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February 18

A greenhouse on Mars

If you put a large air-tight greenhouse on Mars and fuilled it with air, how warm would it get? Would earth-plants grow in the Martian "soil"? Maybe with some human 'compost' mixed in? 78.146.206.38 (talk) 00:00, 18 February 2010 (UTC)

To answer the first part, there is no theoretical barrier to designing a solar greenhouse on Mars that operated at comfortable temperatures of plant life. Dragons flight (talk) 00:14, 18 February 2010 (UTC)

The temperature would depend on the latitude. Climate of Mars#Temperature indicates that temperatures do get up to nice warm temperatures (27 °C max), presumably on the equator. With the (literal) greenhouse effect, the temperature inside your greenhouse would be higher than the surroundings, so anywhere reasonably close to the equator should be fine temperature-wise. The Martian soil might be suitable for the growth of Earth-plants - it seems from that article that more research is required. --Tango (talk) 00:18, 18 February 2010 (UTC)

The temperature would drop quite low, especially at night in the winter, unless there was sufficient thermal mass and a high insulation level. Mars gets on average only 43% the solar intensity or "insolation" received on Earth per [1]. The mean surface temperature (outside the greenhouse) is only -63C (per the NASA site, varying with season and latitude). The air pressure is very low, so a greenhouse dome would have to be quite strong to hold in enough airpressure for earth type plants. See also a NASA project looking at a Mars greenhouse. Apparently 1/4 of earth normal pressure would suffice. Actually sounds doable. The Mars Society also has some suggestions how to build the greenhouse. Edison (talk) 00:21, 18 February 2010 (UTC)

As that Mars Society link says, you can keep it warm with what is essentially an enhanced greenhouse effect - they suggest a silver compound in the plastic that will allow visible light and near-IR (ie. sunlight) through but stop far-IR from getting out. --Tango (talk) 00:45, 18 February 2010 (UTC)

a little help with equilibrium constants and partition coefficients as they relate to solubility

Admittedly, the whole "mole product / mole reactant" thing goes over my head when I imagine scenarios like adding more reactant or taking away more product. So ... like take these titration results.

There's approx 0.013 mmol of dissolved iodine (without iodide) in 50 mL solution.

When I add 5 mL cyclohexane, approximately 0.008 mmol of it escapes into the cyclohexane, resulting in a concentration-in-cyclohexane figure of 0.0016 M. From the new concentrations I calculate a partition coefficient of approximately 16.

When I add 8 mL cyclohexane, despite a 60% increase in the amount of lipophilic solvent there's only a 25% increase in extraction ... Indeed, the concentration of iodine in cyclohexane is now 0.00125 M, a concentration fall. This sort of makes sense since as the iodine gets less-concentrated in the water phase, iodine becomes harder and harder to extract from aqueous solution. The concentration in aqueous phase is now 5.6 * 10^-5 M and the new partition coefficient is 22.

This is clearly experimental error right? Shouldn't the partition coefficients remain roughly the same? (We measured concentrations by titration.) Let's take an ideal case with no experimental error and say I'm adding more water or more cyclohexane. How would I use the partition coefficient to predict new concentrations? What does the partition coefficient really mean, as an equilibrium constant? How likely is it that mistitration or something like that is the source of my error? Basically, I don't know how to think of "mol product over mol reactant" when I say add more of one type of solvent to the mixture. Are the concentrations going to rearrange themselves such that the ratios of concentrations in the different solvents will somehow remain constant.

I get even more lost when I deal with saturated solutions and there's apparently an equilibrium constant between the pure solid phase and a dissolved phase as a solute, because if I add more excess solute to a saturated solution, clearly the ratios cannot adjust themselves to the equilibrium constant since the solution is already saturated. John Riemann Soong (talk) 01:34, 18 February 2010 (UTC)

To address your last question, you need to remember that the concentration of a pure solid substance (or even a pure phase of a liquid) has a constant concentration. TenOfAllTrades(talk) 04:03, 18 February 2010 (UTC)

Coming back to the first part of your question, what's the precision with which you're measuring the amount of iodine (to start with, and in each phase)? If that 0.008 mmol iodine is ± 0.001 mmol (for example), then what is the range of partition coefficients that you could calculate (based on 0.007 or 0.009 mmol in the cyclohexane phase)? TenOfAllTrades(talk) 13:38, 18 February 2010 (UTC)

Angular momentum in collisions

The angular momentum of any object can be divided into the angular momentum of it's center of mass, and it's angular momentum with respect to the center of mass. When analyzing collisions, the total angular momentum is said to be conserved. However, sometimes both the angular momentum of it's center of mass and it's angular momentum with respect to the center of mass are considered, but other times only the angular momentum with respect to the COM is used. Why would there be this discrepancy? And when finding the rotational kinetic energy, are both types of angular momentum used? Thanks. 173.179.59.66 (talk) 04:24, 18 February 2010 (UTC)

I think the OP really means ITS center of mass and ITS angular momentum, without apostrophes in the ITS. Cuddlyable3 (talk) 12:50, 18 February 2010 (UTC)

Angular momentum will be separately conserved around any point. Often you only need to consider it around one point to get the information you need. --Tango (talk) 18:02, 18 February 2010 (UTC)

I'm sorry, I don't see what mean... —Precedingunsigned comment added by 173.179.59.66 (talk) 20:38, 18 February 2010 (UTC)

Angular momentum around point P will be conserved and angular momentum around point Q will be conserved and around point R, etc. etc. You often only need to apply one of those conservation laws to get the answer you are looking for. Sometimes it will be easier to use P, sometimes Q. That's why you see different ones used in different problems - you just use whatever is easiest. --Tango (talk) 21:08, 18 February 2010 (UTC)

Hmmm, I guess my question wasn't well worded. Let's say two billiard balls are heading towards each other and collide (with some impact parameter). When doing all the calculations, the angular momentum due to their pure rotation (ie due to w=V/r) is used, while the angular momentum due to their net velocities are ignored. However, in a problem where a superball is bouncing off a wall, both the objects rotation and net velocity are taken into accound when detailing how the angular momentum is conserved. Why would we look at both in one, and only look at rotation in another? If I'm not being clear, let me know. Oh, and he talk about rotational kinetic energy, is it just equal to (1/2)I*w^2, or does the angular momentum of the center of mass need to be considered as well?

Ok, I don't understand that either. The rotation of an individual ball about its centre of mass won't be conserved during such a collision, because it isn't a closed system (the ball interacts with the other ball). --Tango (talk) 01:03, 19 February 2010 (UTC)

But if the rotation of one ball slows down, the other will speed up, keeping angular momentum conserved, right? —Preceding unsigned comment added by 173.179.59.66 (talk) 03:48, 19 February 2010 (UTC)

To be strictly correct, the 'orbital' angular momentum always should be included along with the 'rotational' angular momentum, but under certain circustances one or the other might be considered negligible.Dauto (talk) 15:40, 19 February 2010 (UTC)

Math in physics

Out of curiosity, do most physicists (outside of ultra theoretical fields like string theory) spend time manipulating equations and such by hand, or is basically all the math handled by computers. The reason I'm asking is that I'm sifting through a textbook on quantum mechanics, and it's evident that to make any progress, physicists back then had to be able to work out equations and model situations by pen and paper...is that still true today? —Preceding unsigned comment added by 173.179.59.66 (talk) 04:40, 18 February 2010 (UTC)

Yes, even experimental physicists still need to keep the pen and paper at arms length. But computers play a increasingly indispensable role on more advanced calculations. Dauto (talk) 05:04, 18 February 2010 (UTC)

To turn the (usually continuous) models of classical physics into discrete computer-solvable models which have the same properties as the original continuous model (conserved quantities, stability, ...) is mathematics in itself. In short: you need to use mathematics in order to get the computer to do the mathematics you want it to do. http://en.wikipedia.org/wiki/Numerical_methods —Preceding unsigned comment added by 157.193.173.205 (talk) 10:57, 18 February 2010 (UTC)

