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May 7

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degree of benefit

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Where might I obtain the data to determine if the installation of Internet capable water meters might reduce fuel consumption by various utility suppliers? 71.100.14.205 (talk) 01:08, 7 May 2008 (UTC)[reply]

You could try asking the suppliers. Otherwise, try googling for some study into them. --Tango (talk) 12:41, 7 May 2008 (UTC)[reply]

Infinity of space

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Today my Physics teacher was discussing how the universe can go on forever or it can be like a snowglobe, and no one can know what's outside the globe. I'm extremely intrigued by the unknown factors in this and was wondering if someone could point me in the right direction to go read more into this (typing in "past what we think is the universe" didn't get me anywhere haha). Thank you! Evaunit♥666♥ 02:09, 7 May 2008 (UTC)[reply]

We've got an article on Shape of the universe, but it gets a bit technical. There's also the issue of the observable universe versus the "whole" universe. The problem about talking about things "outside the universe" or "before the beginning of the universe" is that all of our physics is based on observations inside the universe, so it's hard to say that they apply to things outside that. Confusing Manifestation(Say hi!) 03:22, 7 May 2008 (UTC)[reply]
I think your teacher is talking about the Ultimate fate of the universe and whether the universe will continue to expand such that the energy density approaches zero (infinite universe i.e. Heat death of the universe), or will it contract such that the energy density approaches infinity (reaches a maximum "snow globe" i.e. Big Crunch). Information cannot travel faster than light so the snowglobe is the size of light traveling since the big bang and there should always be a boundary, even in the collapsing model. --DHeyward (talk) 06:04, 7 May 2008 (UTC)[reply]
Data in the past decade seem to be pointing to a hyperbolic, infinite universe that does not contain enough mass to stop expansion. Imagine Reason (talk) 20:00, 7 May 2008 (UTC)[reply]
One weird idea that alot of people seem to have trouble with is that although the universe may have a contained shape, that doesn't mean that anything lies outside the shape. When physicicsts talk about "the universe," they are attempting to describe everything that exists. If the place where everything exists is in some dimension spherical (which it's probably not, by the way), then nothing lies outside of that sphere because the sphere just is everything that exists. Contemporary scientific theory has raised the possibilty of multiple universes, so maybe this should be qualified as "everything that exists for us" or as Confusing Manifestation pointed out "the observable universe." --Shaggorama (talk) 20:09, 8 May 2008 (UTC)[reply]

chemical, mineral, vitamin make-up of menstrual blood?

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I have read that menstrual blood contains all the contents of regular blood, such as plasma, but doesn't it also contain vitamins and minerals and growth hormones? Also I just read of an English firm that is recruiting women to submit their menstrual blood because it contains stem cells. A landscaper friend of mine regularly uses birth control pills to make all his bedding plants look spiffy, which implies that mammalian hormones stimulate plants- and perhaps plant hormones stimulate people?- (Spring fever and all that) Really, I am just looking for a layman's list of the contents of menstrual blood.People777 (talk) 05:34, 7 May 2008 (UTC)[reply]

Menstrual "blood" is a misnomer: what is shed during menstruation is the lining of the endometrium. "Menstrual fluid" is closer to correct. There's blood mixed in there, but it's not the main constituent. I don't know where you'd find an analysis of it, but I'm sure it contains as many vitamins or minerals as you'd find in any body tissue. It certainly contains endometrial cells, cervical mucus and vaginal secretions. As to the birth control pills: plant hormones and human ones are similar; plant sterols and estrogen are synthesized by similar pathways. - Nunh-huh 07:14, 7 May 2008 (UTC) (actually, there are papers, but I don't know that there's online access: e.g. Am J Obstet Gynecol. 1965 May 15;92:183-8 14281826 (P,S,E,B) THE CONTENTS OF MENSTRUAL FLUID: AN ANALYSIS OF 260 SAMPLES FROM HUMAN FEMALES. M S BURNHILL, C H BIRNBERG[reply]
Have you looked at our Menstruation article? Blood as the body's transport medium contains many substances including vitamins. Phytoestrogens are plant substances that mimic estrogen. 71.236.23.111 (talk) 07:37, 7 May 2008 (UTC)[reply]

A new Vector Energy Interpretation of Quantum Mechanics - what to do?

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This derives from a very thorough paper (external) that examines relativity if time were treated as a vector. To the best of my ability, the derivations look valid. Combined with the transactional interpretation one can derive an "offshoot" interpretation that treats a negative energy vector as a photon traveling backwards in time. The interpretation's main premise is that these photons are absorbed by empty space and since energy is added, must be conserved as in pair production (other things conserved too); or the tunnel effect being a particle in an energy well converting all mass/energy into a negative vector energy photon emission and disappears. Outside the energy barrier empty space absorbs a negative vector photon and the particle appears there.


