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December 26

Orbit of a particle

Does the orbit of a particle (whose radius is of course greater than zero) turn into a spin when the radius of the particle's orbit is reduced to zero? 71.100.6.152 01:10, 26 December 2006 (UTC)[reply]

What are we talking about here, classical mechanics or quantum mechanics? —Keenan Pepper 01:49, 26 December 2006 (UTC)[reply]

How about an answer for both. 71.100.6.152 03:20, 26 December 2006 (UTC)[reply]

Quantum mechanical spin is an internal degree of freedom of a particle which is unrelated to motion in space. It is not related to orbital motion, and moreover the total spin of a particle cannot change (total angular momentum can be affected by external forces). Part of the strangeness of QM spin is that one cannot think of a QM particle as a body with finite radius. Classical spin is also somewhat different from spatial angular momentum. A particle which is in an orbit with some angular momentum does not necessarily spin. If every point in a solid body is at rest with respect to all other points in the body, it is not spinning. --Bmk 16:11, 26 December 2006 (UTC)[reply]

Assuming for the moment that the core or the Earth and Moon are fix relative to their surfaces they still spin. The spin of the Earth is not a core to surface relation but rather an Earth to Sun, Moon and Stars relation while its orbit is a relationship to the Sun. The Moon orbits the Earth yet does not appear to spin by observers on the Earth but in relation to the Sun and stars it does spin in sync with its orbit around the Earth so your classical physics answer puzzles me. It seems like what you are saying is that if the surface of a body moves in relation to its core then it spins and that this is not the same as its surface orbiting its core. Maybe the problem is just semantics. 71.100.6.152 21:16, 27 December 2006 (UTC)[reply]

Mmm - i see why that would be puzzling. However, there is something measurably different about a body that is spinning and a body which is not spinning; it actually is not a semantic difference, but quite quantifiable. You can get a feel for the difference if you consider a wet basketball; if you spin it, the water will be thrown off the ball - this is clearly not a matter of perspective, but something intrinsic to the system. Consider reading about inertial reference frames, and angular momentum. Hope this helps --Bmk 05:34, 28 December 2006 (UTC)[reply]

May then I conclude that anything having Centripetal force is spinning and that this is not a characteristic force of an orbit? (...i.e, that the Moon is not moving away from the Earth because of centripetal force.) 71.100.6.152 18:01, 28 December 2006 (UTC)[reply]

How to seep and steep herbal teas and remedies

I have encountered the terms “SEEP” and “STEEP” (I may have misspelled the second term) in relation to the creation of herbal teas and remedies. I am a bit confident that “seep” is the process used in my coffee maker. These terms are often used in hiking and camping books in the editable plants section. These books never seem to explain the method. I have been unable to find a source explaining the processes (both in modern terms and ancient methods). Also the terms always seem to be two very different processes.

To "steep" is to leave an object (such as a tea bag) in (usually hot) water to disperse flavor, etc. Your coffeemaker is doing an action similar to steeping. To seep is for water to permeate an object (usually used in a way that the "seep" is unfavorable, like "wow, the rain seeped through my tent"). Seep is usually a general term, and steep is a term used in making drinks. I'm guessing your recipes are telling you to steep tea bags or loose tea - this is just a term for sticking it in hot water for a few minutes. --W o o ty Woot? contribs 02:23, 26 December 2006 (UTC)[reply]

Agreed. Also, since many people are unfamiliar with the above usage of "steep", they may mistakenly substitute the word "seep", since that word, they know, does have something to do with liquids. StuRat 13:55, 27 December 2006 (UTC)[reply]

Budgie video on YouTube

Just found this: http://www.youtube.com/watch?v=Xe2Jb3U21vA

There's some logic at play here but I'll be damned if I know what it is. Any suggestions? --Kurt Shaped Box 02:06, 26 December 2006 (UTC)[reply]

Bored housepets enjoy play too? Vranak

Yeah, I know it looks like he's just messing around but the bird's glances back and forth between the string of beads and the lone bead make me think that he's actually trying to do something with them position-wise. Sometimes he'll walk halfway to one or the other, stop to think for a moment, then turn back. --Kurt Shaped Box 23:46, 26 December 2006 (UTC)[reply]

Fairly simple. Budgies, being parrots, are quite intelligent as birds go, and like to play. They like to play with novel things especially, and it is quite natural that the temptation of the larger object grows while playing with the similar, smaller object. Thus, the budgie is enticed to give the string of beads a try. But the accessible string is too short to handle comfortably with such a small bill, and it is apparent that they cannot be thrown around with so much vigor. They look as if they might be a better fun object, but actually are lamer. The budgie finds it increasingly hard to decide whether to give the string of beads a try or not, finally does, and on finding them lame walks away. It is like the decision a four-year-old has when having to choose between a large, light box and a small, heavy one. The one bead is known to be fun to the budgie, but eventually, it becomes boring enough to giuve the other stuff (which lokks enticing and not at the same time) a try. Dysmorodrepanis 04:52, 28 December 2006 (UTC)[reply]

Capacitance of batteries.- its a hard one this!

