Wikipedia:Reference desk/Archives/Science/2009 July 29

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July 29

Is this true or made-up? Thanks.

"The concept that light appears to travel faster than the speed of light to an outside observer is known as super-luminous motion....If you have a charged particle moving close to the speed of light at angle 1/gamma (where gamma is the Lorentz factor) with respect the the observer. the particle can appear to be moving faster than the speed of light in the reference frame of the observer. However, the particle isn't actually moving faster than the speed of light. The speed of light is an Absolute. This effect is also known as relativistic beaming and is common in many Active Galactic Nuclei galaxies." [1] Imagine Reason (talk) 01:37, 29 July 2009 (UTC)

Well Relativistic beaming is real - our article is pretty clear and although it only has one reference, it's a good one. However, neither our article, nor it's reference talks about things appearing to move faster than the speed of light. Weird. SteveBaker (talk) 02:29, 29 July 2009 (UTC)

The speed of light may or may not be an absolute. There is a hypothetical particle knows as the tachyon which travels at superluminal speeds.CalamusFortis 03:08, 29 July 2009 (UTC)

That comment is super misleading. There is no reason to believe that the speed of light is not absolute, and every reason to believe it is. Sure it is always a possibility that our physics is wrong, but this question is asking a specific question about our current understanding of physics. It is not asking for wild ass speculation and ungrounded bullshitting. The question also has nothing to do with tachyons. APL (talk) 04:13, 29 July 2009 (UTC)

Indeed - there is a hypothetical space craft called "the starship enterprise" which travels at superluminal speeds too...and it's just about as real as tachyons. Just because we give something like that a name - doesn't bestow any additional legitimacy upon the hypothesis! SteveBaker (talk) 22:25, 29 July 2009 (UTC)

No, no particle can "appear" to travel faster than light. It's a cornerstone in relativity that all observers can agree on the laws of nature independent of how fast they'r travelling relative to each other. No observer should therefore observe a violation of the speed limit. Here's how I think it's meant to be interpreted:

Imagine that you'r shining a flashlight on a faraway wall. You move your flashlight a little and the point of light on the wall zips sideways. If the wall was further away, the point of light would move faster along the wall. Eventually, if the wall was far enough away, the point of light would move faster than the speed of light. However, no photons in the ray of light exceed the speed limit and it's impossible to transfer information at superluminal speed using the blob of light on the wall, as noone at the wall can affect its path or any other property. EverGreg (talk) 11:05, 29 July 2009 (UTC)

yes. I'm mentioning the light-ray because I'm guessing that's the effect in the relativistic beaming, though from what BenRG is saying, I'm not so sure anymore. :-) EverGreg (talk) 17:01, 29 July 2009 (UTC)

See also Faster-than-light#FTL phenomena. There are all sorts of cases where something can be made to exceed the speed of light relative to something else. The deal is, no information can travel at greater than the speed of light, so such actions are not useful. --Jayron32 11:15, 29 July 2009 (UTC)

See Cherenkov radiation. 98.234.126.251 (talk) 07:06, 30 July 2009 (UTC)

I think that quote is referring to the following fact: suppose you're at (0,0) and an object at (D,0) is moving with velocity v = (v_x, v_y), so after a time dt it's at (D + v_x dt, v_y dt). The light from (D,0) will reach you after a time of D/c and the light from (D + v_x dt, v_y dt) will reach you after a time of, to first order, dt + (D + v_x dt) / c = D/c + (1 + v_x/c) dt. The object's angular displacement over that time, meanwhile, is Δθ ≈ (v_y/D) dt. If you (wrongly!) approximate the object's tangential velocity as distance times angular displacement divided by time then you get D [(v_y/D) dt] / [(1 + v_x/c) dt] = v_y / (1 + v_x/c), which is off by a factor of 1 / (1 + v_x/c) and can exceed c. That's not surprising since the calculation is wrong (and would be wrong even in a Newtonian universe), but for some reason astronomers think it's interesting and call it "apparent transverse velocity" or some such. This doesn't agree with the paragraph's claim that the critical angle is 1/gamma, so it may be referring to a more sophisticated wrong calculation that gets a more sophisticated wrong answer. This "effect" is discussed briefly at Faster-than-light#Astronomical observations, which cites a paper that unfortunately doesn't seem to be available online. -- BenRG (talk) 16:08, 29 July 2009 (UTC)

That explains a lot. Thanks! Imagine Reason (talk) 00:10, 31 July 2009 (UTC)

Hypothetical math! ~~ Volume of water required for "the Great Flood" in Genesis?

Genesis 7:19 ~ 21 says that even the tallest mountains in the world were underwater. So, assuming that we need enough water to submerge Mt. Everest, just how much water do we need?

Bonus points for calculating the proportion of needed water against the world's current water volume! 95.172.239.38 (talk) 04:45, 29 July 2009 (UTC)

You only need enough water to submerge the "known world" to the writers at the time. That may have been as little as submerging Mount Sinai. 71.236.26.74 (talk) 05:46, 29 July 2009 (UTC)

Mean radius of the Earth is 6371.0 km and elevation of Mount Everest is 8.848 Km.

Approximate volume of water required to submerge the Mount Everest is = volume of sphere of elevation - volume of mean radius = 4/3*(6371+8.848)^3 - 4/3*PI(6371)^3 = 4521140066 km3

From this value, subtract volume of all mountains and land above mean radius. I guess 5 % of it. That gives us 4,295,083,063 km3

There is approximately 1,360,000,000 km3 water on earth.