The problem is that this can be used in several ways, all IMHO have merit.

One of these appears to be absurd at first; it is much deeper, though not many I feel will see the deeper significance. So I don't know what to do?

any help appreciated

reference user:fx303 page for the 'absurd' view. —Preceding unsigned comment added by 67.174.224.78 (talk) 06:07, 7 May 2008 (UTC)[reply]

Hello. Can you supply us with anything at all to read on the subject, or at least point to where/under what name this paper has been publicised? Scaller (talk) 14:42, 7 May 2008 (UTC)[reply]
I think the OP was referring to User:Fx303#Vector_Energy_Interpretation_of_Quantum_Mechanics. -- JSBillings 01:01, 8 May 2008 (UTC)[reply]

Amount of zinc gluconate in Zicam

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I understand the formula is propietary, and I have been looking for the patent, but I was wondering how much zinc is actually in it in mg. This is due to problems of zinc destroying nasal tissue, although I'm not sure of the mechanism.

Thanks, JD139.225.81.128 (talk) 14:57, 7 May 2008 (UTC)[reply]

Could time be expanding?

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Hi. Have physicists considered that time could be expanding like space is? Or does the standing definition of 'time' and the physics around it somehow make this impossible? What if time were expanding (somehow), but we just haven't noticed it (either in real life or theoretically)? I wish I could give more 'punch' to this possibility, but I'm no physics expert... Thanks in advance, Kreachure (talk) 16:44, 7 May 2008 (UTC)[reply]