There is really three parts to this - there is the business of arithmetic - plugging actual numbers into the equations to get answers in practical, numeric form - and for that, a simple calculator will sometimes suffice - but computers are better for repetitive stuff. Another aspect is taking raw data and performing statistics or fitting equations to those numbers (for which computers are pretty indispensable). But the other part is in the development of the equations themselves. Computers are making inroads into that too - we have 'symbolic math' packages that can manipulate equations symbolically - but it still takes a human mind to spot some of the subtle changes that can change a sheet of paper covered with hieroglyphics into something small, elegant and memorable. SteveBaker (talk) 15:52, 18 February 2010 (UTC)

In my experience, "it depends whether the physicist knows how to tell a computer to solve their problem." That depends on the individual physicist's level of interest and formal training with computers. I hang around mostly with "numerical physicists", which is sort of techno-jargon for "computer programmer". The types of problems we know how to solve with computers reach pretty far and wide. Categorically, we use more computational power than a theoretical physicist. For example, this last week I have been solving the wave equation for residuals in a numerical inversion scheme. Now, when a theoretical physicist "solves the wave equation," they write down some variables on paper and call it "solved." When I "solve the wave equation," that means that I arrange the equation to the simplest form that I can use to represent. Then I formalize an algorithm, and write computer code to represent that scheme. When the computer "solves the wave equation," it reads input data, performs calculations on that data following my instructions (that hopefully represent some physical process), and spits out numerical output data representing a model of experimental observations. So in some sense, both me and the computer are doing math - but I save the redundant calculations for the computer, and the mathematical formalism for myself. We blur the line as far as where the mathematics is actually being "done," and me and my team of powerful computers really solve the wave equation together.

Obviously, the simplest case to consider is basic arithmetic. I'm mathematically inclined, but even I have limits - so when I need to determine a value like (36 + (50*2)/1024)/4, it's a waste of my time to do that in my head or on paper. A computer gives me the correct answer, the first time, as quickly as I can type that in. But in my brain, I have done math to estimate the acceptable range of answers I expect. In that case, the computer has pretty much done most of the math.

Non-arithmetic calculation is a little harder to type into a computer. If I want to know the instantaneous phase of a function or a scale-factor for a fourier transform, I often have to decide whether it's easier to use a symbolic algebra system, a numerical approximation, or a paper-based analytic solution. It depends on the problem. But I'm a physicist, not a "number-cruncher" - so it'd be a mischaracterization to say that I spend all day typing out equations and hitting "enter." So let me diverge a little and explain a bit about what a physicist like me actually does.

People come to physicists with quantitative problems that they would like to answer. In my particular field, they come to us with field-recorded geophysical survey data, and ask us to generate images of the Earth from it. We want to perform this process faster and better, creating clear pictures even if the source data is crappy. So, I look at the physics that represents the field-data collection; I model the physical phenomena, observe the effects of unknowns and interferences, and use those insights to design an algorithm to convert input data into output data, while preserving the physical constraints that we know apply.

I would categorize the design of algorithms as a subcategory of "mathematics." Now, whether we design a particular algorithm with a computer or not depends on its complexity. Having some training in formal software engineering, I find UML diagrams to be a great way to set forth a large numerical physics scheme - especially since I like modular code. So, I can use a CAD tool to help me draw out the mathematical operations I plan to do. But, I also keep a stack of blank paper at my desk to scratch work on - diagrams are easier drawn by hand than by mouse. When I have to do geometry, I do it on paper with a pencil. When I need the value of a tangent, I get that answer from a computer (or desk calculator). When I want to do a coordinate transform for an integral kernel, I often use a computer algebra system, but more often than not, I need to do it by hand anyway. In this way, I am "doing math" - I am applying the structure and formalism of analysis to solve a physics problem. The overwhelming majority of this stage of work is by hand and in my brain.

By the time I have formalized my problem, I inevitably write it out as a series of program statements, in the form of a standalone subroutine, a full-blown application program, or a short script for simple arithmetic - all depending on the scale of the problem. The repetitive calculations are performed by machine - and sometimes, intelligent mathematical decision-making is programmed in as well. I would say that almost everybody (physicist or not) knows how to perform arithmetic by computer; almost all physicists know how to perform algebra and calculus by computer; and many specialized physicists and mathematicians (and others) know how to perform matrix mathematics, optimization (mathematics), and so on.

One of the turning points in formal physics education is being able to distinguish the subtle, qualitative difference between "math" and "physics." In other words, when a physicist sees a new form of the wave equation, they aren't looking at the values of the constants - they're looking at the qualitative interactions and relationships between components. (Ask a real physicist whether Schroedinger's equation, which defines a wave-function, is actually a wave equation! The way they approach that question will astound you). And if you sit in on enough physics seminars, you'll inevitably hear some stodgy old guy grumbling something to the tune of "That's all great, but where's the physics!?" What they mean by this is that despite a load of impressive mathematical maneuvering or experimental observation, the presenter has not identified any qualitative physical principle. In the same way, when you ask whether a physicist manipulates math by hand or computer, they can really do either - whether their physics needs a numerical result will modulate the way that they interact with their computers. Nimur (talk) 22:35, 18 February 2010 (UTC)

Great, thanks for the detailed response. PS Is the Schroedinger's equation a wave equation? —Preceding unsigned comment added by 173.179.59.66 (talk) 00:03, 19 February 2010 (UTC)

Yes. See the article Schroedinger's equation. Please sign your posts. Cuddlyable3 (talk) 17:02, 19 February 2010 (UTC)

Actually it is a diffusion equation ([2], [3]), because the time derivative is first-order and complex (a complex parabolic partial differential equation); while a wave equation is a hyperbolic partial differential equation). Ultimately, you can define a "wave equation" a lot of ways - but most commonly, the defining factor is whether you can construct an invariant of the form (x_i +/- v*t) - in other words, propagation - which cannot be done for Schroedinger's equation! It's the defining state-equation for the wave function - but it is not a wave equation! To some extent, this is semantics and a matter of definition - but the idea is that we care about the physics that the mathematics represents - in other words, the algebraic term that represents wave propagation is decidedly absent in the Schroedinger solution. This means a lot of things - there is no effective velocity to propagate perturbations of the wave-function. This has implication to quantum entanglement, because in the absence of a term to define a velocity, technically there is no speed limit on the propagation of quantum information - hence the paradox of faster-than-light propagation of quantum entanglement information! So, by playing with the maths analytically, we can try to peel away at the actual physics implications. Nimur (talk) 18:14, 19 February 2010 (UTC)

Your explanation for quantum entanglement is fishy. The Schroedinger equation is a non-relativistic approximation. A fully relativistic wave equation such as the klein-gordon equation is a hyperbolic equation with a speed limit associated to it and yet(!) quantum entaglement exists. Dauto (talk) 05:07, 21 February 2010 (UTC)

Of course, there's always room for a more complicated model. Klein-Gordon_equation#Derivation explains the extension/conversion to the relativistic model. Anyway, I'm not really sure how to use the Klein-Gordon model, but as I understand it, it does not correctly solve for the observables in a simple case such as a hydrogen atom. It seems that the Klein-Gordon model only works for certain spinless particles, pions. (I really haven't ever used it). As far as I know, the Dirac Equation is the standard form, relativistic-corrected wavefunction equation for atomic physics - and it too is a diffusion equation. Nimur (talk) 06:13, 21 February 2010 (UTC)

Nuh-uh. Dirac's equation is not a parabolic equation. Don't let the first order time derivative fool you. Parabolic equations are second order differential equations while the Dirac's equation is a first order differential equation. It can be easily shown that every solution of the Dirac's equation also is a solution of a second order differential equation but it turns out to be a hyperbolic equation. That's not surprising at all because it is the limited speed characteristic of hyperbolic equations that make them consistent with relativistic speed limit. Schroedinger's equation is not a relativistic equation so it is free to violate that principle.