Any one got any idea of the inherent ac capacitance between the terminals in batteries and whether this capacitance bridges the internal resistance?? In other words, what determines the battery's impedance? 8-)--Light current 02:27, 26 December 2006 (UTC)[reply]

Well, looking at the battery purely as a black box, there must be capacitance both between the two terminals of the battery and also between the battery and the surrounding environment. I'd guess that the capacitance to the environment is pretty small. I'd further guess, given the construction of most batteries, that the terminal-to-terminal capacitance is pretty small also, although some NiCd and lead-acid gelled-electrolyte batteries use a rolled-up construction that probably has quite a bit of inherent capacitance.

If you're really interested and you have a capacitance meter, it'd be pretty easy to measure: just put a known-value blocking capacitor in series with the battery, short the combined battery-capacitor series circuit (so the blocking cap charges up and no inrush current damages your meter), attach the series circuit to your meter, and unshort the series circuit. Take the reading you get from the meter and, using the known value of the blocking capacitor, solve the series circuit for the unknown capacitance of the battery. If the blocking cap is of large value compared to the reading of the meter, then just take the reading of the meter directly as the capacitance value of the battery.

Atlant 15:42, 26 December 2006 (UTC)[reply]

Needless to say, I have never measured this nor heard of anyone doing so and of course normally we decouple the batteries from the circuit with a large capacitor. So one question is, assuming a baatery with low internal reistance, like a lead acid job, is it really neccessary to put large electrolytics across it for low freq decoupling. I am ignoring the inductance of the battery leads as this will only be important at the higher frequencies.

And yes of course Atlant, the method you suggest should work well at the spot frequency of the meter. I suspect, though, that the 'capacitance' actually rises with decreasing frequency (due to the chemical reaction effects) such that, at DC, you actually end up with the 'capacity' of the battery Q =it =CV. So C = it/V which is going to be quite large even for a 1 A hr cell. --Light current 18:23, 26 December 2006 (UTC)[reply]

You are correct, of course, that at low frequencies classical capacitance becomes meaningless for a battery; thanks for pointing that out. Also, apropos a topic you and I were discussing quite some time ago, folks who build real neighborhood-thumping car stereos often put large "ultracapacitors" in parallel with their car batteries (or at least, near the welding cables with which they feed their power amplifiers). They do this to provide the maximum transient power to their window-breaking amplifiers. And the manufacturers of hybrid cars are also looking at this, seeing as how ultracapacitors can be very rapidly (and efficiently) charged and discharged without the cycle-life issues that affect classical batteries. So they're investigating a two-level energy storage hierarchy with ultracaps first and batteries second.

Atlant 18:43, 26 December 2006 (UTC)[reply]

Actually, I think the capacitance of batteries is fairly huge, up in the Farads realm, since it arises from the atomically-thin insulating "Double layer" on the surface of the battery plates. The electrolyte is a conductor, and so are the battery plates, so the only insulator of significance is the surface double-layer where the charge-pumping effect originates. If we construct a battery where the two plates are made of the same metal, then we end up cancelling out any charge-pumping process, and this produces zero output voltage. Yet in that case the capacitance of the double-layer still exists, and as long as we avoid driving chemical reactions by applying voltages above ~1v to the plates, we can measure the battery's capacitance directly. It's the capacitance of two double-layers in series, and can be as high as tens of Farads. Don't forget that supercapacitors and ultracapacitors are based on just this same effect. They use carbon-powder electrodes immersed in sulfuric acid solution, and both of these materials are conductors. The dielectric in the ultracapacitor is the double layer, but in order to be useful at voltages > 1V, these capacitors use many double-layer capacitors wired in series. --Wjbeaty 05:55, 27 December 2006 (UTC)[reply]

If I get some time, I'll wander off to our precision LCR bridge. Alternatively, a capacitance of Farads ought to show up as a huge pulse of current if the battery is short circuited; that ought to be pretty easy to detect with a 'scope, current probe, and low-value series resistor (so I don't magnetize the current probe too badly; ours is rated 30A peak).