I'm not an expert either but i'm pretty sure time is part of the whole space idea. If space expands, time should expand proportionally and it'd make no difference? Probably wrong, but hey, worth a shot! Regards, CycloneNimrodTalk? 16:59, 7 May 2008 (UTC)[reply]
We mostly define distance in terms of the time it takes light to travel over that distance. I suspect you could describe the expansion of the universe as mathematically equivalent to a static universe with a changing time dimension. As we live in the universe, if the two descriptions are equivalent (as I suspect) then it is probably easier for us mortals to assume the nature of time is not changing. Dragons flight (talk) 17:09, 7 May 2008 (UTC)[reply]
The big bang cosmological model doesn't contain anything that I'd describe as "expanding time". The concept doesn't make much sense because expansion is something that happens over time. When you say that something is expanding you mean it's smaller at an earlier time and larger at a later time. You could probably come up with some sense in which "time is expanding"—for example, the time you've been alive is "expanding" at a rate of one second per second—but probably nothing very interesting or meaningful. -- BenRG (talk) 17:28, 7 May 2008 (UTC)[reply]
Well, let's pull an Einstein and think about what time is: a measurement of how long it takes for some regular interval to happen. Let's say we had a light clock with a 1 second interval—that is, a photon bouncing between two mirrors and it takes it one second to get from one mirror to the other. Now let's say the space between the mirrors expands with space so it takes longer than 1 second for the photon to go over the space: do we redefine the second, or do we say that that particular clock no longer measures what it used to measure? I would think the latter. But I think the rub is that if all space was expanding evenly, you'd have no way of knowing, though: how would you know your clock no longer worked the same way it used to, without another clock to compare it to? If the same expansion of space affected that clock too, in exactly the same way, then in effect there'd be no way to determine the difference. If, however, the expansion was detectable in one way or another (which our current expansion is, because it affects different scales more than others), then it would be easy to say that the clock had just gotten inaccurate. Someone correct me if I'm reasoning totally wrong here. --98.217.8.46 (talk) 18:23, 7 May 2008 (UTC)[reply]
First you have to define time. In an attempt to be overly general (so don't respond with quotes out of your physics book to clarify the specifics), there are two views on time. One view is that time is a dimension (like length, width, and height) that contains events. Objects travel through the time dimension and measure it by the time distance between events. Another view is that time is not a dimension, nothing travels through it, and it does not contain events. It is merely a value placed by humans to make formulas work properly. If you use the first definition, it is possible to claim that the time dimension is expanding since it has the same properties as the other three dimensions. If you use the second definition, time cannot expand because it doesn't exist. -- kainaw 18:45, 7 May 2008 (UTC)[reply]
The definition of the term "expanding" breaks the symmetry between time and space. Even if time can somehow expand, it can't expand in the same way that space does. It's like asking whether space expands with distance. There's a weird circularity in the idea. You can think of time as an implicit dimension that parametrizes change, or you can think of it as an explicit dimension like space, but you can't have both at once. In the spacetime picture there's no implicit change. Objects don't move along their worldlines, they simply are their worldlines. Changing the past isn't a physical impossibility, it's a logical contradiction. (Actually that's true of either notion of time.) If the original question was whether the time dimension can expand over the course of some other (implicit) time interval, then the answer is no, that doesn't make sense. If the question was whether it can expand over the course of the same time that it itself measures, then I think the answer is still that it doesn't make sense. -- BenRG (talk) 20:12, 7 May 2008 (UTC)[reply]
To expand on my comment above. You generally describe the structure of the universe as a whole through the FLRW metric of spacetime:
Where the scale factor a(t) describe the expansion of space as a function of time. However, I am fairly confident that one could construct an equivalent metric:
where b(t) is a new scale factor, and yet I suspect that if you apply the Einstein equations and follow the construction through it could be designed to give rise to the same dynamical structure. The second metric would be described as having static spatial dimensions and a dynamic time dimension, but from the point of view of any quantities observable by people living in the universe they would give exactly the same behavior. Dragons flight (talk) 20:32, 7 May 2008 (UTC)[reply]
P.S. I also think it is possible to construct a system with both dynamic time and dynamic space, but I think there no physically useful reason for doing so as one can always condense all of the variation into one or the other. Dragons flight (talk) 20:39, 7 May 2008 (UTC)[reply]
More importantly, if time was expanding in this way (which I think it probably is), do you think we would be capable of noticing it? in relative terms, the amount of time we experience probably would stay the same, for the same reason we don't notice out bodies expanding as the universe does. relativity is a pretty elegant theory. --Shaggorama (talk) 10:11, 8 May 2008 (UTC)[reply]
Our bodies don't expand as the universe does. The expansion is defined in terms of real physical distances that we can measure; our measuring devices tell us that the universe is expanding at large scales but human bodies aren't. The trouble with the b(t) metric above is that it doesn't describe a measurable expansion the way the FLRW metric does. -- BenRG (talk) 18:05, 8 May 2008 (UTC)[reply]
No, this metric won't work. Make the substitution and it becomes
,
which is the FLRW metric with a(t ') = 1. The observed geometry of the universe requires a non-constant a. Also, b isn't actually a parameter, since you get the same geometry regardless of b. You could try making b a function of r instead of t. I think this would get you a family of geometries distinct from the FLRW geometries (in which case they still don't describe the real world), but I'm not entirely sure. -- BenRG (talk) 18:05, 8 May 2008 (UTC)[reply]
This does work, though: , if you take , i.e. the derivative of the inverse of a. (I'm writing for the big parenthesized part of the metric.) The only caveat is that a has to be invertible (and have no critical points), which rules out a recollapsing universe but doesn't rule out the Lambda-CDM model. For k > 0 the "picture" you get from this is a universe with the big bang at the center and later cosmological times represented by concentric spheres. Distances within the spheres are what you'd expect from a background Euclidean geometry, but distances from the center are timelike and scaled by b(t). So this is something like an expanding/contracting time picture of the FLRW universe. You're free to think of the universe this way, but I think the usual expansion picture is intuitively clearer. (The usual picture can even be understood in Newtonian terms—see this long thread.) -- BenRG (talk) 14:15, 10 May 2008 (UTC)[reply]

The Doolittle Raid

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Why wouldn't the U.S. navy install booster rockets to the B-25s of the Doolittle Raid? Wasn't it cheap and proven technology? -- Toytoy (talk) 17:29, 7 May 2008 (UTC)[reply]

