Wikipedia:Reference desk/Archives/Science/2010 November 26

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November 26[edit]

Swallowing one's tongue[edit]

I read the following sentences on the Wikipedia: "Teammates Ivica Dragutinović and Andrés Palop immediately ran to his side as he lost consciousness. Moments later, club medical staff and other players followed suit, as Dragutinović stopped Puerta from swallowing his tongue." What is "swallowing ones tongue"? Can you give some explanations? Thank you! —Preceding unsigned comment added by 72.198.195.96 (talk) 05:43, 26 November 2010 (UTC)

When you lose consciousness your tongue can lose muscle tone and (especially if you're flat on your back) can fall back in your mouth occluding the airways. It's not that you actually 'swallow' it in the way that you swallow food, but merely that if can block the airway. That's why one of first steps in first aid with an unconscious patient is to turn them on their side (along with preventing choking on things like spontaneous regurgitation). --jjron (talk) 08:24, 26 November 2010 (UTC)

RFID[edit]

   Can RFID tagging (i.e. of products exposed for sale in a retail store) be used to defeat aluminium-lined bags used by shoplifters? In addition, what do RFID scanners which are to be placed at a store's entrance look like; are they conspicuous enough to deter would-be shoplifters from even trying? Alternatively, can RFID scanners placed at store entrances be made inconspicuous enough to aid in apprehending shoplifters? Rocketshiporion 05:50, 26 November 2010 (UTC)

Aluminum foil will attenuate RFID signals in the same way it attenuates most security tags. The amount of foil required to prevent a detection will depend on the details of the system, but for many RFID uses the signal may be undetectable with only 1 to a few layers of foil. Dragons flight (talk) 07:30, 26 November 2010 (UTC)
Note that current security measures try to detect an object in the vicinity of the door, but a better approach is to continuously monitor the position of every item in the store, and alarm when any of them disappear, noting the site of the disappearance. This requires greater range, bandwidth, and computing power to track all those objects, but it is possible right now. However, the cost of the system and size of the tags makes it not yet practical for most items. Perhaps very expensive items, like jewelry, might be the first to get this treatment. StuRat (talk) 17:46, 26 November 2010 (UTC)

mining asteroids[edit]

Hi, I came across this page where they say an asteroid (3554 Amun) that is a mile wide contains 30 times as much metal as Humans have mined throughout history (ever). Surely this is wrong, or am I understanding it incorrectly? Sandman30s (talk) 06:45, 26 November 2010 (UTC)