Three times more water is required, approximately :) - manya (talk) 06:01, 29 July 2009 (UTC)

Apparently there is an entertaining book by someone called Bernard Ramm who tried to do all the flood maths and stuff on Polar bears getting back to the North Pole. James Barr reviews it in his book "Fundamentalism" but the review is short: "much good fun can be had reading Bernard Ramm". --BozMo talk 08:48, 29 July 2009 (UTC)

We don't know how tall the moutnains were at that time, that's the problem. Mount Sinai, as noted by 76., may have been as tall as any mountain was anywhere in the world. there could have been violent movements of the earth as well during that time. (And, I thought I read somewhere where Mt. Everest was getting higher., as the one continental shelf was pushing against another.) Some Christians believe that Pangea existed before the Flood.Somebody or his brother (talk) 13:52, 29 July 2009 (UTC)

Come on folks, Mount Ararat, which is mentioned in the flood story as the first land to emerge, is way higher than Mount Sinai. (Ararat is over 5000 meters, Sinai less than 2300.) Looie496 (talk) 16:19, 29 July 2009 (UTC)

If we introduce the extreme beliefs of some Christians that the continents have moved around dramatically around in the time humans have been on Earth, and that the heights of mountains changed dramatically in that span, then the volume calculation might as well set pi equal to three, since that is the value given by Kings 7:23, where a round font was said to be ten cubits across and thirty cubits around. Edison (talk) 18:31, 29 July 2009 (UTC)

You've touched on a small hobby-horse (hobby-pony?) of mine, Edison. If the figures of ten and thirty are considered to be not precise, but rounded to the nearest whole number, they're not incompatible with the correct value of pi. 87.81.230.195 (talk) 03:01, 30 July 2009 (UTC)

Clearly God manipulated the geological record to test our faith in the Bible and decide who should be saved! (More faith=more gullible=less worthy of immortality.) --99.237.234.104 (talk) 02:37, 30 July 2009 (UTC)

Oblation

While re-reading Kim Stanley Robinson's Red Mars, I encountered the unfamiliar term "oblation:" The sun was...small and round even though it was near setting; there wasn't enough atmosphere for oblation to enlarge and flatten it. I know what he means, but in trying to read about the actual science behind it, I can only find a religious meaning with Google, including our own oblation article. Does this phenomenon have another name, and how do I read more? - Draeco (talk) 04:59, 29 July 2009 (UTC)

From the context I would guess that what he meant is that the atmospheric refraction was insufficient to make the Sun look oblate (meaning here - looking like an ellipse rather than a circle) close to the horizon because the atmosphere was too thin to perceptibly refract the sunlight. --Dr Dima (talk) 05:39, 29 July 2009 (UTC)

We have nice examples of the atmospheric refraction distorting the apparent shape of the sun disk near the horizon in the green flash article. At any rate, the proper name of the phenomenon is "atmospheric refraction" :) . By the way, I wonder if KSR was really right about Mars atmosphere being too thin to do that. It's not just density, it's the density gradient that counts. And, Mars atmosphere being colder than ours, the gradient may be not too low... --Dr Dima (talk) 05:49, 29 July 2009 (UTC)

The speaker was very near the Martian north pole at that time, if that helps justify him. - Draeco (talk) 15:45, 29 July 2009 (UTC)

this is an image of Martian sunset taken by Mars Pathfinder. The colors are real. The color variation is due both to the light scattering on the dust and to the atmospheric refraction. Note that the Sun disk looks distorted (elongated vertically). So yes, Martian atmosphere can distort the apparent shape of the Sun near the horizon. --Dr Dima (talk) 16:18, 29 July 2009 (UTC)

Low pressure and low gravity suggest a weak density gradient; is the cold enough to counter these? —Tamfang (talk) 23:31, 2 August 2009 (UTC)

On the other hand, the first part of the circumstance ("enough atmosphere for oblation to enlarge... it") smacks of the moon illusion. Additionally, our article there notes that flattening occurs in the vertical, not the horizontal. As such, I expect the vertical elongation in Dr Dima's image is a lens artifact or a squashed image, not a true representation. — Lomn 18:18, 29 July 2009 (UTC)

Note: the vertical distortion of the image may also be a function of the three exposures used to create the color image. Any motion of the sun between exposures would cause apparent vertical elongation when combined. — Lomn 18:29, 29 July 2009 (UTC)

Yes, Lomn, you may certainly be right about the vertical elongation being the artifact of the several exposures used to acquire the color image. Indeed, according to jpl.nasa.gov, the imager on Mars Pathfinder used a set of filters on both of its optical channels. That means that it required several consecutive exposures to take real-color images, pausing between exposures to rotate the filter wheels. However, I do not think that the atmospheric refraction always causes a vertical flattening; not even on Earth. Indeed, there is a well known "Etruscan vase" atmospheric effect where the apparent Moon disk is dramatically distorted and elongated vertically. So, lest we -God forbid!- dare bring the Original Research into this Temple of Wiki, we should probably leave the verdict on the question of whether the Martian atmospheric refraction is strong enough as "inconclusive". --Dr Dima (talk) 19:49, 29 July 2009 (UTC)

Lomn, how could elongation of the image due to movement between colour exposures occur without causing coloured edges on the sun? Cuddlyable3 (talk) 20:13, 30 July 2009 (UTC)

Conservation of energy

Let's say we have a collision between two atoms, where energy cannot dissapear in the form of internal kinetic energy or whatnot(let's pretend that, in this particular case, a photon isn't released). What theoretical argument justifies kinetic energy being conserved(in other words, how do we know that the work done on one atom = the negative work done on the other)?

If no energy is dissipated, the collision of fully elastic, i.e. in a frame of reference moving with the barycenter, both will have the same speed after the collision (but different directions). "Negative work" does not make sense here - "work" and energy are undirected scalars. Hence the kinetic energy before and after the collision will also be the same. --Stephan Schulz (talk) 08:25, 29 July 2009 (UTC)

Stephan, the question did not specify that you had to work on the reference frame of the center of mass. Negative work makes perfect sense on a different reference frame. Dauto (talk) 12:18, 29 July 2009 (UTC)

Conservation of energy is shown to hold empirically. Any physical theory is based on a set of axioms, which have to be accepted as true (hopefully because they match experimental evidence), and the rest of the theory is derived from there. Energy conservation I think is usually taken to be an axiom, but there are equivalent ones. Conservation_of_energy#Noether.27s_theorem mentions that time translational invariance of physics implies conservation of energy. Rckrone (talk) 15:45, 29 July 2009 (UTC)

To say that the reason the total kinetic energy remains constant is because energy is conserved, followed by saying that energy is conserved because it's an elastic collision, is circular logic. My question is, why is energy conserved in such a situation? And I would like to hear an explanation other than Noether's theorem, because conservation of energy was well understood before her time. I doubt a derivation using Newton's laws would be hard, and I have a rough sketch of what it would look like in my head (the atoms slow down because of the electric force bewteen them, hence energy is stored as potential energy, which should the push the atoms away with the same work), but the details are lacking.