Starting with Dirac's equation

$\left(\beta mc^{2}-i\hbar \sum _{k=1}^{3}\alpha _{k}\nabla _{k}\,c\right)\psi (\mathbf {x} ,t)=i\hbar {\frac {\partial \psi (\mathbf {x} ,t)}{\partial t}}$

And making use of $\beta ^{2}=1\,$, $\{\alpha _{i},\alpha _{j}\}=2\delta _{ij}$, and $\{\alpha _{i},\beta \}=0$ we can find an expression for $(i\hbar )^{2}{\frac {\partial ^{2}\psi (\mathbf {x} ,t)}{\partial t^{2}}}$

$(i\hbar )^{2}{\frac {\partial ^{2}\psi (\mathbf {x} ,t)}{\partial t^{2}}}=(i\hbar ){\frac {\partial }{\partial t}}(i\hbar ){\frac {\partial \psi (\mathbf {x} ,t)}{\partial t}}=(i\hbar ){\frac {\partial }{\partial t}}\left(\beta mc^{2}-i\hbar \sum _{k=1}^{3}\alpha _{k}\nabla _{k}\,c\right)\psi (\mathbf {x} ,t)$

$=\left(\beta mc^{2}-i\hbar \sum _{k=1}^{3}\alpha _{k}\nabla _{k}\,c\right)(i\hbar ){\frac {\partial }{\partial t}}\psi (\mathbf {x} ,t)$

$=\left(\beta mc^{2}-i\hbar \sum _{k=1}^{3}\alpha _{k}\nabla _{k}\,c\right)\left(\beta mc^{2}-i\hbar \sum _{l=1}^{3}\alpha _{l}\nabla _{l}\,c\right)\psi (\mathbf {x} ,t)$

$=\left(\beta ^{2}m^{2}c^{4}-i\hbar \sum _{k=1}^{3}\{\alpha _{k},\beta \}\nabla _{k}m^{2}c^{3}+(-i\hbar )^{2}{\frac {1}{2}}\sum _{k=1}^{3}\sum _{l=1}^{3}\{\alpha _{k},\alpha _{l}\}\nabla _{k}\nabla _{l}c^{2}\right)\psi (\mathbf {x} ,t)$

$=\left(m^{2}c^{4}+(-i\hbar )^{2}\sum _{k=1}^{3}\sum _{l=1}^{3}\delta _{kl}\nabla _{k}\nabla _{l}c^{2}\right)\psi (\mathbf {x} ,t)$

$=\left(m^{2}c^{4}+(-i\hbar )^{2}c^{2}\nabla ^{2}\right)\psi (\mathbf {x} ,t)$

so

${\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}\psi -\nabla ^{2}\psi +{\frac {m^{2}c^{2}}{\hbar ^{2}}}\psi =0.$

Which is the Klein-Gordon equation!!

Dauto (talk) 22:21, 21 February 2010 (UTC)

Hydrogen bond position transition

I was trying to understand the File:Glycine-zwitterion-2D-skeletal.png variant of Glycine but then I realized that File:Glycerin Skelett.svg might be subject to the same hydrogen bond position transition.

What reasons are there to believe that glycerin hydrogen bonds are stationary? 99.60.3.241 (talk) 05:02, 18 February 2010 (UTC)

In glycine (the neutral form), there is an "acid" part (a hydrogen can be released easily) and a "base" part (has a high affinity for free hydrogen). Acids and bases react with each other pretty well, and the result is the zwitterionic form. Glycerin does not have any part that is particularly acidic or basic, so it does not change to an alternate hydrogen attachment pattern. DMacks (talk) 18:25, 18 February 2010 (UTC)

Black hole

Hello i have read article on black hole but i do not understand how in some documentary space time is shown to be warped so much that it is such one layer is underneath or over lap with another layer such that a large enough distortion of gravity though to another place in space time that is space can be warped by mass but how can it tunnel be created or what causes space to warp in such that the infinite steep sides of a black hole gravity impression comes out on space time instead of just going forever (Dr hursday (talk) 06:29, 18 February 2010 (UTC))

It sounds like an Einstein-Rosen bridge. Check out that article and see if it answers your questions.  :) Mac Davis (talk) 06:54, 18 February 2010 (UTC)

Hello yes this is what I am talking about but i do not understand how the "U" curve at the left of this picture http://en.wikipedia.org/wiki/File:Worm3.jpg occurs. what causes this? (Dr hursday (talk) 07:01, 18 February 2010 (UTC))

The U is only there because of the way the image is drawn. The same thing is happening in this image (imagine the plane "above" the wormhole extending on forever instead of only in two spots). The only difference is the way it was drawn. The U is there because two places in spacetime are connected by a jump through a higher dimension. If you were on any part of the "U" you would not notice any bends and it would appear perfectly "flat." Mac Davis (talk) 07:43, 18 February 2010 (UTC)

The "higher dimension" has no physical relevance, it may be useful to "visualize" the situation but it need not "exist" in any physical sense. The last paragraph in http://home.fnal.gov/~skent/cosmo/cosmo3.pdf makes this point too. —Preceding unsigned comment added by 157.193.173.205 (talk) 08:28, 18 February 2010 (UTC)

Strange bug

Omg what is this. Can someone tell me? --✶♏‭ݣ 06:31, 18 February 2010 (UTC)

Belostomatidae. -- kainaw™ 06:35, 18 February 2010 (UTC)

Ew ew ew. But thank you. Ew. *shudders* --✶♏‭ݣ 06:37, 18 February 2010 (UTC)

Sometimes fear of the unknown is the scariest thing of all!!! What the Jesus God Hell ever happened to curiosity. 86.4.186.107 (talk) 06:58, 18 February 2010 (UTC)

Given that the sting of the Belostomatidae is considered "one of the most painful that can be inflicted by any insect" and thus is beyond the highest on the Schmidt Sting Pain Index (which is limited to Hymenoptera) "4.0+ Bullet ant: Pure, intense, brilliant pain. Like fire-walking over flaming charcoal with a 3-inch rusty nail in your heel" perhaps Belostomatidae should be rated "5.0+ Jesus God Hell that hurts!" 58.147.58.28 (talk) 08:33, 18 February 2010 (UTC)

Why the profanity? Did your question get a better or quicker answer because of it? Just curious. Kingsfold (talk) 14:51, 18 February 2010 (UTC)

Did the profanity ofend you?Dauto (talk) 15:27, 18 February 2010 (UTC)

No. The answer was quick because all it took was a simple web search. Instead of trying to make a highly juvenile joke by seeing how profane I could be, I went to http://tineye.com and pasted in the URL of the photo. It showed multiple results. The second one was to a Russian site that had a link below the photo right back to the Wikipedia page for the bug. Then, all I had to do was put a link to the article here. I didn't complain because many people seem to prefer to ask questions instead of searching for themselves. -- kainaw™ 14:58, 18 February 2010 (UTC)

Thanks for that link. I didn't know that site. Dauto (talk) 15:27, 18 February 2010 (UTC)

There's a Firefox extension that lets you just right-click on an image and search for it in Tineye. Unfortunately, Tineye seems to have a very tiny index. I wish Google would just buy them and do it right. --Sean 16:10, 18 February 2010 (UTC)

I think it's obvious that the OP was humorously incorporating the sort of reaction one might make upon seeing this beast into his/her question. "Jesus", "God", and "Hell" seem pretty mild profanities for such a sight. --Sean 16:13, 18 February 2010 (UTC)

If you think that's bad, you should meet my Italian girlfriend. Imagine Reason (talk) 16:59, 18 February 2010 (UTC)