I agree it is wrong, though perhaps not as wrong as one might guess. According to list of countries by iron production, global iron mining was 2.3 billion tons / year in 2009. As raw metal, that would have volume of 0.3 km3. 3554 Amun is only about 7 times that annual volume. So definitely not 30 times all metal ever, but still a large amount on the scale of iron mining. Also, iron stands out for its very large production volumes. Most other metals we mine are in much smaller quantities (e.g. copper and aluminum are only a few percent of the iron values), so a concentration of those metals would be comparatively more significant if one existed. Dragons flight (talk) 07:50, 26 November 2010 (UTC)
Thanks, I would never have imagined those figures... Sandman30s (talk) 09:48, 26 November 2010 (UTC)
Hang on a moment. I could not find a web page giving the volume of Amun (I did find multiple sources giving its "diameter", but since small asteroids are not spherical, this gives little idea of its volume). But Wikipedia's page shows its mass as 1.6e13 kg, which is 16 billion metric tons. This accords with Dragon's figure of 7 times the 2.3 billion tons of iron mined in 2009, if those are metric tons; if they're short tons, as one might expect from a US source, it would be nearer 8 times. But several other Internet sources give Amun's mass as 30 billion metric tons, which (if iron) would be equivalent to 13 or 14 years' terrestrial production rather than 7 or 8. Still nowhere near 30 times the world's all-time production; perhaps someone slipped a factor of 1,000 in their original calculation. --Anonymous, 02:02 UTC, November 26, 2010.
Hangh on another monent, list of countries by iron production actually gives iron ore mining statistics, not refined iron production. Thar page needs to be fixed. 75.41.110.200 (talk) 05:01, 27 November 2010 (UTC)
Good catch! I've put a "needs expert attention" flag on it. Still, judging by the some of the numbers mentioned in one of the linked references, the proportion of iron in iron ore is often pretty large, between say 25% and 40%, so this only introduces an error of a factor of 3 or so. Maybe something like 40 years' production, then. --Anonymous, 05:50 UTC, November 27, 2010.
Hmmm, that's 40 years CURRENT production. If production grows at 7% a year which is not unreasonable to assume, this equals to a doubling of production every decade. I don't know specifically about iron, but this is what has been happening to Oil for over a century. If production doubles every decade, it means that in the last ten years, you have actually produced more resource then ALL THE PRECEDING production combined. Think about it: If you mined one "unit" of iron in the 1st decade, and you mined 2 units in the next decade, in the decade after that you'll mine 4 units, this is more then 1 and 2 combined. The next decade you will mine 8 units, this is more then 1, 2 and 4 combined. And so on. So saying 40 years of current production could very easily be orders of magnitude more then what has ever been produced, if the rate of growth is just 7% annually. Scary but true. Hubbert curve, Hubbert peak theory, Peak oil. Vespine (talk) 23:30, 28 November 2010 (UTC)
Perfectly true, but we don't hear people talking about "peak iron", so I doubt there has been such a rate of increase. Data? I couldn't readily find anything by googling. --Anonymous, 03:17 UTC, November 29, 2010.
I wasn't trying to say anything about peak iron specifically. My main point is that even with growth rates of 7% a year which is not considered "aggressive" in our consumerist culture, 40 years worth of current production can easily be far more then has ever been produced in the past, regardless of how much time is "in the past" or the resource. Vespine (talk)
Right sorry, i think i misunderstood your question a little, this article has a LOT of data which I find a bit confusing because it talks about "exports" and all sorts of other "percentages" which i'm not sure how to interpret. One pertinent point near the beginning states : "World production of iron ore grew by 12% in 2006 to reach 1.5 billion mt (See Table 1). This was a fifth consecutive record high. " Vespine (talk) 04:28, 29 November 2010 (UTC)

How can a solvent be non-polar but be made of polar molecules?[edit]

My organic chemistry textbook makes the distinction between polar molecules and polar solvent. It says that polar molecules are identified by the high dipole (u) seperation of the molecule, while polar solvents are identified by a high dielectric constant.

It says that all polar solvents are made from polar molecules, but the opposite is not true and provides the example of formic acid vs acetic acid to demonstrate this.

Both formic acid and acetic acid are polar molecules by virtue of their dipole moment. However, formic acid, with a dielectric constant of 59, is also a polar solvent, while acetic acid, with a dielectric constant of 6.1, is not a polar solvent.

Why is this? Would both solvents be adequate for dissolving ionic compounds such as NaCl? Acceptable (talk) 08:05, 26 November 2010 (UTC)