Atlant 12:28, 27 December 2006 (UTC)[reply]

Okay, I did some research. I used the test rig I recommended above (with a capacitor in series with the battery) and our HP 4284A Precision LCR bridge. My series capacitor was a "1000" μF alumin[i]um electrolytic which was actually running on the wimpy side at about 900 μF. I took readings at 20 Hz, 100Hz, 10KHz, and 10 KHz, stopping there because my rig was apparently starting to resonate and had gone noticeably inductive at 100 KHz. For batteries, I used a brand new alkaline "AA" (LR6) cell and a stone cold dead alkaline AAA cell. The resulting battery capacitances:

20Hz: AA=2991 μF AAA=946 μF
100Hz: AA=2907 μF AAA=287 μF
1KHz: AA=2332 μF AAA=161 μF
10KHz: AA=1086 μF AAA=99 μF

I think it's safe to say that ordinary alkaline cells have "hundreds to thousands of microFarads" of capacitance, but there's no evidence that this design of batteries has "Farads". And I also think it's safe to say that Light current's hypothesis is also born out: the dead AAA showed a very noticeable rise in "capacitance" as the frequency dropped, quite possibly due to electrochemical storage rather than real capacitance. It's interesting that the fresh ("fully-charged") battery didn't seem show much of this effect. Now, who wants to do the same experiment with a "rolled-construction" battery, where it wouldn't surprise me to see much higher capacitance values, including the "ultracapacitor" effects hypothesized by Wjbeaty?

Atlant 15:45, 27 December 2006 (UTC)[reply]

Atlant these results are extremely interesting! And thanks so much for taking time to do these tests (especially at work!). I just wondered if your last frequency reading should be 10 kHz.

Its intersting to note the reduced capacitance of the 'dead' cells cf the new ones. This may be due to the incresed esr of the 'dead' ones. Actually if your bridge does Q factor (or Dissipation) that would indicate the series resistance!

The results do indeed seem to confirm my gut feeling that healthy batteries do indeed have sufficient capacitance for general low frequency decoupling purposes. (which maybe is how designers of cheapo transistor radios got away with out big storage caps. More thoughts later.8-)--Light current 19:35, 27 December 2006 (UTC)[reply]

"Hz" versus "KHz" -- right you are. Fixed now.

Also, please note that the dead cell was physically smaller (a "AAA" versus a "AA" live cell), so the comparison must be taken with several grains of salt. I didn't have live and dead cells of the same size at hand.

I was reading ESR rather than Q, but nothing about those results was startling. The data's at work; I'll post it tomorrow if you like.

Atlant 23:13, 27 December 2006 (UTC)[reply]

Yeah thats really good of you to do that! OK understand you were measuring esr as well: so your bridge resolves the 2 components? I look forward to the results! --Light current 23:18, 27 December 2006 (UTC)[reply]

Yes, the bridge measures the vector impedance and will then display it to you in your choice of units (so C+Q, C+ESR, L+Q, L+R, etc.) As I write, I'm finishing the discharge of a mostly-dead AA cell; although it's not the same brand as my "new" AA, this ought to give us a better comparison between "new" and "dead" than the new AA -- dead AAA comparison I offered earlier. Later, I'll take and post the data.

Atlant 12:36, 28 December 2006 (UTC)[reply]

Riskiness Rankings

Just about every country has some sort of environment conservation legislation. This legislation often ranks species by how close they are to becoming the next Dodo or Pyrenean Ibex. So there are many systems for ranking plants and animals. Often independent bodies also do ranking (such as the World Conservation Union).

I'm trying to flesh out a list of as many of these ranking systems as possible with:

What are the categories for the ranking system?
Who's the group or organisation who assigns the rankings?
What country or countries are involved?
What's the Act that makes it official? (if any)
Does the rank apply to the global population or only the one in that country?

Just about every country has a system, so please check if your country, or your favourite country, to the list on the Conservation status page. If you don't know the answer, try googling. There are a lot of Acts and systems and it's a monumental task putting them all together. Any help appreciated. —Pengo ^{talk · contribs} 05:08, 26 December 2006 (UTC)[reply]

Some more specific questions:

Do threatened species of Australia protected under the [[Environment Protection and Biodiversity Conservation Act 1999

EPBC Act]] have to be threatened globally, or just in Australia? (the legislation looks vague: see part 7, but perhaps there are specific cases or something?