And I wouldn't consider conservation of energy to be an axiom (at least not in classical theory), but rather Newton's laws to be the 'axioms', because from them, conservation of momentum and the like are able to be explained.

Let me turn your question around (and maybe let you clarify what you're really asking): where do you propose happens if energy is not conserved via work/change-in-velocity during the collision? Either the energy goes "somewhere else" (not into kinetic energy in the particles--conserved, but in some other variable in the system) or it disappears entirely (not conserved at all in the whole closed system). DMacks (talk) 19:30, 29 July 2009 (UTC)

Conservation of energy is not a consequence of Newton's laws. Rckrone (talk) 20:47, 29 July 2009 (UTC)

Are you sure? The interaction bewteen two bodies is mediated by the Coloumb force, and apparently (I remember reading this somewhere) because the Coloumb force can be represented as a gradient of a potential, energy is conserved.

Newton's laws alone only give you conservation of momentum. Coulomb's law together with Newton's laws does imply that energy is conserved when the only force is the electrostatic force. In other words, the electrostatic force as defined by Coulomb's law is a conservative force. (That article also goes into how you would go about proving whether a given force is conservative.) If you had a full description of all the fundamental forces, you could prove from that the conservation of energy generally. However, I don't think there's an accurate self-consistent theory like that right now, since quantum gravity is still a problem. Rckrone (talk) 03:40, 30 July 2009 (UTC)

The article says that a conservative force is one where the work done by it on an object does not depend on the path taken, only the change in position of the object...what I don't see is how this means that, in a hypothetical collision between two atoms, kinetic energy should be conserved.

You're right, you need a few more steps to show that things work out when both objects are accelerating. If you want to assume that only the electrostatic force is in play and that the objects can be approximated by point charges, then the Two-body problem article has what you need. Specifically, the displacement r between the particles follows the equation $\mu {\ddot {\mathbf {r} }}=\mathbf {F} (\mathbf {r} )$, and the total kinetic energy in the center of mass frame can be shown to be ${\frac {1}{2}}\mu {\dot {\mathbf {r} }}^{2}$ where μ is the reduced mass, so it behaves like the case of a single particle in a stationary electric field. Rckrone (talk) 17:28, 30 July 2009 (UTC)

Trees showing the undersides of their leaves before a summer storm

When I was a lil guy, I always heard that you could tell a summer storm was coming if the trees (especially silver maples and tuliptrees) showed the undersides of their leaves. I was thinking about this yesterday when driving home from a job:

1. at one point on the highway, I noted that many of the trees had upturned leaves
2. a few miles up the road, I went through part of a storm
3. past the storm, the leaves were turned up again, then after a few more miles they weren't

...so there's at least some anectdotal evidence from yesterday's commute :-).

I'm wondering though if this really does work, and if so, why? Is it something to do with an updraft? Buildup of electrical charge before a storm? --SB_Johnny | ^talk 09:32, 29 July 2009 (UTC)

We have an article on the weather stick. The web page that article uses as a reference says that the branch is responding to relative humidity.[2] Other web pages claim that it’s barometric pressure.[3] So far, I haven’t found a really reliable source that would give a definitive answer. Red Act (talk) 12:29, 29 July 2009 (UTC)

this link (can only access Google cached version for some reason) goes to a discussion among some university professors who conjecture that it has to do with a shift in winds. this link (warning .pdf!) concurs, saying that the leaves grow according to prevailing winds, and the approach of a storm, being a non-prevailing wind, will turn them over. Some jerk on the Internet (talk) 12:45, 29 July 2009 (UTC)

The experiment reported by this web site referenced by the weather stick article does make it look like it’s the relative humidity that’s affecting the stick. However, the experiment doesn’t control for other possible changes, such as barometric pressure, so it’s not quite conclusive. Red Act (talk) 12:54, 29 July 2009 (UTC)

So I'm getting from this that the "folklore" is true (leaves turned up do indeed mean rain coming), but it's not sure how or why it happens? Is there a name for the effect? Might be nice to start an article on it if there is.

Bummer that there doesn't seem to be any non-sales-oriented material around on the weather sticks. I'll look and see if maybe they're discussed in one of the Foxfire books. --SB_Johnny | ^talk 10:06, 30 July 2009 (UTC)

Erm... has anyone noticed that the answers given so far have nothing to do with trees? A weather stick is a dead branch, that is expected to expand/contract according to humidity. But a living tree is regulating the humidity in its branches, so that it is nearly independent from short-time fluctuations (like a storm). Did you notice that many light, loose things, e.g. tends, will preferentially show their underside before and in a storm? This is mainly an effect of wind. I would think that leaves, being very lightweight, react to the increased wind heralding a storm. Come on, everyone knows that a storm is coming when the wind starts to blow stronger and in gusts. I can't see anything special in the observation that wind turns leaves. Or did you really see the leaves standig still, in an upright position? Or were they "dancing in the wind"? --TheMaster17 (talk) 08:43, 31 July 2009 (UTC)

There’s more going on than just a branch expanding and contracting according to humidity. If that were the case, then a straight-ish branch isolated from the wind wouldn’t be expected to change its shape with variations in atmospheric conditions, due to symmetry. There’s also something more going on than just leaves blowing in the wind. I believe what’s going on is an example of hydronasty, which we unfortunately don’t have an article on, although we do have an article on nastic movement in general. My guess is that in some way the bottoms of the branches are more hydrophilic, and the tops of the branches are more hydrophobic, which causes the unequal swelling to bend the branches. I think the survival advantage of evolving hydronasty in this case is a matter of taking on a branch configuration that maximizes water making it down to the roots when it’s raining, and taking on a branch configuration that maximizes exposure of the leaves or needles to the sun when it’s not raining. Unfortunately, the first 50 Google hits on “hydronasty” don’t turn up a good web page about it, instead just turning up one-sentence definitions of the word, or listing it in a list of nastic movements. Red Act (talk) 10:09, 31 July 2009 (UTC)

