one more: Your "Grashopper" may be a migratory locust [10] the wings are a bit short which would indicate it's not an adult or another species.

This large bird is an Australian Brush-turkey.
119.12.168.6 (talk) 12:36, 24 May 2009 (UTC)[reply]