Same question for the US's EPA. Do species need to be threatened globally or locally to qualify? (Canada only considers how threatened a species is within its borders)
What categories or protection is there in ~~New Zealand~~, Argentina, Belgium, South Korea, Cyprus, Seychelles, Mauritius, Libya, Barbados, Tonga, Chile, Uruguay, Puerto Rico, (your country here)...? —Pengo ^{talk · contribs} 05:42, 26 December 2006 (UTC)[reply]

- I am pretty sure that the EPA is governed in this respect primarily by the Endangered Species Act. In this act, species are considered threatened if they are close to global extinction. Species can be either "threatened" or "endangered", the latter defined as closer to extinction. Though the law's mandate is primarily for the USA there are also some stipulations in it about international cooperation which I have not parsed through. --24.147.86.187 14:39, 26 December 2006 (UTC)[reply]

electroscope

A zinc plate is connected to a electroscope. if the electroscope is negatively charged, when ultraviolet radiations falls on the zinc plate, the leaves of electroscope fall down. if the electroscope is positively charged, nothing happens to the leaves. why? —The preceding unsigned comment was added by Skanthckd (talk • contribs) 06:52, December 26, 2006 (UTC).

What does it mean for something to be negatively charged? Read up on electric charge. And what does ultraviolet radiation (no plural please; "radiation" is a mass noun) do? Read up on the photoelectric effect. Combine the information you find in these two articles. By the way, it was Einstein's explanation of the photoelectric effect won him the Nobel Prize in physics. --Lambiam ^Talk 09:59, 26 December 2006 (UTC)[reply]

Well, the electroscope is obviously losing negative charge somehow. How do you think this could happen ? How does illumination with ultraviolet light cause this ? How quickly is charge lost ? What happens if you change the brightness of the UV light ? Would light of other wavelengths have the same effect ? The best way to find an answer to your question is to do some experiments with a real electroscope. If this is not possible, you can always read our article on photoelectric effect instead. Gandalf61 10:02, 26 December 2006 (UTC)[reply]

Hey, what about electroscope for additional info ?--Light current 21:32, 26 December 2006 (UTC)[reply]

Eyes

Are the eyes of a near-sighted person smaller than the eyes of a regular person? And does it look weirder because when i take my glasses off and i took a picture off me i look sooooooooo weird.

Farsightedness (hypermetropia) can be caused (one of the reasons) by the eyeball being too small but regarding near sightedness (myopia), nothing has been mentioned. But I suppose that myopia can be caused (one of the reasons) by the eyeball being too long -- WikiCheng | Talk 08:56, 26 December 2006 (UTC)[reply]

Near-sighted people may have a tendency to squint when trying to focus on something in the distance without corrective glasses. This may look strange if you're not used to seeing it, but you would almost certainly be aware of doing this. The glasses make your eyes appear slightly smaller, so taking them off should actually have the converse effect to people (possibly including you) used to seeing your face with glasses on. But aren't you used to seeing yourself without glasses in the mirror? --Lambiam ^Talk 09:46, 26 December 2006 (UTC)[reply]

I think I saw some stuff on Howstuffworks. Here's some. [[1]] - Nearsightedness. And [[2]] about corrective lenses. And a comparison of refractive vision problems in general - [[3]]. Hope that helps!! FruitMart07 03:04, 27 December 2006 (UTC)[reply]

Friction

Please help me solve this problem-

A small block of mass 'm' is kept at the left end of a larger block of mass 'M' and length L .The system can slide on a horizontal road.The system is started towards right with an initial velocity 'v'.The friction coefficient between the road and the bigger block is 'u' and that between the blocks is u/2.Find the time elapsed before the smaller block separates from the bigger block.

Thank you. —The preceding unsigned comment was added by 59.93.67.55 (talk • contribs) 10:03, December 26, 2006 (UTC).

Is this literally how the question is phrased? Letting o indicate the smaller block and X the larger block, I can imagine (at least) two possible initial situations:

  XXXXXXXXXX
  XXXXXXXXXX   ------>
_oXXXXXXXXXX__________

Situation 1: smaller block strictly to the left of the larger block

  o
  XXXXXXXXXX
  XXXXXXXXXX   ------>
__XXXXXXXXXX__________

Situation 2: smaller block at the left on top of the larger block

In situation 1, it is not clear how the friction between the blocks plays any role. The coefficient of friction for the smaller block is not given, so it would be impossible to tell what will happen. If this is a homework assignment, then I assume the intention is that there is a definite solution, ruling out this interpretation of the question. (Also, homework problem statements tend not to provide irrelevant data, like L and the u/2 would be).

In situation 2, the smaller block will start decelerating due to the force of friction with the (also decelerating) larger block. The relevant mass for determining this force is m. The large block will start decelerating due to the force of friction with the road. Note that the relevant mass for determining this force is M+m as long as the smaller block is on top. However, that decelerating force will partly be countered by the smaller block: in conformance with Newton's laws (action = reaction), the same force of friction that pulls on the smaller block towards the left, pulls on the larger block towards the right. So the net force acting on the larger block is the difference between these two forces. Now you can give the equation for the right end of the larger block as a function of time, and likewise for the position of the smaller block, which I guess should be idealized as a point particle with zero width; when the latter exceeds the former, the smaller block will fly off (or fall off) at the right end.