It was clear to me that just contraction wouldn't bend the branch, but as branches are not symmetric (they work against gravity, which is really asymmetric), this is easily explained by the distribution of different types of fibres in the dead wood. But my point was: Is there really repositioning of leaves before a storm? Is there a reference available for this behaviour, which did account for wind and indeed did measure the position of the leaves? If it is hydronasty as suggested because of air humidity, how can the tree differentiate between an upcoming thunderstorm and a "normal" humid day? And I must say that I have never observed the phenomenon described in the OP's post, and I have been camping all my life, occasionally watching trees in thunderstorms. I have never read about this in any botany book (and I have studied biology at a university, with botany as one of my major subject). And what do you mean by "maximize the water making it to the roots"? All water comes to the ground (for examples as drops that travel over the leaves, finally dropping). Only a small amount of water sticks to the leaves as they have a hydrophobic surface of waxes, and as rain does seldom fall vertically in a storm, I cannot even see how a certain position would minimise the amount of droplets that remain on the leaves (which would probably anyway be blown of by the wind in a storm, which shakes the drops of the leaves). And as a remark, most trees don't depend on the water that falls just near their stem, they have a big system of roots, either collecting deep underground or in a wide area (or both). So in summary, I'm sceptical to the existence of the described movement, and even if it exists in certain plants that I'm unfamiliar with, I'm rejecting the explanations so far on scientific grounds. I think this question is really in a need of references. --TheMaster17 (talk) 13:26, 31 July 2009 (UTC)

It’s not surprising if you haven’t observed this behavior, since it doesn't occur with most species of trees, and since it’s not a motion that occurs fast enough that you’d notice it if you were to sit there and watch a tree for a couple minutes. As to the exact nature of how this behavior works, or what the survival advantages might be, or even if it’s even a survival advantage vs. just a side effect of other effects, I don’t claim to actually know the answers, which is why I identified my hypotheses with words like “my guess is” and “I think”. But some small branches of some kinds of trees do bend considerably within a short period of time due to atmospheric conditions even when isolated from the wind, as you can see in this video.[4] Or is what you are denying that this branch would exhibit this behavior if it was still attached to the tree it came from? Red Act (talk) 15:34, 31 July 2009 (UTC)

Yes, this is exactly what I am denying. The hydration state of healthy plant tissue is (nearly) independent from short time fluctuations in the environment, as I stated before. The plant has evolved countless tricks to regulate this, because it is so important for all kind of processes (the most important ones being photosynthesis and nutrient transport). And I have not watched trees "a couple of minutes", more like hours, before, during and after every imaginable wheather in middle europe. What I know from this is that trees move in the wind, but I have never seen a branch of a living tree changing its shape because of humidity. Let me state my question again: Why don't the leaves always change their position when humidity changes, but only before a storm? What could be different? My answer is: the wind. And: What kind of advantage would repositioning confer? I didn't get your point when you talked about "rain coming to the roots". I have done experiments with plants(not exactly trees) under different conditions (one of those was humidity), and I have not seen a change of shape, just a change in physiologic rates and processes. I'm perfectly sure that dead wood does bend when you change its hydration state, but what about a video of a living tree, under varying conditions? So far this is only guessing, not giving references. And I have looked for references, and couldn't find a single one in my university textbooks. --TheMaster17 (talk) 11:45, 1 August 2009 (UTC)

While not a paper in an academic journal, nor a scientific study, an article entitled "The Gardner's Weatherman" in the newsletter of the Muskegon County Master Gardner Association, published in cooperation with Michigan State University Extension, discusses the phenomenon:

All the previous plants [list of plants purported to predict weather changes] show a response to increasing

humidity in the atmosphere. As you will recall from previous discussions, the relative humidity usually rises before the onset of rainy weather. Since the internal water pressure of a plant is regulated by the evaporation rate balanced against the rate of water uptake from the soil, any change in the evaporation rate will also affect the internal water pressure. It is then easy to see that when a plant cannot get rid of excess water because of high humidity, the tissues will swell and change their shape, closing flower petals, and straightening stems. When the humidity falls after a storm passes, the plant can get rid of the excess water, and the flowers return to their normal condition.

Silver maple: the silver maple shows its undersides before a rain.... Trees: when the leaves of trees curl during a south wind, it will rain.

Here, the responses are not so easily apparent. The curling of the tree leaves may be due to increased tissue water pressure as above, but it may also be due to a strengthening wind caused by the pressure gradient of a deepening low pressure system approaching the area. Silver maple leaves showing their undersides is definitely due to a wind shift away from the prevailing wind. The tree gets “used” to the wind blowing from one direction, and when the wind shifts ahead of a storm, the leaves will flip onto their

backs, showing their undersides.