I hope you're talking about the use of colorful language and not about the Belostomatidae! (I also hope she doesn't read this page!) SteveBaker (talk) 20:17, 18 February 2010 (UTC)

I'll point out, just because nobody else yet has, that the stuff on the back is a mass of eggs. Looie496 (talk) 00:42, 19 February 2010 (UTC)

Although Ferrofluid smeared on the back of a magnetic beetle would appear similar. I have changed the question title for clearer reference. Cuddlyable3 (talk) 16:52, 19 February 2010 (UTC)

I agree with that but for clarification for any future readers, the profanity discussed above was largely in the title [4] Nil Einne (talk) 22:44, 19 February 2010 (UTC)

What would happen to photon

Hello if I shine a flash light out into space in such a direction that it never encounters anything what happens to the photon over time? (Dr hursday (talk) 06:55, 18 February 2010 (UTC))

It keeps going and will get redder due to the metric expansion of space according to Hubble's law. Mac Davis (talk) 07:20, 18 February 2010 (UTC)

If the photon gets redder where does the energy go? Ariel. (talk) 07:44, 18 February 2010 (UTC)

Nowhere. Hubble flow implies that the farther away one looks, the faster the local matter is moving away from you. By extension, if one travels to those distant places, then you have to subtract the effect of the average local velocity when considering your motion. Hence the farther the photon travels the more it will appear doppler shifted with respect to the local standard of rest. Dragons flight (talk) 08:15, 18 February 2010 (UTC)

You may be concerned that energy does not seem to be conserved in this scenario. That is correct. In general relativity, energy is not conserved. There's a new blog post at Cosmic Variance today explaining this. -- Coneslayer (talk) 17:21, 22 February 2010 (UTC)

Put another way, the photon appears redder because whoever is observing it happens to be moving away from us. Someone moving towards us would see a bluer photon. I think this effect is indistinguishable from that of the space itself between the objects having expanded.. EverGreg (talk) 09:22, 18 February 2010 (UTC)

It's not only indistinguishable. It is the same thing. Dauto (talk) 13:32, 18 February 2010 (UTC)

No. The redshift due to the expansion of space depends on how much space has expanded during the photon's free flight. The usual picture of this is that it's the light-wave that's stretched along with the space it travels in. These are distinctions with real consequences. Consider for instance that distant objects are more red-shifted than nearby ones, so their apparent speed is greater, but an object does not feel acceleration as it recedes from us in this way. Redshift due to expansion also allows an object to speed away from us faster than the speed of light, which is impossible with doppler shift. (Yeah, I read up on the redshift article :-P) EverGreg (talk) 19:32, 18 February 2010 (UTC)

This difference you are pointing out is just an illusion. All redshift derives from the same principle no matter whether it is a gravitational redshift, doppler redshift or cosmological expansion redshift. In fact those distinctions are not as relevant as they sound since what is a doppler redshift for an specific choice of coordinates will be a gravitational redshift for a different choice of coordinates and vice-versa. The cosmological expansion redshift is an artifact from our choice of coordinates. specifically our choice of using comoving coordinates. Dauto (talk) 20:49, 18 February 2010 (UTC)

You cannot do a Lorentz transformation such that an object is accelerating in one choice of coordinates but not in another. EverGreg (talk) 09:13, 19 February 2010 (UTC)

Yes, true but

• A) Lorentz transformations explain the bulk of the effect. See that[5] paper for an explanation on how to build an expanding universe without matter or cosmological constant.
• B) There is no reason to restrict yourself to Lorentz transformations - they do not hold any special status within General Relativity.

Dauto (talk) 15:21, 19 February 2010 (UTC)

All of this fancy color shift stuff only happens from the point of view of the observer at the source of the light. From the point of view of the photon - nothing whatever happens - it just keeps on going. If space is infinite (we're not 100% sure of that) then it'll keep going forever completely unchanged. If space is finite then perhaps it 'wraps around' and comes back towards you from the opposite direction - but it's still completely unchanged. Photons can't "degrade" over time because they are travelling at the speed of light - and for them, the whole of eternity passes by in zero time. If they don't hit something (which seems highly implausible), then nothing can change because for them, time isn't advancing at all. SteveBaker (talk) 15:43, 18 February 2010 (UTC)

If by "never encounters anything" you mean your beam of light misses large objects like stars, galaxies and space rocks, its fate will likely be extinction in the interstellar medium. --Sean 16:19, 18 February 2010 (UTC)

Perhaps its information content will be absorbed into the Omega point. Graeme Bartlett (talk) 03:24, 19 February 2010 (UTC)

OK, I know I'm an ignoranimus about physics, but the OP stated "it never encounters anything". If you observe it, hasn't it encountered something, namely your eyeball? ←Baseball Bugs ^{What's up, Doc?} carrots→ 05:40, 19 February 2010 (UTC)

Not sure, but I suspect all photons are only produced with a matching absorbtion elsewhere.114.75.18.3 (talk) 06:48, 19 February 2010 (UTC)

Not sure what that means. In any case, the earlier discussion seemed to be confusing photons with galaxies. Obviously, galaxies can produce gazillions of photons, and those photons can appear red or blue depending on whether the galaxy is approaching or receding. But talking about photons that way implies that photons are ejecting other photons that would be visible without seeing the original photon. Maybe I'm wrong, but I don't think it works that way. ←Baseball Bugs ^{What's up, Doc?} carrots→ 06:54, 19 February 2010 (UTC)

How much sodium in 1g of Baking powder?

How much sodium is there in 1g of baking powder, after it has been used to make a cake for example? Thanks 89.243.151.96 (talk) 14:50, 18 February 2010 (UTC)

this suggests that there are 520mg per teaspoon. (In that brand, anyway.) I don't know how many teaspoons are in a gram of baking powder, though. Sorry. APL (talk) 15:29, 18 February 2010 (UTC)

It says on your picture that 1/8 teaspoon = 0.6 grams, so about 1/5 teaspoon per gram. --Sean 16:24, 18 February 2010 (UTC)

Most baking powdersoda is "sodium hydrogen carbonate" which is NaHCO₃ - so there is one atom of sodium, one hydrogen, one carbon and three oxygen. You have to figure out the atomic weight of each atom - add them up and then you know the ratio of sodium to the rest of the elements. Then you can figure out the fraction of a gram that is sodium. According to List of elements: Roughly, Hydrogen is 1, Carbon is 12, Oxygen is 16 and Sodium is 23. So one molecule of this compound has a total atomic weight of 23+1+12+3x16 = 84. So for every 84 grams of baking soda, 23 grams is sodium which means that about 27.4% of this stuff is sodium. There is therefore 0.274g of sodium in every gram of sodium hydrogen carbonate - if your brand of baking powder mixes that with some other 'stuff' (which is possible, for example to stop it clumping) then the answer will be different - check on the ingredients list on the packet. SteveBaker (talk) 15:35, 18 February 2010 (UTC)

Steve, that's baking soda. Baking powder is different. -- Flyguy649 ^talk 15:50, 18 February 2010 (UTC)

Baking powder says "Most commercially-available baking powders are made up of an alkaline component (typically baking soda), one or more acid salts, and an inert starch". So I guess we need to account for the other stuff too. SteveBaker (talk) 17:22, 18 February 2010 (UTC)

(EC)It would depend on the brand. In addition to the sodium bicarbonate (baking soda) and acid, like cream of tartar, baking powders contain various amounts of cornstarch. The acids used seem mostly not to contain sodium. So the answer is in how much baking soda is in baking powder. While this article gives some suggestions on how to make your own baking powder, it uses teaspoon measurements (not weight) for the ingredients. That article gives ratios ranging from 1 1/4:1 to 2:1 cream of tartar:baking soda. You should be able to figure out a rough answer by using the densities of each reagent. -- Flyguy649 ^talk 15:42, 18 February 2010 (UTC)