However, there is a complication. If the velocity of the larger block reaches zero before the smaller block flies off, the larger block will come to a halt, and not start moving in the opposite direction (to the left). Then the smaller block will continue, but the equation for determining the fly-off moment has changed. And, finally, if L is large enough the smaller block may then too come to a halt before it reaches the right end and flies off, and in that case the two will remain together forever.

I hope I got this right, what with all the interpreting and the complications, and I hope this helps. If you get stuck in solving this, tell us where you got to and how you are stuck, and we'll see if we can help you go on. --Lambiam ^Talk 11:21, 26 December 2006 (UTC)[reply]

Physics

hey are there any shortcuts to be knowing all the stuff your physics teacher knows? any good sites or books? Thanks (sorry but im an ambitious man!) —The preceding unsigned comment was added by 219.78.44.181 (talk • contribs).

Brain transplantation/Mind transference? (But watch out for Abbie Normal's brain.) Visit Rekall or the teacher on Star Trek?

Seriously, I don't know how much your physics teacher knows, but if you want to understand physics, then you will have to be willing to invest work in your studies, either now or as a result of accumulated life experience. It will also help you greatly if you already understand basic trigonometry and have studied or are simultaneously studying calculus. It also helps if your mind is oriented towards logic, so after being told about principles "A" and "B", you can generalize and extend that knowledge to principles "C", "D", and "E".

I really don't think there are any "shortcuts".

Atlant 17:43, 26 December 2006 (UTC)[reply]

Hi, User 219... I don't have a concrete answer for you, but perhaps checking THIS google result might get you started. Are you looking for info on High School or University level physics? That would help with the answers, to be sure. Good luck! Anchoress 17:27, 26 December 2006 (UTC)[reply]

I dont think there are any shortcuts. Your physics teacher has worked hard to go thro school, taken one or two degrees and put in many hours hard thinking over the years working out the best way to teach physics. The latter part we call experience. There is no real substitue for it. Although you may be able to memorise a load of facts, that does not mean you understand things and without understanding, I think prospects are limited. However, if you purely want to bluff your way through, there are many jokey books, like bluff your wa thro computers etc. Another alternative is to buy a dictionary of physics and learn the words and their meanings. 8-)--Light current 17:57, 26 December 2006 (UTC)[reply]

You mave have heard the phrase: You cant put an old head on young shoulders

Thats what it means.--Light current 18:00, 26 December 2006 (UTC)[reply]

Try The Physics Classroom, Free High School Science Texts, Textbook Revolution, and Free-Ed.net. The last one has a couple of semesters' worth of university physics lectures on video. BenC7 00:47, 27 December 2006 (UTC)[reply]

There certainly are big shortcuts! I've relied on three of them throughout my entire career. The first one is: search for errors in your textbooks, sensitize yourself to contradictions between separate textbooks, then learn to "un-teach yourself" the information you've found to be wrong. This is incredibly effective because introductory textbooks are full of misconceptions that their authors believe and never question. These misconceptions become barriers which prevent your further learning: like trying to build a brick building on a pile of rocks. So, if you manage to remove the mistakes from your head, you'll quickly gain unusual expertise. You'll learn faster than everyone else because you've defeated barriers which prevent speedly learning in all others who still harbor those misconceptions and can only learn very slowly. Defeating one major misconception is worth weeks of normal study (weeks, if not years!) Yet it doesn't exactly speed up your learning. Instead it lets you learn at the pace which SHOULD be normal, but is actually far faster than the slow crawl of everyone else who has been taught those misconceptions. (Here are a few I've found: http://amasci.com/miscon/)

Here's another: don't ever try to thoughtlessly "record" or memorize information from books or relayed by teachers. Instead, always assume that books are badly written and full of errors. Try instead to "own" the incoming information by taking it apart and finding alternate ways to describe it to yourself. Be a skeptical textbook editor, not a gullible textbook reader. This does make for very slow going in school, and makes you a "backwards student," but later it allows you to go far beyond all other students. You'll REALLY UNDERSTAND physics, as opposed to just having a head full of disconnected and possibly erroneous facts. And if you're studying on your own, the process becomes easy because there's no classroom pressure to race through the material in lock step with others. I finally heard about another person who learned physics in just this way: Richard Feynman, who said "what I cannot create, I do not understand." That's it! That's exactly it: you cannot memorize someone else's physics explanation, instead you have to disassemble it and then use the parts to construct your own version. This is an ENORMOUS shortcut, since it puts you far ahead of all students who are just trying to memorize facts without understanding, and who will forget all of it after the next exam. Here's another way to say it: it turns you into a physics teacher. You'll never forget any of it no matter how many decades go by. And as you study more physics, it all starts connecting together into a vast mental machine which starts functioning on its own. Those who don't do this, they end up with a big pile of parts inside their heads, but no "machine."