Solubility of polar substances (and by extension, ionic) substances is dependant almost solely on dielectric constant. Your textbook is pretty much spot-on on this one. The solubility of something like NaCl is dependent on the ability of the solvent to solvate the ions; that is to make bonds to the ions which are stronger than the bonds the ions would make to each other (strictly speaking, it is defined thermodynamically; the substance is soluble if the free energy released in the formation of the solvent-ion bonds is greater than the free energy required to break the ion-ion and solvent-solvent bonds). Dielectric constant takes this ability into account, whereas dipole moment does not. --Jayron32 16:32, 26 November 2010 (UTC)
But why does formic acid have a higher dielectric constant than acetic acid when both have a carboxylic acid functional group? They differ only by the fact that one has a methyl group and the other has a hydrogen. Acceptable (talk) 21:24, 26 November 2010 (UTC)
The methyl group is MUCH more "electron donating" than the hydrogen atom is. You can think of this in two ways; either you can think of it as the methyl group donating electrons to the carbon part of the COOH dipole OR you can think of the methyl group as acting like a "positive charge sink", again to the same effect. The dipole moment is calculated for the bond, while the dielectric constant is calculated across the whole molecule; while the dipole moments will be similar (but not the same, due to the methyl's effect described above), the dielectric constant will be much lower on the acetic rather than the formic acid. You can see the effect even greater on very large molecules; take something like stearic acid; the dipole moment on the COOH will not be that much smaller than it will be on the acetic acid, indeed after propionic acid, longer chains of carbons do not markedly affect the dipole moment in that part of the molecule. However, the dielectric constant continues to go down to nearly nil. I am pretty sure that after 4 or 5 carbons, the bulk substance is considered essentially non-polar; despite the acid group. --Jayron32 21:33, 26 November 2010 (UTC)
Ah, ok, thanks a lot for the helpful explanation. Acceptable (talk) 02:08, 27 November 2010 (UTC)

Is it possible to vomit in your sleep and die by suffocation without being under the influence of anything?[edit]

In other words, could a perfectly normal, sober person who hasn't had anything to drink//taken any mind-altering substances vomit and thereby asphyxiate during slumber, or would the body's involuntary control measures trigger countermeasures like gagging, rolling over to one side, etc. before the person awoke?

Or perhaps it's not even possible to vomit in one's sleep? Sign me "curious" The Masked Booby (talk) 09:21, 26 November 2010 (UTC)

Did a few google searches on it, most of the results appear to be people asking about it on various forums - One answer on this forum referred to GERD - Although perhaps not vomiting in the sense that you mean, perhaps mildly along the same lines. All other mentions that I've come across thus far (the last few minutes), seem to point to factors enducing the vomiting while asleep: alcohol; obesity; underlying illness; directly related illness.. etc. Darigan (talk) 09:54, 26 November 2010 (UTC)
I think a better question is can you do that and stay asleep, and I doubt it. If I vomited I'd wake up for sure. Same goes x1000 for not being able to breath. Ariel. (talk) 11:15, 26 November 2010 (UTC)
I think the best anyone could say is it is unlikely. The human body is a fantastically unpredictable thing, so I would not be surprised to find, if I looked hard enough, isolated reports of one or two people who have died by choking on their own vomit, in their sleep, without any complicating factors. --Jayron32 15:22, 26 November 2010 (UTC)
One thing of course if we are thinking there is no illness or other factor as someof the above answers are discussing, vomitting itself is not common asleep or not. Nil Einne (talk) 16:50, 26 November 2010 (UTC)
I recall a report or two of people suffocating to death from their vomit while very drunk and thus probably unconscious. 92.24.178.149 (talk) 22:05, 26 November 2010 (UTC)

The suffocation reflex from carbon dioxide buildup in the lungs is very painful if you aren't used to it, and very powerful. Vomit is a fluid which is almost always easy for a conscious person to expel. A conscious person in shock may be effectively paralyzed, however, which is why CPR and related forms of first aid instruct the person administering the aid to check to see that the airway is clear, even when the victim is conscious. Ginger Conspiracy (talk) 01:03, 27 November 2010 (UTC)

Copper containing biomolecules[edit]

Why did copper containing biomolecules appear later in evolution than their iron based analogues that do the same job? —Preceding unsigned comment added by Blackmetalgrandad (talkcontribs) 10:57, 26 November 2010 (UTC)

I can't answer your question, but you should know that there is something like 100,000 times as much iron on earth as there is copper. Ariel. (talk) 11:22, 26 November 2010 (UTC)
Oops, I meant why did the copper based ones appear earlier!144.32.126.11 (talk) 11:33, 26 November 2010 (UTC)
It could be that hemocyanin would have been more efficient in the early seas which where oxygen poor, than iron based oxygen carriers. Also, it allows (or favours) simpler body organs. --Aspro (talk) 16:44, 26 November 2010 (UTC)