Perhaps a better question is "why would they?" Based on our JATO article, it doesn't appear the US had them at this point, and I doubt that they'd be classified as "cheap and proven" at that stage even if they did. Training for the raid had already established that B-25s should be launchable from a carrier, so why would the Navy and Army add extra unnecessary complexity? — Lomn 19:01, 7 May 2008 (UTC)[reply]
The rocket boosters, had they been a proven and available technology, might have allowed greater takeoff eweight, thereby allowing more fuel load (possibly via aux tanks) with greater range and thus the possibility to launch farther from Japan with greater safety for the carrier and less chance of premature detection, or with a greater bombload, or with a greater chance of making it on to a safe landing in China. Edison (talk) 19:43, 7 May 2008 (UTC)[reply]
As I recall (from the book version of Thirty Seconds Over Tokyo), the B-25s were already doing stuff like carrying jerry cans of fuel internally. While I concur that in principle with the higher weight -> more range -> safer mission principle, I don't know that any practical payoff was likely (bomb load vs range is a different matter, I guess, but meh -- it was a propaganda raid. 5% more bombs doesn't affect that). Regardless, a large part of the range problem was the premature launch of the mission due to possible detection, though my personal opinion from reading about this as well as the Flying Tigers is that the US was overly optimistic to expect these aircraft to successfully rebase out of China. — Lomn 19:59, 7 May 2008 (UTC)[reply]
Retrospective analysis can always find superior solutions. The Doolittle raid was cobbled together from available pieces in a very short time. With the benefit of hindsight, we can find all sorts of clever improvements. In your scenario, JATO units would have increased the bomb payloads, resulting in much more damage to Tokyo. but in an equally likely scenario, one or more JATO units would have failed catastrophically, causing distruction of a B-25 and renderingthe carrier incapable of launching other B-25s. -Arch dude (talk) 02:33, 8 May 2008 (UTC)[reply]

I can hardly think of a reason why they could not modify a military rocket for a mission like this. The army has been using ground-based rockets for many years. This proven technology shall be adapted for launching bombers on an aircraft carrier's deck. It can decrease the risks.

Installing jerrycans on a bomber seems to be ... well ... stupid. These cans are heavy and not very efficient. Why couldn't they make a large aluminium oil tank?

I find the jerrycan article funny. Allied scientists had to reverse engineer steel oil cans? Was it rocket science to make cheap and durable oil cans? Hitler ordered jerrycans secretly? Was it really such a big deal to make some oil cans for people? Couldn't the Germans make up a good excuse for making oil/water cans? e.g., "These cans are for delivery of cooking oil to the hospitals." -- Toytoy (talk) 12:44, 8 May 2008 (UTC)[reply]

Methinks you overestimate the ease of something like this and the likelihood of success. Getting rockets to work reliably was not an easy task; to do it with manned flight and without causing danger to the launching platform was simply not a risk worth taking in 1942. In any case, thinking backwards like this is a poor historical methodology—all you learn about is yourself, not the past.
As for the jerrycan, I think you also underestimate the materials science and good engineering that goes into that particular steel oil can. Think about it this way: if it was easy to do, they wouldn't have needed to reverse engineer it. They weren't any stupider than people today are. --98.217.8.46 (talk) 13:33, 8 May 2008 (UTC)[reply]
What ground-based rockets did the U.S. have in 1942? The German and the Russians deployed things like the Katyusha but these probably didn't have the power to accelerate a bomber significantly. The development of large rockets by the Germans during the war was a major advance. (And so we acquired their technology and personnel after the war.) See Rocket artillery#World War II, V2 rocket. Rmhermen (talk) 14:36, 8 May 2008 (UTC)[reply]
Digging deeply through our poorly organized material on Wikipedia, Aerojet tested a U.S. design for a JATO rocket in August 1941 but it didn't go into production until 1943. Elsewhere I found, that in the 1941 test, it took 12 rockets to launch a 1,260 lb max weight plane. A loaded B-25 weighed 33,510 lb so we only need to attach about 320 rockets to each bomber! In April of 1942 (the same month as Doolittle), the U.S. began testing a liquid-fueled JATO on a bomber smaller than the B-25; however, development was abandoned in 1944. Rmhermen (talk) 14:54, 8 May 2008 (UTC)[reply]

Proving Technology for Aerospace Applications

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Hi all, does anyone know what sort of tests are required to prove a new technology or material for use on aeroplanes or spacecraft? And what issues need to be considered? So far I've considered:

  • An ability to be implemented consistently, reliably
  • Withstanding repeated exposure to the pressure and temperature changes likely to be experienced
  • A knowledge of how to detect flaws and failings in the material or technology.

I have also found, British Aerospace EAP an example of testing aerospace technology.

This is part of revision for an exam coming up. If anyone has any thoughts, knowledge or ideas I'd be very grateful. Many thanks, LHMike (talk) 18:43, 7 May 2008 (UTC)[reply]

Technology Readiness Level has some information on American standards Mad031683 (talk) 19:49, 7 May 2008 (UTC)[reply]
Thanks, that's really helpful. Also found Mature technology linked. LHMike (talk) 00:26, 8 May 2008 (UTC)[reply]

Biology

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Can human polyspermy end as a viable embryo (e.g. by mixoploidy etc.)? Eliko (talk) 19:52, 7 May 2008 (UTC)[reply]