Many university horticultural websites describe silver maples' leaves readily turning in even slight winds. It's how the tree gets its common name. Some jerk on the Internet (talk) 17:40, 31 July 2009 (UTC)

This seems to go into the right direction. I don't know why I didn't find this with my google search, but this at least also states, although not in a reliable text, that it is the wind. I wouldn't count this as real reference, because it seems to be a cite itself, coming from a folklore text. And also its understanding of the influence of humidity is flawed: Just because the rate of evaporation changes in high humidity, doesn't mean that the plant is "helplessly" siting around, waiting for its tissues to swell. Almost all "modern" plants can regulate not only their evaporation rate by regulating the size of their stomata, they also actively regulate the uptake of water and nutrients from their roots (btw: what a poor article for this important topic of botany). --TheMaster17 (talk) 11:57, 1 August 2009 (UTC)

Chlorhexidine

Chlorhexidine is widely used in medicine and dentistry as several % solution. What is chlorhexidine itself, is it solid or liquid? What is its appearance? Renaldas Kanarskas (talk) 10:50, 29 July 2009 (UTC)

Our article on Chlorhexidine has many external links to pages at other sites, such as in the infobox. According to this one: [5] from Chemspider, the melting point is 134 Celsius, which would make this a fairly low-melting solid in native state. --Jayron32 11:08, 29 July 2009 (UTC)

It's solid at room temperature. It's "white to pale yellow" in color.[6] Red Act (talk) 11:19, 29 July 2009 (UTC)

Thank you both! Renaldas Kanarskas (talk) 19:59, 29 July 2009 (UTC)

One more question about chlorhexidine. Cant find its solubility in water, anybody knows? Renaldas Kanarskas (talk) 14:58, 1 August 2009 (UTC)

Can't find solubility data either, but based on the structure, it looks like it's slightly soluble in water at neutral pH, and fully soluble at acidic pH. FWiW Just looked up the MSDS, it says chlorhexidine is insoluble in cold water (solubility is only 0.08% w/v). 98.234.126.251 (talk) 18:43, 1 August 2009 (UTC)

This is at 20 C? Renaldas Kanarskas (talk) 12:23, 2 August 2009 (UTC)

That's right. 98.234.126.251 (talk) 05:07, 3 August 2009 (UTC)

Charging an 18V battery for a drill

Hi all, I have two questions about charging up my cordless drill:

• I have an 18V battery for my Black&Decker drill. The drill came with a plug (wall wort and 5-6 mm jack) which, if you you have the battery inserted in the drill and the plug connected to the drill, you can charge the battery. The question is: I may or may not have lost the plug. I have one that has an output of 5.0V and 2.6 A with the right size jack, but am I right in thinking that this can't be the plug, because the transformer would need to supply at least 18V to charge the battery?
• Apart from this one plug (which has to go through the drill, making it annoying), why do all my power tool batteries need to be charged with some stupid expensive shoe like this or this? Why can't they just have a jack that lets them plug into the wall? Are they just trying to make me spend more money?

Thanks! — Sam 76.24.222.22 (talk) 12:25, 29 July 2009 (UTC)

I am not an electronics expert but I would have expected a much more powerful charger, I think that charger would take a long time to charge a battery of even only a few AH. As for the charging stations, the intention is to have 2 batteries, one on charge with the other in the tool for constant usage. 81.144.241.243 (talk) 13:01, 29 July 2009 (UTC)

See if it's not on this page [7]. If not you'll probably have to do a 2 step process here [8] First enter the model number of your cordless drill in the "need a manual" window (unless you are more organized than most and actually still have that) That will tell you what they call the wall wart/battery charger/power adapter /etc. and if you are lucky also a part no. Enter either the part no. or the description in the search window on top of the page (underneath the DeWalt logo). If that doesn't get you anywhere call their customer service [9] (USA 1-888-678-7278). Good luck. 71.236.26.74 (talk) 14:03, 29 July 2009 (UTC)

There does not appear to be standardization of voltage and plugs. The wall wart you have with a 5 volt output would not be able to recharge your 18 volt battery. It is a good practice to get a white paint marker, and write on each adapter what device it works with, because eventually they are bound to get separated. The adapters often just have the name of a factory in China on them and not the brand or model of device they are supposed to power or charge. Edison (talk)

High performance batteries and their chargers have specific voltage/current characteristic and charging time. They are designed together as a system and the manufacturer has to accept liability of overheating, explosion, chemical leakage or damage. This means it is to your advantage to have the charger and battery that the supplier guarantees for use together. It is inconceivable that any honourable company could dream of making a profit from your purchase. Cuddlyable3 (talk) 16:13, 29 July 2009 (UTC)

A "shoe" to hold the drill can be attached to the wall, and it harder to lose than wall warts. Many tools also have removable batteries, so one can be charged while the other is in use. I've even seen a boxed set of power tools that were cheaper because they shared the one battery pack. The "shoe" style connectors may also be stronger in some way than the tiny plugs on wall warts (something to do with resistance in the wires? all I know about electricity is to keep my fingers off it). - KoolerStill (talk) 09:24, 30 July 2009 (UTC)

Calcium Chloride in Damprid?

Is the calcium chloride in Damprid really just regular ole table salt that I can buy in the grocery store? If so, wouldn't it be cheaper for me to buy a bag of salt instead spending a lot of money buying the refills? --Reticuli88 (talk) 13:18, 29 July 2009 (UTC)

Sodium Chloride is "just regular ole table salt", more or less, but no that's not Calcium Chloride (which is nastier). On damp removers the reusable ones are Silica based but have lower capacity. There may be cheaper places to buy Calcium Chloride--BozMo talk 13:26, 29 July 2009 (UTC)

I think you can buy CaCl2 at a hardware store in large bags for use as ice melter. Coolotter88 (talk) 13:36, 29 July 2009 (UTC)

Bunsen Burners and Gas stoves

This has been puzzling me for long.Consider that you are lighting a gas stove or bunsen burner.When you light it,the gas gets ignited and undergoes combustion and hence it burns.The burning process takes place only outside the outlet.What is the mechanism that prevents the fire from spreading inside along the tube and into the gas cylinder and the final KABOOOOOM ???gdsrinivas 13:35, 29 July 2009 (UTC) —Preceding unsigned comment added by Gd iitm (talk • contribs)

probably because the gas pressure inside the pipe is higher than outside so the flames are blown outwards. also, there is no oxygen in the pipe to support combustion so the gas will extinguish any flame that manages to make it inside. Coolotter88 (talk) 13:38, 29 July 2009 (UTC)

In a bunsen burner,there is a hole at the bottom for mixing of air and gas.So it is a mixture of air and gas along the entire length of the bunsen burner.Now how do you explain it??gdsrinivas 13:43, 29 July 2009 (UTC) —Preceding unsigned comment added by Gd iitm (talk • contribs)