Here's the third: stop being an egotist. Get into a "zen" state where you don't take your mistakes personally, and where you aren't threatened by information which demonstrates that your current understanding is embarassingly faulty. I stumbled into this particular shortcut by my early religious upbringing which had some eastern concepts. Over decades I saw that many other students were greatly slowed down in learning physics because they had something I didn't: a huge need to protect their egos, protecting their self image of having perfect error-free knowledge. For them, admitting their mistakes and going back to revise their understanding was a huge deal. They'd go into denial and fight fiercely against admitting their errors, fight even more fiercely against ever letting them be discovered by others, and perhaps remain trapped in obvious misconceptions which they couldn't stand to find or fix. Now this would be fine if teachers and textbooks were totally accurate, because in that case our tendency towards errors would be much reduced. If someone needs to be right all the time, there'd be no problem if they actually WERE right all the time. But since teachers and textbooks are extremely imperfect, and misconceptions are the norm, emotional intolerance of personal errors becomes a huge learning barrier. Learning physics involves lots of trial and error, so in a very real sense you have to cultivate a taste for error: to seek out your embarassing personal errors. Do just that, and as with the other shortcuts, you race far ahead of everyone who isn't doing this stuff. --Wjbeaty 05:19, 27 December 2006 (UTC)[reply]

I have a friend who manages to get by without knowing or using a single physics formula, by instead substituting purer maths, using imaginary numbers and integration, for example, when working with projectile motion. Seems to work for him --124.243.155.3 09:02, 27 December 2006 (UTC)[reply]

Rather than thinking in terms of 'shortcuts', go to a library and skim a few physics book and see if you can find one that you personally will find the easiest to learn from. Everyone has a different style. Peter Grey 04:32, 29 December 2006 (UTC)[reply]

North Atlantic oscillation

Can anyone point me to theories explaining the causes of the North Atlantic oscillation? Thank you! Marco polo 16:20, 26 December 2006 (UTC)[reply]

Brian Fagan, in his book "The Little Ice Age", has a lot of information about it but no clearly defined theories as to ultimate cause, as far as I can see. Geologyguy 16:30, 26 December 2006 (UTC)[reply]

You can find some information on http://www.met.rdg.ac.uk/cag/NAO/Models.html. --Lambiam ^Talk 00:06, 27 December 2006 (UTC)[reply]

Squirrels

We were wondering why a squirrel's tail is so bushy. No doubt there is more than one advantage (and disadvantages), but has science reached a concensus on the primary reason? --Bob K 18:08, 26 December 2006 (UTC)[reply]

Added insulation in cold weather (when squirrels will fold their tails up along their backs, essentially adding another layer of fur). Distortion of their body image as seen by predators (so the predator gets confused and grabs at the tail and not the squirrel's body; you often see "short-tailed" squirrels where this strategy obviously paid off at least once). Perhaps a little bit of aerodynamic resistance (counterbalancing) as the squirrels bound along.

Atlant 18:32, 26 December 2006 (UTC)[reply]

I thought that the tail also brought some reproductive advantage. For example, I thought that I had read that squirrels prefer to mate with other squirrels with bushy and attractive tails, which are a sign of health and good genes. Marco polo 18:41, 26 December 2006 (UTC)[reply]

Absolutely for balance. Ever seen a squirrel hop from one tree to another? It's quite a feat, and no doubt a tail-less squirrel would be much more prone to plunging to the forest floor — which I have also seen. Vranak

That explains all of the mass and the flexibility, but none of the bushiness. (Squirrel: It's none of your bushiness.) --Lambiam ^Talk 21:19, 26 December 2006 (UTC)[reply]

But squirrel tails are not massive! You can see this if you've ever seen a backlighted squirrel: their tails are just like rat tails except for more fur. If they're used as counterbalancing, at least some of the effect must be the aerodynamic resistance of all of that fur in the tail.

Atlant 12:42, 27 December 2006 (UTC)[reply]

I've noticed that squirrels like to shake their bushy tails at you when they're feeling defensive. The bushier the tail, the more of an impression it makes on interlopers. Vranak

If they didn't have big bushy tails, they 'd be called rats! 8-)

Speculating: they might be like the detachable tails of lizards: when a big preditor makes a grab for the squirrel, all it gets is a mouthful of bushy fur. --Wjbeaty 04:38, 27 December 2006 (UTC)[reply]

Cation and anions

How do you determine the Cation an Anions of a compund? Dragonfire 734 18:02, 26 December 2006 (UTC)[reply]

If it's a positive ion (missing electrons), it's a cation. If it has extra electrons, it's an anion. --W o o ty Woot? contribs 18:19, 26 December 2006 (UTC)[reply]

(Perhaps beating the point to death...) The cations (those missing the electrons) will go towards the cathode, the negatively-charged electrode where they can re-acquire that missing electron. The anions (those with an extra electron), conversely, go towards the anode, the positively-charged electrode where they can give up that extra electron.