Sort of. Such an egg will convert into a pair of diploid zygotes, however. It was never clear to me how that actually happened. Someguy1221 (talk) 23:52, 8 May 2008 (UTC)[reply]
Let me quote from the article Polyploidy:
Triploidy, usually due to polyspermy, occurs in about 2-3% of all human pregnancies and ~15% of miscarriages. The vast majority of triploid conceptions end as miscarriage, and those that do survive to term typically die shortly after birth. In some cases survival past birth may occur longer if there is mixoploidy with both a diploid and a triploid cell population present.
So I'm really eager to know about any source for your statement: Such an egg will convert into a pair of diploid zygotes.
Eliko (talk) 07:28, 9 May 2008 (UTC)[reply]
I'll just say that Nature said it. I could hunt down their own two references, but it's kind of late, so I think I'm out for the night. Someguy1221 (talk) 09:27, 9 May 2008 (UTC)[reply]
And reading the Nature article again, I'll take that partially back. It is never stated that the egg splits into a pair of zygotes, merely that it forms a chimeric diploid embryo that later produces twins. Someguy1221 (talk) 09:29, 9 May 2008 (UTC)[reply]
Thank you for the source and for the news, it really helped me. By the way, this source does not indicate that polyspermy "forms" a chimeric diploid embryo - i.e. as a usual result of polyspermy; it just says that a chimeric diploid emryo - may be a "way" (i.e. a possible "way" out of some possible different results) of polyspermy. Please pay attention to the word "way" (in the report summary). Anyways, thank you again for the news: My original question - whether polyspermy can end as a viable embryo - has been answered. Eliko (talk) 13:10, 9 May 2008 (UTC)[reply]
Oh yes, I realized that after reading it again, last night. But why yearn for ultra-accuracy with my wording when I trust you're going to double-check the source anyway ;-) Someguy1221 (talk) 17:19, 9 May 2008 (UTC)[reply]

Swings and insurance liability

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I've seen in other countries swings on which children stand up. I see only swing seats here in America. Does this have to do with the rate of injuries? I find the seats rather boring. Imagine Reason (talk) 20:08, 7 May 2008 (UTC)[reply]

That doesn't sound like a science question to me... just google it and see if you can find some studies. I doubt anyone here just happens to be an expert on playground accident statistics (although anything is possible...). --Tango (talk) 22:16, 7 May 2008 (UTC)[reply]
You can stand up on any swing with a hard seat. You've never seen a swing with a wooden plank for a seat? --Shaggorama (talk) 10:12, 8 May 2008 (UTC)[reply]
Not in public in America you won't (see any swings with hard seats). The legal threat posed by such a deadly item is just too much! So very uncomfortable, not very swingable "sling seats" are all you'll see.
Atlant (talk) 13:14, 8 May 2008 (UTC)[reply]
I never had any trouble standing up on a sling seat. --Carnildo (talk) 20:52, 8 May 2008 (UTC)[reply]
I didn't know if any desk would be appropriate for this question, and I'd watchlisted this page, so...anyway, if I've seen planks in America, I've forgotten about it. I never saw any seats in Asia, either. I did try standing on the seats, but couldn't stand on their round bottoms. Imagine Reason (talk) 01:02, 9 May 2008 (UTC)[reply]
Hm. Hard seats certainly existed in America forty years ago. Did something happen? —Tamfang (talk) 22:28, 13 May 2008 (UTC)[reply]

Are old-growth forests being cleared faster than they regenerate?

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My question is specifically directed towards the forests in British Columbia, where environmentalists (namely Greenpeace) have targeted most. I know that the UN's Forest and Agriculture Organization (FAO) reports that Canada has maintained over 400,000 hectares of forest cover, but how much of that is old-growth? Does that amount of cover provide old-growth forests to regenerate by the time they are logged again? —Akrabbimtalk 22:06, 7 May 2008 (UTC)[reply]