(edit conflict) Baking powder will certainly be mixed with other stuff; that's what makes it different from baking soda. Most baking powder is sodium hydrogen carbonate as well as an acid, like cream of tartar or monocalcium phosphate. They also typically contain an inert filler, like cornstarch. Look on the ingredient label, for both the serving size (hopefully they give a value in grams) and the amount of sodium per serving. Divide the second by the first to find grams sodium (or probably milligrams sodium) per gram powder. Buddy431 (talk) 15:49, 18 February 2010 (UTC)

This particular brand, for example, appears to contain 65 mg of sodium per 0.6 g serving, which is 108 mg sodium per gram baking powder. Buddy431 (talk) 15:53, 18 February 2010 (UTC)

The brand in my kitchen (Magic baking powder)has 45 mg of sodium per 0.6g (1/8 tsp) serving or 75 mg per gram of baking powder. This plus Buddy431's contribution above shows that there is a huge variation by brand. -- Flyguy649 ^talk 15:59, 18 February 2010 (UTC)

Yes and given all this, I think it's clear that unless the OP provides the precise brand and name of baking powder, we can't answer the question. It would surely be easier for the OP to just look on the label which is probably on his/her product anyway Nil Einne (talk) 16:38, 18 February 2010 (UTC)

None of the brands of baking powder I have in my possession or have seen in supermarkets have mentioned how much sodium they have in them. I presume they are trying to hide that they have a lot. 89.243.197.22 (talk) 15:19, 21 February 2010 (UTC)

File:refraction.jpg

what should be the minimum distance of the sun from the horizon so as to enable the observer to see it's image? —Preceding unsigned comment added by 117.201.65.74 (talk) 16:37, 18 February 2010 (UTC)

I converted your header into a proper subject header with the image as a link Nil Einne (talk) 16:38, 18 February 2010 (UTC)

Even only a part of the Sun need be above the sea horizon for its image to be visible. Cuddlyable3 (talk) 16:47, 19 February 2010 (UTC)

Monarch Butterflies

Where do Monarch butterflies go in the winter? —Preceding unsigned comment added by Dredfern (talk • contribs) 17:10, 18 February 2010 (UTC)

They migrate. "The most famous Lepidopteran migration is that of the Monarch butterfly which migrates from southern Canada to wintering sites in central Mexico. In late winter/early spring, the adult monarchs leave the Transvolcanic mountain range in Mexico for a more northern climate. Mating occurs and the females begin seeking out milkweed to lay their eggs, usually first in northern Mexico and southern Texas. The caterpillars hatch and develop into adults that move north, where more offspring can go as far as Central Canada until next migratory cycle." --Mr.98 (talk) 17:19, 18 February 2010 (UTC)

Addiction

Hi,

I'm very worried. In fact I think I'm addicted. It's a highly dangerous drug; an overdose can lead to death and the substance constitutes 98% of all cancer cells. So, should I be going to Waterholics Anonymous? —Preceding unsigned comment added by 86.150.210.228 (talk) 18:04, 18 February 2010 (UTC)

It's not as funny if you call it water. Try Dihydrogen monoxide - that sounds really scary. --Tango (talk) 18:06, 18 February 2010 (UTC)

Oh my God, you're not inhaling it, are you? AlexHOUSE (talk) 18:54, 18 February 2010 (UTC)

Think this is bad? I'm far more concerned about oxygen being a mutagen. Regards, --—Cyclonenim | Chat  21:39, 18 February 2010 (UTC)

Ah, but the antidote is red wine - have enough of that, and you won't be concerned at all! --Tango (talk) 23:05, 18 February 2010 (UTC)

I hate wine. I'm f**ked, but I guess I could just start on those vitamins... Regards, --—Cyclonenim | Chat  23:52, 18 February 2010 (UTC)

Bear in mind the kind you take intravenously is usually cut with salt. AlmostReadytoFly (talk) 09:57, 19 February 2010 (UTC)

waters not a drug —Preceding unsigned comment added by 67.246.254.35 (talk) 05:56, 19 February 2010 (UTC)

Fish reproduce in it (as per W.C. Fields). ←Baseball Bugs ^{What's up, Doc?} carrots→ 06:34, 19 February 2010 (UTC)

On the contrary, you could very broadly define water as a drug. It can alter bodily functions. If you're dehydrated, it's a drug which can relieve symptoms. Regards, --—Cyclonenim | Chat  11:28, 19 February 2010 (UTC)

Water intoxication is real. DMacks (talk) 11:38, 19 February 2010 (UTC)

Twin Cities

What is the most twin city to New Bedford, MA in Europe, geographically and climate-wise? Note I am not referring to sister cities. --Reticuli88 (talk) 18:49, 18 February 2010 (UTC)

The article about New Bedford does not give any info about its climate. I'm not sure if places having the same Hardiness zone would feel as if they had the same climate from a human point of view, since the seasonal daylight, summer temperatures, and rainfall may differ. What hardiness zone is it please? 92.24.96.55 (talk) 22:21, 18 February 2010 (UTC)

I think it is 6a. --Reticuli88 (talk) 22:51, 18 February 2010 (UTC)

Assuming New Bedford has a Hardiness Zone of 6 and a Heat Zone of 4, then the places listed in Europe with the same figures are Bratislava Slovakia, and Vienna Austria. They are both inland. Kaliningrad in Russia is more coastal, but it is only heat zone 2 and I think its latitude is higher so the seasonal daylight will be different. Do not know about the rainfall. 89.240.61.50 (talk) 23:51, 18 February 2010 (UTC)

Of course there is nothing close to a perfect match, but in terms of setting, population, and climate my choice would be La Rochelle, France. Looie496 (talk) 00:33, 19 February 2010 (UTC)

The climate would be much warmer in the winter than it would be in New Bedford, and probably cooler in summer. I assumed that the "geographical" similarity specified by the OP did not include population. 78.146.181.195 (talk) 00:49, 19 February 2010 (UTC)

Thanks everyone. Just wanted to know if I traveled to Europe today from New Bedford, MA, what city would seem like I never left MA - meaning the landscape (hills n such) and weather-wise. --Reticuli88 (talk) 13:20, 19 February 2010 (UTC)

WAG here, but maybe Sheffield? --TammyMoet (talk) 15:14, 19 February 2010 (UTC)

No, that would be one of the least similar places in all of Europe. 78.147.225.78 (talk) 20:36, 19 February 2010 (UTC)

If you are willing to accept somewhere with a warmer winter and a cooler summer then you have much more choice. With this in mind, if you want to limit yourself to Britain, then Nairn in Scotland would be a possibility, but with a smaller population. Inverness has a similar population. But even in Scotland, the urbanisation of the surrounding area is probably going to be much more than I expect it is around New Bedford. 78.147.225.78 (talk) 20:43, 19 February 2010 (UTC)

It's really unlikely that you'll find a place with similar weather in Europe. New Bedford has that vast continental landmass off to the west and only ocean to the east. Nowhere in Europe has that much continental mass behind it and such as there is tends to be to the east - not to the west. All of that results in very different climates in Europe and the USA. Also, the prevailing ocean currents are quite different. New Bedford gets the gulf stream bringing warm water up from the south. There isn't an equivalent thing in Europe. SteveBaker (talk) 01:01, 20 February 2010 (UTC)

I'm shocked - wernt you brought up in Britain? I thought every schoolboy knew about the Gulf Stream coming across the Atlantic and keeping Britain, plus Scandinavia and western Europe, much warmer than it would otherwise be for its latitude. When I was at school it was called the Gulf Stream, now people call it the North Atlantic Drift. Where the OP is does I believe have what meteorologists call a Continental climate, while Britain for example is a maritime climate (Oceanic climate) with less extremes of temperature. Given that climate difference, plus the different latitudes, and the far greater population density, makes it as people have said impossible to match. 78.149.241.220 (talk) 16:54, 20 February 2010 (UTC)