Iodine heptafluoride[edit]

What is its appearance? Colorless gas? (just a guess) --Chemicalinterest (talk) 15:02, 26 November 2010 (UTC)

Yes, good guess! It sublimes at 4.8 °C under atmospheric pressure. Physchim62 (talk) 15:12, 26 November 2010 (UTC)

Iodine from caliche[edit]

The article does not state how iodine is extracted from the iodide and iodate in caliche. How is it extracted? All of these iodine questions are because of this. --Chemicalinterest (talk) 15:25, 26 November 2010 (UTC)

It can be reduced with sodium bisulfite but there are more advanced multi-step processes which seem to be more popular these days.[1] Ginger Conspiracy (talk) 00:54, 27 November 2010 (UTC)

Why kilogram as a base unit rather than gram?[edit]

Why was the kilogram chosen as a base unit of the International System of Units rather than the gram? The kg is the only base unit with an SI prefix as part of its name. Our article on the Kilogram says, "Since trade and commerce typically involve items significantly more massive than one gram...[the standard became] one thousand times more massive than the gram—the kilogram." Although this seems common sense, wouldn't the desire for consistancy in the new system be more important? Doesn't having one base unit that already contains a prefix in its name cause confusion when multiplying it by a another prefix? It seems that the term "kilokilogram" would be necessary to describe a mass of 1000 kg, which is awkward. --Thomprod (talk) 15:26, 26 November 2010 (UTC)

(1) "Since trade and commerce typically involve items significantly more massive than one gram...[the standard became] one thousand times more massive than the gram—the kilogram."
(2) No.
(3) No.
--Shantavira|feed me 15:45, 26 November 2010 (UTC)
The OP is making a common mistake; the confusion between the metric system and SI. The SI is a subset of the metric system chosen for convenience in the widest possible applications. The other thing about SI is that its "base" units are used to derive the so-called "derived units"; thus the Newton and the Joule and the Pascal and other units are always expressed as ratios or products of things like kilograms, meters, seconds, etc. There are other systems besides the SI which use different metric units, see cgs system, which uses units like the erg. In summation: The SI is not the metric system. It is a set of units speficially chosen from the metric system, chosen to be convenient for use in certain applications. --Jayron32 15:52, 26 November 2010 (UTC)
See grave (unit). Basically a grave (from gravity) was going to be the base unit; a gram was an alias for a milligrave; just as a ton is an alias for 1000 kg. However, this was just before the French Revolution, and Grave is also a French title; similar to the German Graf, or English Count. Grave as a title has the same etymology as Graff, which is different from Gravity. After the revolution it was felt that this would be contrary to égalité so grave as a unit was dropped. BTW, this came up a few months ago. CS Miller (talk) 16:00, 26 November 2010 (UTC)
Jayron: So the kilogram was chosen over the gram as the base unit of mass in the SI because the former was (and is) more "convenient" in most applications. I understand that. But didn't the kg stick out to the designers of the SI as the only base unit name to include a prefix? If I was designing a new system (based on mathematics and powers of ten and such) with base units and modifying prefixes (somewhat analogous to nouns and adjectives in grammar), why would I build in a point for possible later confusion by including a modifier in the name of a base unit? There is no color named "lightred", for example.
CS: I understand the politics of not using the word "grave". But, why didn't they just come up with a completely different word to represent 1000 grams, if that amount was thought to be more convenient than one gram? --Thomprod (talk) 17:00, 26 November 2010 (UTC)
Pass. Perhaps it was just an interim decision use gramme instead of milligrave, and no-one got around to making a new word for grave/killogramme. I made a slight mistake in my previous statement - originally the definition was the gramme - 1cc of water at 0°C, and the grave was the practical physical object to represent it. I don't know why the definition was moved to kg/grave - perhaps it was to make the definition and representation the same. CS Miller (talk) 17:28, 26 November 2010 (UTC)
Ok, now that makes sense. If the unit called "grave" was practical, but now politically incorrect, it would have made more sense to call it something brand new, rather than the hybrid "kilogram". --Thomprod (talk) 18:35, 26 November 2010 (UTC)
Don't expect SI to be this perfect, logical system. It's a system of units like any other. Use it when it's convenient; ignore it when its pronouncements are silly. In particular, ignore completely any recommendation not to use convenient units like the curie, or to avoid the name micron. --Trovatore (talk) 08:20, 27 November 2010 (UTC)