"Back burning" can occur in a bunsen burner. No flame shows and it smells bad. I think it can be provoked by throttling back the gas supply. It can be corrected by thumping the rubber supply pipe. Cuddlyable3 (talk) 13:48, 29 July 2009 (UTC)

yeah, probably the gas pressure pushes the flame outwards. When you light the bunsen burner, it's like a jet. Coolotter88 (talk) 13:52, 29 July 2009 (UTC)

There are two features here. Even in a tube of perfectly mixed methane and air, the speed at which a flame propagates into the mixture (the Flame speed) is limited (by memory about 5 cm a second I think) so if the flow speed exceeds this the flame will be being blown away not traveling down the tube. The flame only stays put on a bunsen burner because the double burner edge creates a recirculating eddy: try to light a straight tube of premixed air and you will fail. Gas stoves have similar design features. The second feature to stop this inadvertently happening at low nozzle speed is a Flame trap which is made of wound up nit-mesh: the flame cannot penetrate it because the heat loss to the metal fibres quenches the flame. However do not be fooled into thinking the gas emerges completely premixed; close inspection reveals a rich premixed inner flame (blue, the characteristic colour of Carbon Monoxide flames) and a CO rich diffusion flame outside it. Clever chap Bunsen. --BozMo talk 13:57, 29 July 2009 (UTC)

Your description of Back burning would indicate a cheap version of a Bunsen with no flame trap I guess. --BozMo talk 14:00, 29 July 2009 (UTC)

Or an old one! Most Bunsens don't easily do back burning anyway, but they certainly did when run on coal gas, as I remember well, when I started work. Not that we have Bunsen burners in the lab any more (sigh)  Ronhjones  ^(Talk) 19:14, 30 July 2009 (UTC)

Thats a nice question.But when I asked one of my friends he said it was one of the important concepts in fluid dynamics.Any info based on this???thanks117.193.144.206 (talk) 16:29, 29 July 2009 (UTC)

As an add on to the question why does the flame occur only exactly at the top of the bunsen burner opening???Why not it occur a few centimetres above the bunsen tube or a few centimetres inside the tube???117.193.144.206 (talk) 16:36, 29 July 2009 (UTC)

BozMo please explain this :The flame only stays put on a bunsen burner because the double burner edge creates a recirculating eddy.117.193.144.206 (talk) 16:38, 29 July 2009 (UTC)

I am not very good at explaining things. Flames are just reaction fronts which eat into flammable gas mixtures. To ignite the gas coming rapidly out of a pipe end you need an ignition source where the gas comes out (or with very fast turbulent flow ignition further downstream). Otherwise the flame will blow itself out. For a burner this could be a pilot light by the pipe end but if you can induce an eddy by the pipe end (a recirculating piece of flow) then the gas which is already burning keeps the flame going by reintroducing a piece of flame into the flow. This is the purpose of the double skin on a Bunsen. Unfortunately the Wikipedia article on Bunsen burners is not very good apparently because there isn't a very good reliable source on Bunsen's. However I have written a few peer reviewed papers on gas combustion [10] [11] and I do have some idea what I am talking about. --BozMo talk 19:42, 29 July 2009 (UTC)

If you understand flow dynamics I should add that a stagnation point is sometimes sufficient but not as stable and that just having a thick tube wall works at some flow rates. However if you get a pair of pliers and squeeze the two walls of a Bunsen together (so that you just have a circular pipe end) the flame will blow itself out. --BozMo talk 19:50, 29 July 2009 (UTC)

A really cool fact I heard which I think is relevant here, if you were in a pure methane atmosphere and had a burner connected to a bottle of oxygen, it would essentially behave the same way as the other way around here on earth. Vespine (talk) 05:03, 30 July 2009 (UTC)

Assuming the gas is leaving the 'tube' at room temperature, it has to be heated by the existing flame until it reaches 'flash point' or ignition temperature. For propane, that's 920-1020°F. While that's happening, the gas keeps moving. Twang (talk) 06:19, 30 July 2009 (UTC)

Yes, good way to look at it although technically ignition by gas inside a premixed cloud is not just by heat, but by a combination of heat and a load of free radicals from the ongoing combustion process (the radicals diffuse forward just as the heat does), which makes the temperature needed a little lower. --BozMo talk 06:45, 30 July 2009 (UTC)

Hey guys this question hasnt yet been answered..thought you didnt notice this " Thats a nice question.But when I asked one of my friends he said it was one of the important concepts in fluid dynamics.Any info based on this???As an add on to the question why does the flame occur only exactly at the top of the bunsen burner opening???Why not it occur a few centimetres above the bunsen tube or a few centimetres inside the tube???" —Preceding unsigned comment added by 117.193.137.0 (talk) 07:03, 30 July 2009 (UTC)

I think that the flame don't occur in the tube cause the gas flow blows it out of the tube faster than the flame front travels back into the tube (as BozMo said earlier). As for it not occurring above the tube, that's because the flow pattern recirculates part of the burning gas-air mixture back to the burner opening, so it ignites the rest of the mixture just as it comes out of the tube. FWiW 98.234.126.251 (talk) 07:20, 30 July 2009 (UTC)

By adjusting the fuel flow and air-holes, one can usually get the flame to partially "detach" from the rim of a bunsen burner, or even entirely "lift off" of it. If you have supplies higher than what the burner is designed to handle (perhaps mis-adjusted regulator or flow valve, or using a pressurized-air or oxygen source on a burner designed for simple air-holes), you can easily cause the flame to "blow out". Googling "bunsen burner" along with any of the other quoted terms I used can get you some fluid-dynamics and other studies related to these phenomena (and conversely, the conditions required for a stable flame). DMacks (talk) 07:27, 30 July 2009 (UTC)

Sorry I should add a strong warning not to try using pressurized oxygen as a source unless you really know what you are doing, he means a pressurized air source: changing to an oxygen cylinder or other oxygen increases the flame speed considerably and you may get something nasty happening (including conceivably gas detonation). As for the rest, I thought it was all answered. It is all determined by flow. Flow does tend to recirculate but not steadily and not necessarily at the right speed. A stable flame requires a stable re-feeding of burning gas into unburnt gas. The rest is all said...--BozMo talk 08:04, 30 July 2009 (UTC)

Concur on the safety issue. The ideas of flame detachment, burn-back, etc aren't limited to bunsen burners and others that use a simple inspirator/venturi-tube to get the oxidant. The same questions (and flame stability issues) apply to torches with compressed oxidant feeds (and are rated to handle them!). The oxidizer and fuel gases mix and then flow through a few inches of pipe before the tip where the flame is. DMacks (talk) 16:15, 30 July 2009 (UTC)

Volume to pressure ratio?