Not quite all true. As the Gulf Stream article to which you linked says, the Gulf Stream proper does indeed warm all of the USA's eastern seaboard, and the North Atlantic Drift is not a recent synonym for it, but an offshoot from it. 87.81.230.195 (talk) 00:40, 21 February 2010 (UTC)

You've got the wrong end of the stick. I was referring to the North Atlantic Drift which was, in a classroom in the UK many years ago, referred to as the Gulf Stream. 78.146.167.216 (talk) 01:44, 21 February 2010 (UTC)

Freezing Point Depression and Boiling Point Elevation

After reading one of the above posts I came across this picture. I understand what the unbroken segments of the lines are. But how can one understand the dashed segments of the lines? For example on the right hand side of where the black "solid" line crosses the dark blue "Liquid (pure solution)" line? •• Fly by Night (talk) 22:55, 18 February 2010 (UTC)

It appears that the dashed lines are merely the continuation of the trends for the chemical potential versus temperature. That is, if the solid didn't melt at its melting point, its chemical potential vs. temperature curve would follow the dashed black line. I guess that the dashed lines are just in there to make the intersections more clear. Buddy431 (talk) 04:49, 19 February 2010 (UTC)

February 19

Altered taste buds after toothbrushing

Why does milk and orange juice taste so terrible after brushing? --70.167.58.6 (talk) 02:56, 19 February 2010 (UTC)

The Sodium lauryl sulfate in the toothpaste supresses the "sweetness" sensors in your tongue and decomposes phospholipids which normally inhibit your "sour" tastebuds. Net result, the usually sweet and subtly tart orange flavor becomes totally unsweet and super-sour. I don't know why milk would taste bad - I wasn't really aware that it did...but if you're right then it's probably the same kind of reason. SteveBaker (talk) 03:08, 19 February 2010 (UTC)

You can buy toothpaste without the Sodium laureth sulfate foaming agent. EDIT, Sodium lauryl sulfate is different to Sodium laureth sulfate and probably not found in many toothpastes... 188.221.55.165 (talk) 13:38, 19 February 2010 (UTC)

That makes me wonder... on a similar topic... Is there a toothpaste that enhances taste of "healthy" food? For example, they could throw some miracle fruit juice in the toothpaste. -- kainaw™ 04:21, 19 February 2010 (UTC)

You're supposed to brush after you eat, not vice-versa. APL (talk) 05:51, 19 February 2010 (UTC)

That supposes the purpose is to remove food particles. Ew. I brush before breakfast to remove the layer of plaque that builds up overnight, so the sugars in my breakfast have nothing to stick to. If I were brushing after eating, I would wait half an hour to allow my mouth pH to return to normal, as otherwise the teeth are softer and you can end up (over cumulative brushings over the course the years) thinning them, gradually brushing off the hard layer. Since I don't have half an hour to spare after breakfast, and the thought of brushing bits of food out in the froth makes me gag, I brush before eating. 86.182.38.255 (talk) 15:56, 19 February 2010 (UTC)

Hmm... That's pretty complicated. I had no idea that by skipping breakfast I was saving myself so much mental effort. APL (talk) 18:17, 19 February 2010 (UTC)

We're on the Science desk, and you think that's complicated? Maybe I should have just said "You're supposed to brush before you eat, not vice-versa." as if I had access to some ultimate truth, then left people to suppose that it is only cultural with no advantages or disadvantages either way. 86.176.185.157 (talk) 12:40, 20 February 2010 (UTC)

Toothpaste that enhances the taste of "healthy" food is called mayonnaise. --Dr Dima (talk) 10:06, 19 February 2010 (UTC)

Whether teeth are brushed before or after breakfast depends primarily on a cultural difference. And not eating breakfast simply makes one fat. ~AH1^(TCU) 00:01, 20 February 2010 (UTC)

In concert with Steve's response above, orange juice contains sugars as well as naringin, a bitter component of the peel and other parts of the orange. After sodium lauryl sulfate (essentially a detergent) destroys one's sweet-detecting taste cells, the only taste of the orange juice that can be detected is the bitter naringin. DRosenbach ^{(Talk | Contribs)} 13:56, 19 February 2010 (UTC)

"destroys" is a rather strong term for what this detergent does. It doesn't damage the cells - it merely inhibits them in some way. If it destroyed them it would be days to weeks before you'd get your sweet taste sense back rather than tens of minutes. SteveBaker (talk) 00:57, 20 February 2010 (UTC)

Actually, it destroys the cells. It does not take days-to-weeks because the subjacent progenitor cells are called upon and are very soon available to replace them. DRosenbach ^{(Talk | Contribs)} 03:08, 24 February 2010 (UTC)

Integrating Partition Functions

When computing the partition function or doing other such calculations in statistical thermodynamics, one needs to sum (integrate) over all the possible microstates. How do you determine the appropriate variable over which to do the integration? An example to clarify: if one of the degrees of freedom of your system is a rotation around a molecule's bond, one possibility is to integrate over the dihedral angle. However, that degree of freedom can be quantified by any number of other equivalent formulations (e.g. something based on, say, the dot-product of the normal of the plane containing atoms A-B-C and that containing B-C-D). How can we tell what's the appropriate variable to use in the integration, and what is the property of that measure that makes it the appropriate one to use? -- 174.21.247.23 (talk) 04:20, 19 February 2010 (UTC)

For the outcome, it doesn't matter which representation you choose; all of them are quevalent by definition. Of course, choosing a different representation can make the integral easier to solve. But there no general rules which one to choose. --baszoetekouw (talk) 10:05, 19 February 2010 (UTC)

I believe it does matter which one you choose. I'll give an example: You have two different way to parametrize a single degree of freedom, x and α, related by the expression α = x², each of which varies between 0 and 1. Say your energy function is E = (1 - x²)/β = (1 - α)/β. The two different formulations for the partition function give

$Z(x)=\int _{0}^{1}\!e^{-\beta E(x)}\,dx=\int _{0}^{1}\!e^{-(1-x^{2})}\,dx\approx 0.538,$

$Z(\alpha )=\int _{0}^{1}\!e^{-\beta E(\alpha )}\,d\alpha =\int _{0}^{1}\!e^{-(1-\alpha )}\,d\alpha \approx 0.632,$

So in this case (and I believe most other cases, especially when the two are not linearly related) the two formulations do not give equivalent results. My question was, how do I know which formulation is the correct one to use? That is, which is the "natural" variable over which to do the integration, and how do we know which one it is? -- 174.21.247.23 (talk) 16:41, 19 February 2010 (UTC)

In general, the correct way of looking at the problem is actually:

$Z(x)=\int \!e^{-\beta E(x)}\rho (x)\,dx$

Where ρ(x) is the density of states in the neighborhood of x. For many practical problems, we tend to engineer the variable of integration such that ρ(x) = 1. In a sense that gives you a preferred choice for the variable of integration, but it is not a required choice. In particular, one can change as in a regular integration be realizing that ${\rho (x) \over \rho (\alpha )}={\partial \alpha \over \partial x}$.