Atlant 18:35, 26 December 2006 (UTC)[reply]

Hmm maybe worth an extra few strokes!

The term CATion derives from the fact that these ions are attracted to the CAThode (negative electrode) in an electrolytic cell. Cations are therefore positvely charged.

The term ANion derives from the fact that these ions are attracted to the ANode (positve electrode in the cell). Anions are therefore negatively charged.

--Light current 21:20, 26 December 2006 (UTC).[reply]

BUt how do you determine it for example the compound CO2? Dragonfire 734 22:18, 26 December 2006 (UTC)[reply]

What happens when CO2 dissolves in water? You get an acid, right? Which acid? What are the properties of acids in general, and this acid in particular, when it comes to forming ions? --Lambiam ^Talk 00:02, 27 December 2006 (UTC)[reply]

Well I suggest you look up carbonic acid. THen if you have any further Qs come back here.

heres the dope from that page anyway:

Carbonic acid has two acidic hydrogens and so two dissociation constants:

H2CO3 ⇌ HCO3− + H+

Ka1 = 2.5×10−4 mol/L; pKa1 = 3.60 at 25 °C.

HCO3− ⇌ CO32− + H+

Ka2 = 5.61×10−11 mol/L; pKa2 = 10.25 at 25 °C.

--Light current 01:25, 27 December 2006 (UTC)[reply]

If you are looking for a more simple answer, cations are usually the first part of an ionic compound, or generally the metal component of an ionic compound. For example, in NaCl, Na is the cation, and it is also written first and is the metal component of this ionic compound.

Two things the introductory student needs to watch out for: whenever you see NH₄, it is almost always going to be a cation; and when you see leading hydrogens in acids, they will always be a cation. The latter statement will need to be explained more.

HCCH is not an acid. HCl is. Both have their first atom as hydrogen, and it is difficult to tell which is which. This is an aspect of chemistry that comes with experience. What you need to ask yourself is if there is any evidence that the compound in question will break into cations or anions. If it does, it is most likely an acid, with H+ as the cation, and everything else in the compound as the anion.--Acewolf359 16:18, 29 December 2006 (UTC)[reply]

Classification of an amoeba

I need to know the classification of an ameba thanks Chris H.

Hi, Chris. Try reading the amoeba article. If you have any further questions, check back here. Anchoress 18:52, 26 December 2006 (UTC)[reply]

Bacteria in the Intestines

The large intestine hosts several kinds of bacteria that deal with molecules the human body is not able to breakdown itself. This is an example of symbiosis.

We can be fairly certain that if we were able to hypothetically remove, all at once, every bit of the bacteria from the intestines of the average human being, the bacteria itself would almost certainly perish without its host. But what about the human? Would they be likely to perish as well, in the complete absence of this intestinal bacteria?

Have the two systems developed complete codependency? Are many of the normal flora present in our bodies likely to belong to similar codependent relationships? And I believe that pregnant mothers pass down their bacteria to the fetus. Is this correct?

Thanks ahead of time...71.246.38.187 21:34, 26 December 2006 (UT

I imagine the substances (such as cellulose) would simply be excreted with feces. This might prove a problem with excessively large or pointy undigestible material, but by the time it gets to the large intestine it's probably done its damage anyway. --W o o ty Woot? contribs 03:16, 27 December 2006 (UTC)[reply]

Gut bacteria are, in fact, quite important for our health. Even besides the assistance in various digestive tasks, including absorbtion of broken down materials, they also produce much of our vitamin K₂, without which we are unable to perform such vital tasks as clotting. We would live without them, very likely, but it wouldn't be an especially pleasant life. Leave your bacteria alone, they are your little, little friends. – Clockwork Soul 03:53, 27 December 2006 (UTC)[reply]

I beleive dysentery has the undesirable effect of killing most of the bacteria in the gut.(although our page doesnt say so)--Light current 06:10, 27 December 2006 (UTC)[reply]

Oral antibiotics commonly do have that bad effect as well.