Force as Gradient of potential[edit]

We can define the quantity U such that dU is to equal -dr. But dU = ∇Udr. So F⋅dr = -∇Udr For this to be true in general, F = -∇U.

If a force can be written as the gradient of a scalar field, then this is taken as a definition that F is conservative. But where in the above derivation was F assumed to be conservative? 70.52.44.192 (talk) 20:54, 26 November 2010 (UTC)

Pictogram voting delete.svg Please do your own homework.
Welcome to Wikipedia. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know.
Having said that, you might benefit from reviewing force, gradient, and reading Newton's laws of motion#Variable-mass systems carefully. Ginger Conspiracy (talk) 23:52, 26 November 2010 (UTC)
No, it isn't a homework question, although looking back it does seem that way. The reason I spelled out the F = -∇U derivation was so that someone could point out more clearly where the assumption that F is conservative was made.
I looked at the articles you mentioned. I don't see what variable-mass systems have to do with conservative forces, and the gradient article doesn't mention potential energy. The force article starts by saying "a potential scalar field is defined as that field whose gradient is equal and opposite to the force produced at every point". But why can't such a potential be defined for any force? You might say, well U isn't a function of just r for all forces, just conservative ones. But if dU is defined as above, then it does seem, incorrectly, that all forces can be constructed to be the gradient of a scalar field. 70.52.44.192 (talk) 01:46, 27 November 2010 (UTC)
Oh, so by conservative you mean subject to conservation laws instead of covering only the particle case instead of the variable mass case? In that case, would the assumption of elastic force transfer be the reason, as opposed to force from an inelastic collision? Ginger Conspiracy (talk) 02:32, 27 November 2010 (UTC)
(ec) Such a U does always exist for a radial force whose magnitude is a function of only the radius, i.e., all such forces are conservative. But it doesn't always exist for more general force fields. For example, take . -- BenRG (talk) 02:37, 27 November 2010 (UTC)
F is assumed to be conservative in the line "We can define the quantity U such that dU is to equal -dr." If F is not conservative, there is no field U that satisfies this definition. Looie496 (talk) 05:35, 27 November 2010 (UTC)
In BenRG's example, wouldn't dU = -ydx + xdy satisfy dU = -F⋅dr? 70.52.44.192 (talk) 07:23, 27 November 2010 (UTC)
Yes, but what U satisfies dU = -ydx + xdy? -- BenRG (talk) 08:47, 27 November 2010 (UTC)
A conservative force is a force for which the work done by the force on a particle that moves from A to B depends only on the start and end points A and B and not on the path taken between them. Equivalently, a conservative force is a force for which the work done by the force on a particle travelling round any closed path is always 0. As has been said, a force that is the gradient of a scalar field is conservative - you can see this by looking at the integral of dr around any closed path. Also, any conservative force can be expressed as the gradient of a scalar field U by picking a fixed point A and defining U at any point B to be the integral of dr along any path from A to B - the conservative property means that this definition gives a well defined value for U at each point. U depends only on the choice of the point A, which is the same as saying U is uniquely defined up to an arbitrary additive constant.Gandalf61 (talk) 09:32, 27 November 2010 (UTC)

Thanks 70.52.44.192 (talk) 16:06, 27 November 2010 (UTC)