By definition, each set of allowed quantum numbers will contribute exactly one term to the partition function sum. In general, this means one can find the ρ(x) = 1 formulation by starting with an explicit sum (or often a multiple sum, with one sum for each quantum number), and then when one extends it into the continuum limit the integrand for ρ(x) = 1 will have the same form as your summand. I'd suggest that a large part of the problem you are having in deciding how to do the integration is that you probably haven't figured out how your system is quantized. Being able to enumerate the quantum states is a necessary precondition to writing out a partition function. Dragons flight (talk) 18:12, 19 February 2010 (UTC)

How does one go about the quantum to classical conversion for complex systems such as large molecules? Going back to my original example of a rotation around a bond, is there a straight-forward way of determining how ρ(x) is constructed? "Enumerating quantum states" is all well and good for something like ethane, but for larger systems like erythromycin (or erythropoietin) it begins to break down. -- 174.21.247.23 (talk) 04:48, 20 February 2010 (UTC)

transformers

i was reading about transformers and came to know that it can easily defy ohm's law ie "V is directly proportional to I". i have a QUESTION what will happen if a appliances is rated to be used at 220V, and let us assume it needs 10A-- is given 220V and 5A? will it work perfectly or just slowly? since no appliance is rated abot current i am unable to conclude any thing. { we got a source of 110V and 10A, using a step up transformer we get 220V and 5A } <<<a request please dont start asking ur questions in this page, many a times it had happened to me, my question is interrupted by someone else>>>--Myownid420 (talk) 07:40, 19 February 2010 (UTC)

You'll blow a fuse. The current-rating on an appliance says how much current it takes at the given voltage. That is, if you push with a certain force (voltage), that's how much/rapidly the electricity flows (current) through its inner workings (consider Ohm's Law for the fixed resistance of the appliance). Your circuit is pushing with a certain force, which will cause that amount of current to flow. But your circuit is limited to supplying less than that. So either there is not enough energy in the appliance for it to do [whatever], and the failure will depend on how it uses the energy, or else your circuit will try to keep up with the flow the appliance is using, and overheat. Current output is a maximum-available, not a constant amount--as you draw more on the transformer secondary, the primary draws more from the mains. If you do not connect the secondary to anything, the primary is drawing nearly zero, not "the rated supply current". DMacks (talk) 08:08, 19 February 2010 (UTC)

Usually the voltage V and power P in watts are specified for a mains appliance. Given these two it is unnecessary to specify the current I in amps because I=P/V. The supply is just a voltage, such as 220V or 110V. The current depends (mostly) on the resistance R of the appliance. Here Ohm's law I=V/R is useful. With one proviso, any appliance rated for 220V can be connected to any 220V supply and any appliance rated for 110V can be connected to any 110V supply. The proviso is that the supply is able to deliver the current the appliance demands because if not, something in the supply will burn up, hopefully just a fuse. Your step-up transformer will work if it has high enough power rating (watts) for the appliance. Thank you for your question which was clearly written. Sometimes we need to ask questions to help us give better answers. Cuddlyable3 (talk) 16:21, 19 February 2010 (UTC)

Transformers are generally given a primary and secondary voltage rating and an amp or voltampere rating on the secondary side. They have some internal impedance to the flow of electricity,so if no current flows to the load, the secondary voltage may be higher than nominal. If more than the specified current is drawn out, the secondary voltage may drop lower than specified. How much current is drawn out of the secondary is up to the connected load, not to the transformer. It is not a pump of electricity. Under a very high load or a dead short, the current flow would be limited mostly by the transformer impedance (and to a slight extent by the impedance of the source feeding the primary) and the transformer might overheat and fail if a primary or secondary fuse or breaker did not blow. The primary fusing often is just to protect against a short in the transformer, and would let the transformer continue feeding a short on the secondary for a long time. I have seen 12kv primary, 480v secondary transformers continue to feed an arcing short on the secondary for over a half hour, providing enough energy to incinerate a metal enclosed switchgear. Edison (talk) 03:18, 20 February 2010 (UTC)

And, of course, your appliance may be fibbing. It may only draw 2 amps in normal operation. But talking about a household power supply: if you plug in a nice simple appliance, say a 220V 10A kettle, into a 110V supply via an appropriate transformer you would either trip a circuit breaker or melt something - hopefully a fuse. --203.22.236.14 (talk) 11:07, 21 February 2010 (UTC)

The above assumes that the 110V supply is unable to deliver the 20A that the transformer+kettle needs. I don't see the relevance of the "your appliance may be fibbing" remark because supplies must be able to deliver the indicated power (current) of an appliance, whether that is required continuously or occasionally. Cuddlyable3 (talk) 00:00, 22 February 2010 (UTC)

The specified 220V 10 A kettle described by 203.22.236.14 would have a resistance of 22 ohms and would use 2200 watts when connected to 220 volts. If connected to 110 volts (lower than standard U.S 120 volt supply), via a transformer of negligible impedance which stepped the 110 up to 220, it still supply 10 amps at the 220 connection to the kettle, but would draw 20 amps from the 110 volt outlet. If the kitchen had a 20 amp circuit for that outlet, with no other load, the breaker would not trip immediately, but U.S. electric codes generally call for circuits and breakers to carry only 80% of their nominal rating, which would be only 16 amps or 1760 watts at the stated 110 volts. (At 120 volts, a more normal U.S. voltage, 16 amps would be 1920 watts, or 97% of the heating effect seen from 220 volts.) If some appliance only drew 5 amps at 220 volts, as the OP stated, then (if a resistive load) it would have a resistance of 44 ohms. The appliance determines how much current is drawn. The outlet or transformer does not imperiously and ruthlessly force a certain number of amperes through the appliance regardless of its resistance. Likewise there can be no stingy outlet or transformer which sees an appliance designed to draw 10 amps at 220 volts, and somehow presents it with 22o volts but only allows it to draw 5 amperes. See Ohm's Law. Edison (talk) 05:56, 22 February 2010 (UTC)

This "no stingy transformer" rule is accurate with the modification (already explained above by Edison) that overloading a transformer will usually lower the output voltage, thus reducing the power drawn, but this is usually a small effect. The transformer will overheat, and this could be dangerous, so you are strongly advised not to use an under-rated transformer. If the appliance is purely resistive (like a kettle) then it can safely be connected to a lower voltage without a transformer, but it will heat much more slowly. (It would be very dangerous, of course, to connect a 110v appliance to a 220v supply.) Dbfirs 09:27, 23 February 2010 (UTC)

washingg labware

how is labware like beakers washed when it had strong acids like 100 % hydrofluoric acid —Preceding unsigned comment added by 67.246.254.35 (talk) 11:14, 19 February 2010 (UTC)

also eye dropers —Preceding unsigned comment added by 67.246.254.35 (talk) 11:20, 19 February 2010 (UTC)

A glass beaker would not contain 100% hydrofluoric acid for at least two reasons. DMacks (talk) 11:36, 19 February 2010 (UTC)

At least, not for long! One of my lecturers used a dilute HF solution when he was washing out all his glassware as an undergrad back in the day just to get the stubborn marks off. I don't think you could get away with that now. Also, for 67: I can't see why you'd use any special procedures; several water rinses then a squirt of acetone would probably do it. Brammers (talk) 14:24, 19 February 2010 (UTC) (Edited 14:26, 19 February 2010 (UTC))

No, first you have to safely neutralize the remaining HF. HF is only handled in fume hoods nowadays but you would wash it 3 times with dilute basic solution and store that in a waste acid container. Then you would wash with polar and nonpolar solvents and scrubbrush depending on what you were cleaning and store that rinsate in a organic (or solvent) waste container. Then you would wash with water (which rinsate may need to be stored in an aqueous waste container). Finally (at least in my college research lab) a 24-hour bath in dilute acid, followed by a 24-hour bath in dilute base, then overnight in a drying oven and you're done. Unless you are doing anayltical chemistry where you may then need to wash it three times in triple distilled dionized water before drying it. (Ah, the memories of washing dishes. The only time I have ever needed to use a fire extinguisher.) 75.41.110.200 (talk) 15:26, 19 February 2010 (UTC)