Although the term Old growth forest is somewhat disputed, from what I know it is generally agreed that once it has been cut down it's gone. What you replant is a managed tree lot usually called Secondary forest. Certain features of old growth forest like natural selection of species growth pattern, influence of dead trees and fallen old logs, established drainage patterns and natural territory boundaries are not re-created once the old growth has been logged. UW researchers studied Mount St. Helens after the blast [1] what is not mentioned in this article is, the study plot that was cleared of fallen and dead logs and replanted actually recovered slower than the plots left to their own devices. Some forestry managers now use limited controlled fires to control underbrush, because they found that dousing all fires just made the next fire bigger and more dangerous. Sustainable forestry is no substitute for old growth because it just means that loggers don't take out more trees than they grow. It's like taking out a creek and putting a concrete lined canal there instead. The water still flows but many other factors can not be re-created. On a positive note, some areas of South American rain forest used to be farmed by natives in ancient times and are now almost indistinguishable from undisturbed growth. We just won't leave a forest alone for a couple of centuries for "old growth" to develop again. Hope this will help with answering your question. --71.236.23.111 (talk) 23:52, 7 May 2008 (UTC)[reply]
Does that mean that if and only we continue sustainable forestry for a couple of centuries, the answer to this question becomes yes? Of course, that doesn't address the amount of old-growth forests there would be. If we continued to cut down trees at our current rate, and always cut down a certain portion of old-growth trees (say, half the ones we cut down are old growth, regardless of how much old growth is left), then I'm willing to bet the answer would be "none". Of course, centuries from now, humanity would probably either be advanced enough for this make an unnoticeable difference, or extinct. Either way, the only practical time frame is short enough for new old-growth to be, for all intents and purposes, nonexistent. — DanielLC 00:01, 9 May 2008 (UTC)[reply]
Nope only if you'd leave your replanted forest alone it "might" happen. What you plant is all the same species and with little genetic variation at that. Trees are planted at regular intervals to enable easy access for machines. They are all the same age. The underbrush is gone. The soil is all stirred up and topsy-turvy. Roads cut through plots depending on local conditions they can either create a wall for water-flow patterns and animal and plant dispersal or a highway for floodwater and opportunistic species (or both). Some modern forestry techniques are trying to copy pages from nature's book, like planting a mix of species, but still while we meddle it won't become old growth. For that to happen you'd have to maybe airlift individual trees out and only cut one every mile or so every 50 to 1oo years maybe. (You might be able to get away with closer together and more often or you might need to do it farther apart and less often.) The trees that you take out would not be available as fallen trunks, but a system can and does usually cope with a little bit of interference. Sustainable forestry just means that there is a tree growing for a tree you cut down. It's as though you'd take the Mona Lisa out of the museum and instead would hang a photograph there. Eventually, in a hundred years or so that photograph might also become art. If it survives a couple of hundred years more it might even become unique art and deserve a place next to the old masters, but if you hang a new picture in the frame every couple of years that's never going to happen. Lisa4edit (talk) 20:48, 9 May 2008 (UTC)[reply]
Based on how things work with farming tribes in tropical rainforests, you could sustainably harvest an old-growth forest by cutting one-acre patches amounting to about 0.25%-0.5% of the forest each year. --Carnildo (talk) 21:20, 9 May 2008 (UTC)[reply]

Question: What is the compound/name of the chelate of calcium produced by plants?

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As chlorophyl is a chelate of magnesium produced by plants, I wish to know any chelate of calcium that is produced during the symbiotic relationship of plant roots and soil microbes. Best answered by a SOIL BIOLOGIST.

I have asked several chemists this question and they always say, "EDTA." We are talking about organic chemistry here. Engintinc (talk) 22:50, 7 May 2008 (UTC)[reply]

Well, EDTA certainly will chelate calcium ions, but it's a synthetic molecule, and not really something made by plants. (And if you're interested in calcium specifically, EGTA may be a better bet for a synthetic chelator.) For natural compounds, you're probably looking for a calcium equivalent to siderophore. Unfortunately, I am unaware of any such molecules, although they certainly may exist. -- 128.104.112.85 (talk) 14:37, 8 May 2008 (UTC)[reply]

Conservation of angular momentum

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During a physics lab on the conservation of angular momentum, we were asked to sit on a chair could rotate effectively frictionlessly. After firmly placing our feet on the ground, someone spun the wheel, and then we were asked to flip the wheel. Now, as per the conservation of angular momentum, it is expected that we would spin in the opposite direction, and that is what we observed. However, what I don`t understand is what causes the person in the chair to spin. To illustrate my point, let us consider linear momentum. Linear momentum, like angular momentum, is conserved if the sum of the external forces on the system is zero. Now, force is by definition what causes a change in momentum, so the fact that momentum is conserved if there are no external forces is fairly obvious. But, when a stationary object is hit by a moving object, the fact that the stationary object moves is because not because momentum is conserved, but because there is an internal force acting on the object. So, my question is this: when considering angular momentum, what causes the person in the chair to spin? It must be a torque (the rotational equivalent of force). But what is the source of the torque? Also, a perhaps related question, why is it hard to flip a spinning wheel? Can how hard it is to spin be calculated?