How can I know the volume of a gas, knowing the capacity of its container (50 L), the amount of moles (64,48) and the pressure (1,621,200 Pa), but not knowing the temperature? Thank you. --83.56.187.237 (talk) 19:58, 29 July 2009 (UTC)

You just defined the volume as 50 L. Temperature is the unknown, as you surmised, and can be calculated easily (assuming an ideal gas). — Lomn 20:02, 29 July 2009 (UTC)

Thank you, what a blunder! Therefore, what should I substitute for $R$ in the following state equation in order to determine the temperature?

$1,621,200\cdot 50=64,48\cdot R\cdot T\,\!$

--83.56.187.237 (talk) 20:20, 29 July 2009 (UTC)

R is the gas constant. Just make sure you use the right value for the units you're using. Rckrone (talk) 20:26, 29 July 2009 (UTC)

Precisely. I'm probably stupid, but I don't know which is the right value I should use. --83.56.187.237 (talk) 21:04, 29 July 2009 (UTC)

Well I'm not going to do your homework but you need to select the gas constant from the list that uses the same units as your problem. The closest one is in m3 Pa K−1 mol−1. Since you have your volume in L instead of cubic meters, you need to convert that 50L into cubic meters. Then you can use the R constant from the list at the article and it will work (because it's all in consistent units). TastyCakes (talk) 21:16, 29 July 2009 (UTC)

First I would recommend leaving the units in your equation. That makes everything much easier. It's not important that you choose R so that all the units cancel out right away. You can use unit conversion where necessary to get things into the form you need. So for example if you choose R = 8.314 472 m³ Pa K⁻¹ mol^-1, you'll be left with L on one side of the equation and m³ on the other, but we know that m³ = 1000 L. Any of the values for R on that page will work, but some will just require more unit conversions than others. Rckrone (talk) 21:44, 29 July 2009 (UTC)

Damps in coal mines

Reading Firedamp, I'm curious — using current technology, would it be possible/likely for coal mining companies to collect these gasses in mines and sell them for fuel? Nyttend (talk) 20:10, 29 July 2009 (UTC)

See Coalbed methane. That seems to be the general idea there. Rckrone (talk) 20:38, 29 July 2009 (UTC)

(EC)The concentration of methane in a coal mine might vary significantly over time and might be too low to run an internal combustion engine, but methane recovered from landfills also varies. Firedamp does not say what the methane concentration is as commonly found. It might be higher in rooms of abandoned mines than in working mines, so some mines might work better than others as possible sources. See Biogas. The solution with gas from landfills is to mix it with sufficient natural gas to properly run an engine such as a modified diesel to generate electricity, if the concentration is too low. The mix can be automatically adjusted as the concentration varies. Edison (talk) 20:40, 29 July 2009 (UTC)

Flufenamic acid

Our article on flufenamic acid, or fenamic acids in general, lists them as NSAIDs. However, I am not at all sure they inhibit COX. Do they? What flufenamic acid actually does is to block the Calcium-sensitive nonselective cationic current in neurons. What does that have to do with the anti-inflammatory action?! Any input is welcome. Thanks in advance, --Dr Dima (talk) 20:31, 29 July 2009 (UTC)

A Google Scholar for "flufenamic acid NSAIDs" gets a bunch of hits, including PMID 10866999 which seems to be relevant. Looie496 (talk) 23:06, 29 July 2009 (UTC)

Thanks Looie. So it does affect both COX expression and COX activity (as well as cation currents, Ca ion balance, and who knows what else...). It figures. Thanks. --Dr Dima (talk) 23:37, 29 July 2009 (UTC)

Dragons of Eden updates

Has Carl Sagan's The Dragons of Eden ever been updated since it was first published in 1977? NeonMerlin 21:02, 29 July 2009 (UTC)

It doesn't look like it. There was a paperback edition published in April of 1978 and a 'mass market' paperback in 1986 - but from what I can see of them in Google books and Amazon - they contain the same text as the 1977 version. Neither of them say anything about being revised - and there is only one author's copyright date. SteveBaker (talk) 21:31, 29 July 2009 (UTC)

Too bad - it's a great book, even though it's over 30 years old. — QuantumEleven 14:48, 30 July 2009 (UTC)

Eurasian plate

is Eurasian plate (western part) moving north or it is moving south in altitude. Eurasian is a large plate. The east plate is China, Japan, Phillipines, surrounding by Yellow Sea. Part of east half of Eurasian plate is moving southeast. Africa is suppose to collide with Europe, closing all the oceans nearby.--69.228.145.50 (talk) 21:19, 29 July 2009 (UTC)

I don't know if this source is valid. The Future Is Wild said Australia is moving north only, no showing of it moving south again. i'm confuse if Europe will be moivng north or south.--69.228.145.50 (talk) 21:07, 30 July 2009 (UTC)

I have watch The Future Is Wild in my classroom once, this is why I don't think Australia will ever move south again. This would help if you give me some suggestions.--69.228.145.50 (talk) 21:10, 30 July 2009 (UTC)

Help idenfying the following yard plants

The first is some kind of berry bush that has some kind of little (Chinese) red jack-o-lantern type fruit.

The second is uncut and thinned out (crab?) grass that when watered every day and cut every week grows very dense to make a lawn.

The third has fronds with very thin and tough stems.