Atlant 12:44, 27 December 2006 (UTC)[reply]

The immediate problem with the use of broad spectrum antibiotics is that they don't kill all the bacteria, leading to overgrowth by species usually kept "in line" by the others, with the possible result of C. difficile enterocolitis, toxic megacolon, and death. - Nunh-huh 13:11, 27 December 2006 (UTC)[reply]

Paternal Age

Paternal Age ==

Might Wikipedia be interested in my research on the connection between advancing paternal age and sporadic cases of these conditions: Ihave found research papers on the following conditions all finding increased incidence with advancing paternal age

Multiple Sclerosis, Diabetes Type 1, Athoid/dystonic cerebral palsy, hemiplegia, Acute Lymphositic Leukemia, pre-menopausal breast cancer, some Alzheimer's, shorter life for women (fathers 45+), prostate cancer, epilepsy, some early childhood cancers, schizophrenia, autism, Multiple edocrine neoplasia type 2B, Hemophila A X linked maternal grandfather, progeria, marfans, achrondroplasia, aperts, some heart defects, other very rare disorders, Down Syndrome if the mother was 35 or older and the father was 40 or older [4] if this information can be used by someone for an entry or for inclusion with entries already in wikipedia I would be happy to give the citationsAnniepema 21:53, 26 December 2006 (UTC)[reply]

Wikipedia has Maternal age effect.....I suspect we are rather weak on this topic. --JWSchmidt 00:59, 25 December 2006 (UTC)[reply]

There has been a recent spate of inclusions/reversions on this topic on the menstrual cycle page, perhaps you should check there to see if your contributions would help. Anchoress 04:41, 25 December 2006 (UTC)[reply]

Anchoress maybe you didn't read that I was asking about interest in paternal age. The paternal age effect has to do with ,for instance, the ground-breaking work of NP Singh et.al. from the Unversity of Washington in Seattle in the research paper "Effect of age on DNA double-strand breaks and apoptosis in human sperm" In this research it was found that the percentage of sperm with highly damaged DNA and DNA break numbers was statistically significantly higher in men aged 36-57 than in those aged 20-35 years, but percentage apoptosis was statistically significantly lower in the older group. This finding was the first to suggest that there was an age related decrease in human sperm apoptosis. This new finding may indicate deterioration of the healthy sperm cell selection process with age. Also the work of James F. Crow on the high rate of spontaneous mutation in the sperm which increases with age is relevant and Andrew Wyrobeck's research on sperm DNA damage with age.. Anchoress I think you read the last part about the maternal age effect and not my question about if there was any interest research on paternal age Anniepema 21:51, 26 December 2006 (UTC)[reply]

No, I read exactly what you were asking about, and as I said, there was just an inclusion on the menstrual cycle page about that exact topic; interestingly enough - since you have expressed an interest in providing references - asking for a reference. I'm curious to know why you thought I had mis-read your question? Anchoress 08:24, 25 December 2006 (UTC)[reply]

I wouldn't have expected menstruation to be dependent on paternal age, myself, but perhaps I'm wrong. StuRat 11:13, 25 December 2006 (UTC)[reply]

As for an article on paternal age, sure, that would be great. However, since Wikipedia has a "no original research" rule, you need to only use sources in published works. If your own research is published, that's fine to use, but you can't include info from any unpublished studies you've done. StuRat 11:13, 25 December 2006 (UTC)[reply]

Sorry Anchoress

I didn't understand that the statement about male sperm be freshly made was what you meant by the new inclusion. The opposite argument is made about the many genetic diseases associated with older paternal age at the time of birth because of all the cell divisions to make new sperm.

Excerpts from geneticist James F. Crow's "The High Rate of Spontaneous Mutation:Is it a health risk" PNAS August 1997

"Paternal Age Effect"

"How can we account for a higher mutation rate in males than in females? The most obvious explanation lies in the much greater number of cell divisions in the male germ line than in the female germ line. In the female the germ cell divisions stop by the time of birth and meiosis is completed only when an egg matures. In the male, cell divisions are continuous and many divisions have occurred before a sperm is produced. If mutation is associated with cell division, as if mutations were replication errors, we should expect a much higher mutation rate in males than in females. "At age 20 the number of(sperm) cell divisions is about 200, at age 30 it is 430, and at age 45, 770."

This makes the strong prediction that the mutation rate should increase with the age of the father, since the older the man, the more cell divisions have occurred. On the other hand, there should be no age effect in females."

Let me interject at this point that there is a well-known maternal age effect for traits that are caused by errors in chromosome transmission. The kind of accident that leads to a child with an extra chromosome is strongly associated with the mother's age (15). There may be a slight paternal age effect, but the far more striking effect is maternal. My concern, however, is with gene mutations which, when those with small effects are considered, are much more frequent." .... Anniepema

Can anyone write any kind of acceptable entry in wikipedia on the paternal age effect that is well written and neutral?Anniepema 06:34, 27 December 2006 (UTC)!Anniepema[reply]