Could you please re-explain the scenario in your lab? I'm having trouble following it. Oh, and please remember to sign your posts using four tildes (~~~~). -mattbuck (Talk) 23:22, 7 May 2008 (UTC)[reply]
You're sitting in a chair that can spin, and you're holding a wheel (the wheel can spin; it has handles of sorts going through the center of the wheel). Someone spins the wheel (it is parallel to the floor), and the you flip it 180 degrees so that it's spinning in the opposite direction. Afterwards, you begin to spin. —Preceding unsigned comment added by 76.69.240.138 (talk) 23:30, 7 May 2008 (UTC)[reply]
I'm pretty sure it's Newton's 3rd law - equal and opposite forces. The spinning wheel has angular momentum, and when you flip it, the torque you provide changes the direction of the angular motion 180 degrees, so you receive an equal force in the opposite direction, causing you to spin, conserving (angular) momentum. —Akrabbimtalk 23:38, 7 May 2008 (UTC)[reply]
Yeah, the force making you spin is the reactive force to you moving the wheel. It's the same basic principle as rocket propulsion - sit on the same chair with a heavy weight and throw it away from you and the chair will move backwards. This is just the rotational equivalent. --Tango (talk) 00:35, 8 May 2008 (UTC)[reply]
Okay, I think I get it. But why is it hard to flip the wheel, and how can this be calculated? —Preceding unsigned comment added by 76.69.240.138 (talk) 00:53, 8 May 2008 (UTC)[reply]
You can think of it as the rotational analogue of Newton's 3rd law - essentially, when you try to flip the wheel, you are applying a torque to the wheel (torque is a change in momentum over time, just like force is a change in momentum over time). The wheel applies an equal and opposite torque to you! You may need a pencil and paper to see this properly - the angular momentum vector of the wheel is initially vertical. As you flip the wheel, you're changing the angular momentum vector, and the torque is in the direction of that change; as you flip the wheel, there is a torque with some component in the vertical direction, and the equal and opposite torque on you and your chair produces your angular acceleration. It is "hard" to flip the wheel for the same reason it's "hard" to throw the wheel across the room to give yourself linear momentum - both you and the wheel have rotational inertia! Using the equation in our article on precession (look in the torque-induced precession section), the net torque causing you to spin will be where is the angular speed of your flipping motion, is the spin angular speed of the wheel, is the angular moment of the wheel, and is the angle of the axis of the wheel to the vertical. It is possible to think of the forces "microscopically", i.e. in terms of the forces on each part of the wheel but it is pretty tedious in my opinion. If you want to see that treatement, there's a decent one in the aforementioned article on precession. --Bmk (talk) 01:30, 8 May 2008 (UTC)[reply]
Got it! Thanks a lot. —Preceding unsigned comment added by 76.69.240.138 (talk) 02:49, 8 May 2008 (UTC)[reply]

Momentum of a photon

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How can a photon, which is massless, have momentum? —Preceding unsigned comment added by 76.69.240.138 (talk) 23:14, 7 May 2008 (UTC)[reply]