All three are located in coastal mid Florida but no one here can identify the first two without the berries and the watered and cut grass roots.

No one here has a guess what the third one is either except that it might also be native to China. -- Taxa (talk) 23:21, 29 July 2009 (UTC)

For number 2, it looks like good-old, garden-variety fescue. It doesn't look like crabgrass, which is more "stalky" than "leafy". Most other lawn grasses in the southern U.S. are low-creeping grasses which don't grow bushy like this; such as Centipede grass, Bermuda grass, St. Augustine grass. I'm not sure which fescue variety it is, but based on your description and location, unless you are watering the heck out of it, it's probably not tall fescue which goes dormant during the dry summer months, but I would not rule that out. A few good thunderstorms and my tall fescue springs back to life. --Jayron32 04:45, 30 July 2009 (UTC)

You don't show the #1's fruit, but could it be Euonymus americanus aka "Hearts-a-bustin" aka "Strawberry bush" (not a real strawberry)? The give away would be the distinctive fruit, which look better in this google search, and the green stems. --Jayron32 04:50, 30 July 2009 (UTC)

Thanks for taking a shot... The grass has a seed bearing stalk that is about knee to waist high with only two branches like a sling shot. The fruit is completely round with a very sour taste unless you get one that's just dropped or is about to drop. They are kind of like little pumpkins only red and with soft skin and innards (when ripe). The ribs are much more pronounced than a pumpkin's. -- Taxa (talk) 06:41, 30 July 2009 (UTC)

Well, I guess I was wrong about the first one being hearts-a-bustin, but #2 still sounds like a fescue variety, of which there are hundreds. Fescue puts out small seeds on knee-high stalks with 2-3 branches lined with the seeds. I'm still going with fescue there. --Jayron32 18:27, 30 July 2009 (UTC)

The third one looks like Cyperus alternifolius.--Shantavira|^{feed me} 08:09, 30 July 2009 (UTC)

The third plant is a sort of Lady Palm, likely in the genus Rhapis.CalamusFortis 22:47, 31 July 2009 (UTC)

Drugs to make me smart

Can I get drugs to make me more intelligent to pass my exams? —Preceding unsigned comment added by 79.75.104.10 (talk) 23:31, 29 July 2009 (UTC)

Not really, no. There are many nootropic drugs, of various legal status, but none of them will really make a difference if you haven't studied for the exams. --Dr Dima (talk) 23:47, 29 July 2009 (UTC)

Sure they can. Try caffeine, preferably as coffee, which will also prevent dehydration. —Preceding unsigned comment added by 84.187.91.197 (talk • contribs) 00:12, 30 July 2009

I'm not sure if the OP is expecting to become somehow become more knowledgeable after eating the drug, as in the case where he/she doesn't study. Rather, I think the question is whether there are any drugs which will make your brain more active and sharper for a limited period of time, much like a steroid does to the body. Such a drug would be highly useful in a competitive exam, where speed and accuracy are essentially more important than knowledge. So I'm interested in knowing some answers, too. Rkr1991 ^{(Wanna chat?)} 00:59, 30 July 2009 (UTC)

There definitely are such drugs, amphetamine is the most widely used. Unfortunately it is illegal and highly addictive. Looie496 (talk) 01:21, 30 July 2009 (UTC)

In my experience, smart drugs seem to make me just smart enough to realise they're not having any effect. Adambrowne666 (talk) 01:38, 30 July 2009 (UTC)

If such drugs existed, were legal, and could be sold without a prescription - you can be 100% sure that you'd know it because you'd be bombarded with adverts for them every time you turned on the TV. Instead, the best we get are exceedingly dubious "food supplements" that can claim just about anything without any proof whatsoever because they fall through a convenient loophole in the law. I agree that caffeine is really the only one with any kind of a legitimate track record. It doesn't make you more intelligent though - it just ensures that you're awake. But even then, you have to be quite careful because if you take too much of the stuff, you'll become super-nervous and jittery - and that's just as bad as being half asleep! IN terms of not-legal-without-a-prescription, I've heard that some people take beta blockers because (as our article says) "Beta blockers protect against social anxiety: Improvement of physical symptoms has been demonstrated with beta-blockers such as propranolol; however, these effects are limited to the social anxiety experienced in performance situations. (example: an inexperienced symphony soloist)" - which can leave a person with exam anxiety feeling calm and relaxed. Not more intelligent - but perhaps calm enough to be able to better use the intelligence you already have. But these things can be dangerous - and taking them without a doctor's advice would be pretty amazingly stupid. SteveBaker (talk) 03:57, 30 July 2009 (UTC)

During the past 5 years or so there has been a large upsurge in the number of "smart drugs" that are sold in schools. The studies I have seen on these drugs say they range from ineffective to ineffective and dangerous. Even caffeine has debated impact on grades. While it would help you stay up the night before to study and help you stay aware during the exam itself, it lacks the ability to give you the mental agility provided by a good nights sleep. Using caffeine to stay up will not help with any exam that takes abstract thought, though could help with other types of exams. Anythingapplied (talk) 04:11, 30 July 2009 (UTC)

:::Polyjuice and then Steve could sit the exam for you (assuming he is cleverer than you for which no warranty is available under the Wikipedia Disclaimer). --BozMo talk 07:42, 30 July 2009 (UTC)

Would Modafinil help me be smarter in class and in my exams? —Preceding unsigned comment added by 79.75.68.48 (talk) 00:10, 31 July 2009 (UTC)

It might help a little if you've been crammin' all night long or if you've recently been sick with the swine flu (or if you're so high on crack that you can't think properly), but other than that, prob'ly not. It can also have some pretty bad side effects if you take it without a prescription. FWiW 98.234.126.251 (talk) 05:11, 31 July 2009 (UTC)

See State-dependent learning (and also Cue-dependent forgetting), attention span, and study skills. Forget the drugs and instead focus on how best to study. This will have a much greater impact on your test scores than any "wake-up" drug or nootropic. 152.16.16.75 (talk) 09:38, 3 August 2009 (UTC)