A photon has zero rest mass. However, under special relativity, accelerating particles gain mass. If you were to accelerate a massive particle to the speed of light, its mass would become infinite. Hence it is necessary for a photon to be "massless" for it to be able to travel at the speed of light. Confusing Manifestation(Say hi!) 23:26, 7 May 2008 (UTC)[reply]
You've probably heard of the equation "E=mc2". It describes an equivalence between mass and energy - that equivalence goes a further than that simple formula (which is the special case for objects at rest - a photon isn't at rest, since it always moves at the speed of light). While a photon has zero rest mass, it has a positive energy, and that gives it a momentum. That momentum (in a vacuum, at least) is where h is the Plank constant and lambda is the wavelength. --Tango (talk) 23:32, 7 May 2008 (UTC)[reply]
I never thought about that. that's messed up. --Shaggorama (talk) 10:14, 8 May 2008 (UTC)[reply]
Relativity can be rather counter intuitive... For example, even though photons are massless, they are still affected by gravity in the same way as anything else (a beam of light on Earth will fall at 9.81m/s/s - it's hard to tell, since it's out of sight before it's fallen a noticeable amount, but it does fall). --Tango (talk) 13:28, 8 May 2008 (UTC)[reply]
[repost from old question] Be careful with that equivalence! As described at general relativity, light suffers twice the deflection that one would naïvely expect from analogy with a fast rock. --Tardis (talk) 15:30, 8 May 2008 (UTC)[reply]
Interesting... what causes the difference between a very light particle travelling at 0.999999c and a photon? --Tango (talk) 20:18, 8 May 2008 (UTC)[reply]
all I can think of is that maybe light is refracted by the gravitational time dilation in addition to the distortion of space.Em3ryguy (talk) 21:26, 8 May 2008 (UTC)[reply]
I'm confused. I was told once that it was one half the expected value. Also you might find this interesting-http://www.experiencefestival.com/a/Newtons_aether_model_-_Modernised_Newtonian_theory/id/5333093Em3ryguy (talk) 20:41, 8 May 2008 (UTC)[reply]
It's twice, but that only applies at large scales where the whole gravitational field is involved. In local experiments where the field is uniform, light falls at 9.8 m/s² just like everything else. (You can see this from the equivalence principle: imagine doing the same experiment on a rocket ship accelerating at one gee.) The extra global deflection comes from the fact that the circumference of a circle around a gravitating object is less than π times the diameter. Here's an analogy: take a circular sheet of paper with a pie-shaped wedge missing. Draw a straight line which intersects one edge of the missing wedge and a second parallel line which intersects the other edge. Pull the edges together, making a cone, and extend the lines across the seam in as straight a fashion as possible. Going around opposite sides of the cone has caused them to converge. This deflection is (in a certain sense) independent of the speed of the object, whereas the "ordinary" deflection of course depends on the speed. At light speed they have the same magnitude; at 0.999999c they have almost the same magnitude; at small speeds the "ordinary" deflection dominates. -- BenRG (talk) 01:14, 9 May 2008 (UTC)[reply]
That makes sense. I thought it odd that a photon would behave significantly differently to a very fast moving light massive particle... --Tango (talk) 18:07, 9 May 2008 (UTC)[reply]
In special relativity the relation between energy and momentum is E² = (pc)² + (mc²)², where m is the rest mass. When people say the photon is "massless" they mean that m = 0, which implies that E = |p| c. Both E and p are always nonzero. -- BenRG (talk) 01:14, 9 May 2008 (UTC)[reply]

centripetal motion of car

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I was given a question where a car of known mass was turning on a circular road of known radius and coefficient of static friction. The question then asks what's the maximum velocity of the car. That wasn't to difficult, but I have a theoretical question. In this question, I treated the centripetal force as the static friction force, but for this to work the static friction force should be pointing to the center of the road. But it's pointing along the path of the car. Then, how can static friction cause centripetal motion. Wouldn't we need the angle of the tires to treat friction vectorally? —Preceding unsigned comment added by 76.69.240.138 (talk) 23:25, 7 May 2008 (UTC)[reply]

The scope of the question seems to be assuming that the tires are rolling ideally, that is to say no slipping. The torque the axle provides causes the tire to push backwards on the road. Since the wheel is not slipping, that force is solely static friction, as the rubber is not necessarily moving in relation to the road, as each infinitesimal point of the tire touches down on the road in an instant, pushes back, and then is removed from the road, as the tire rolls. —Akrabbimtalk 23:44, 7 May 2008 (UTC)[reply]
I don't think you've understood my question; perhaps I was unclear. I know why the car moves. It is a simple application of Newton's third law. My question is why we treat the centripetal force as equal to μ*m*g. After all, isn't the friction force not directed at the center of the circle of motion. —Preceding unsigned comment added by 76.69.240.138 (talk) 00:18, 8 May 2008 (UTC)[reply]
I think it has to do with the fact that the tires aren't pointing in the direction of travel. I'm not sure about the details, though... My attempt to work it out ended up with the car going in the wrong direction, so I'm clearly missing something... --Tango (talk) 00:31, 8 May 2008 (UTC)[reply]
The car is observed to be moving in a circle at constant speed. Therefore, the car is constantly accelerating toward the center of the circle. Therefore, there a is force acting on the car and the force is acting in the diretion of the center of the circle. The only force acting on the car (in a pure gedankenexperiment) is the friction between the roadway and the tires. Therefore, the direction of this force must be toward the center of the circle. Conclusion: the vector of the force is in this case perpendicular to the direction of motion. -Arch dude (talk) 02:03, 8 May 2008 (UTC)[reply]
From this, we can see that your "obvious" assertion (that the friction force is in the direction of motion), is incorrect. -Arch dude (talk) 02:12, 8 May 2008 (UTC)[reply]
There is no motion. In static friction, the two objects touching (the bottom of the wheels and the ground) remain stationary compared to each other. The rest of the car isn't touching the ground, so its motion is irrelevant. The friction force is in the opposite of the force acting upon it (centrifugal force). — DanielLC 23:48, 8 May 2008 (UTC)[reply]