If you have an acid-waste container, then put your acid waste in it and let the waste-collection folks handle that. Few quick rinses with then water, and you're done. Well you were done long ago, because the HF (which cannot be made 100% in a beaker) would have dissolved the glass away. If poured out promptly, the HF would have merely etched the glass to become fluorosilicic acid, one of the fluorinating agents used in public water supplies. Repeated acid and base washes are waaaay overkill unless you actually need something that really clean. And anyway unless you use deionized water you're still going to bake out a bunch of ions and get an unpredictably-reactive glass surface, but for many purposes, it doesn't matter--triple-water-rinse and it's clean enough for non-analytical/non-biochemical purposes. Organic chemists just use a squirt of acetone and say "good enough":) A totally impossible scenario without detail/context, no way to know "how clean" the glass needs to be. DMacks (talk) 18:38, 19 February 2010 (UTC)

Ah right, thanks for the insight. The organic labs are the only ones where we have to wash our own glassware, so I'd assumed the acetone squirt was standard practice. Brammers (talk) 01:35, 20 February 2010 (UTC)

I loved lab sinks with a ring of Litmus paper around the drain. Edison (talk) 03:20, 20 February 2010 (UTC)

Years ago I heard a chemical-safety talk, involving the usual "nothing non-neutral down the drain, our waste-stream is monitored for pH outside a narrow range". Talked all about how one night there was a acidic spike, as usual blame went to the chem labs, rather than the cafeteria kitchen where someone had actually just poured out a bottle of vinegar, but the overall campus flow was low enough that it didn't get diluted to the legal level. DMacks (talk) 06:32, 21 February 2010 (UTC)

Apollo Missions

I think I have a decent grasp of most of the stages involved in the Apollo missions. However what I don't get is how NASA was so confident they could dock the orbiter and the lander AFTER the surface mission. To properly dock two untested components (how could they really test docking in lunar orbit?) seems like a major challenge. How did they know they wouldn't be stuck with two vehicles in non-intersecting orbits? TheFutureAwaits (talk) 11:59, 19 February 2010 (UTC)

Well, it's mostly just "docking in orbit" -- "docking in lunar orbit" adds no particular degree of difficulty. NASA had already done considerable work in Earth orbit to verify rendezvous launches (see Gemini 6 and Gemini 7) as well as Apollo CM/LM docking tests (see Apollo 9 and Apollo 10, the latter conducting lunar orbit operations). So, the vehicles weren't untested, and they'd verified that they could do the math to avoid non-intersecting orbits. — Lomn 12:32, 19 February 2010 (UTC)

By the time of Apollo 11, it was fully tested because Apollo 10 had done exactly the same thing, just without having actually set down on the moon inbetween. I guess Apollo 10 were taking a bit of a risk, but the Apollo programme accepted a significantly higher level of risk than modern space programmes. --Tango (talk) 12:34, 19 February 2010 (UTC)

Apollo 9 tested the lunar module in Earth orbit, flying more than 100 miles from the command module before separating from the descent stage and returning to dock. I assume that they limited the delta-V so that had the lunar module failed at any point, the command module could have caught up with it for docking. So, while these missions did take big steps, it was not done all in a single step. 58.147.58.28 (talk) 13:47, 19 February 2010 (UTC)

I have always assumed that NASA had planned for the contingency of the loss of the descent crew, either due to a crash of the LEM during descent or a failure of the ascent module, but I've never heard it specifically said that the command module pilot would have been able to execute the return to Earth single handed. (I suppose that it was done in Shane Johnson's Christian science fiction novel Ice, but I don't recall him discussing the logistics of the return.) 58.147.58.28 (talk) 13:47, 19 February 2010 (UTC)

Yes, the CM pilot could have returned home on his own. I can't find a reference for that at the moment, but I do remember reading it. I don't know quite how it would work, perhaps ground control would do some of the work remotely. --Tango (talk) 14:24, 19 February 2010 (UTC)

Considering that they had a planned presidential speech in case Neil Armstrong and Edwin Aldrin died [6], I expect they would have had a detailed procedure to bring Collins back. 75.41.110.200 (talk) 15:12, 19 February 2010 (UTC)

Cause of Menopause

Does meno pause happen because of running out of eggs in the avaries, or is there something else that causes it. In that case. are there eggs left over? —Preceding unsigned comment added by 79.76.254.35 (talk) 13:53, 19 February 2010 (UTC)

There are millions of eggs, so that's not the problem. This seems to indicate that you are correct (check out the last paragraph)...but biology class has always taught that this is not true. It's a reflection of lowered hormone production, specifically estrogen and progesterone. You can check out the article on menopause for more information. DRosenbach ^{(Talk | Contribs)} 13:58, 19 February 2010 (UTC)

Running out of eggs in aviaries usually results in a lack of birds! --TammyMoet (talk) 15:12, 19 February 2010 (UTC)

Simply put. The eggs in the women cycle not only contains the gonades, but they also end up producing hormones, (which explains the hormonal variation, most of the time a single at a time, the same one which migrate waiting to be fecunded). That's controled by the putiary gland. During menopause there are no eggs left. That's quite different than the production of androgenes by the testicules and that's why male adropause is more gradual. -RobertMel (talk) 17:14, 19 February 2010 (UTC)

This question is tantamount to asking why people age or die. It's a long story! Vranak (talk) 17:31, 19 February 2010 (UTC)

No actualy, it is not. Menopause is due to no eggs left to produce female hormones and the surrenals conversion of DHEA does not suffice to replace that loss. It's really that simple. -RobertMel (talk) 17:35, 19 February 2010 (UTC)

Symptoms of a higher-level process. Vranak (talk) 19:07, 19 February 2010 (UTC)

Menopause simply means the cessation of menstruation as there is no eggs left which will be answering to LH and FSH. It's really not that long as a story, really. Menopause can be induced by simply removing the ovaries, because you get rid of the eggs all at once. What it means to be old? The question is much more complex. -RobertMel (talk) 19:31, 19 February 2010 (UTC)

~Shakes head~ I think we are at loggerheads here. Vranak (talk) 04:08, 20 February 2010 (UTC)

Im still not clear of the answer to my question. Can someone simplify the answers? —Preceding unsigned comment added by 79.76.244.151 (talk) 10:56, 20 February 2010 (UTC)

Okay - first thing to be clear about is that the ovaries do not contain a supply of mature eggs - they contain immature egg cells called primary oocytes. The supply of immature egg cells in the ovaries is fixed during embryonic development, well before birth - see oogenesis. There are far more than a woman will need in her reproductive span - the ovaries contain around two million immature egg cells at birth, but only around 400 of these will become mature egg cells and be released from the ovaries in ovulation - see folliculogenesis. The remainder die off over time in a continuous process called ovarian follicle atresia. The development of a small proportion of immature egg cells into mature egg cells is triggered by a set of interacting hormones, one of which is follicle-stimulating hormone or FSH. As a woman approaches menopause, her immature egg cells become less sensitive to FSH, ovulation becomes less regular, and eventually stops altogether. At menopause, FSH is still produced (in fact, post-menopausal women have higher levels of FSH than pre-menopausal women, because one of the side-effects of ovulation is to inhibit the production of FSH) and the ovaries still contain immature egg cells at menopuase - up to 10,000 according to this source. The cause of menopause is that the remaining immature egg cells in the ovaries have become insensitive to FSH. Gandalf61 (talk) 13:27, 20 February 2010 (UTC)

Thanks that makes sense. But what causes the immature eggs to become less responsive to FSH? —Preceding unsigned comment added by 79.76.132.10 (talk) 23:47, 20 February 2010 (UTC)

It is assumed by many that only those which are not responsive to begin with are left. -RobertMel (talk) 00:18, 21 February 2010 (UTC)

Assumed by who, exactly ? Do you have a source for that ? The development of immature egg cells into mature egg cells is a long and complex process that takes thirteen menstrual cycles altogether. At the start of this process, many immature egg cells start responding to FSH and developing. Most die along the way. At the start of the follicular phase between five and seven developing egg cells and surrounding structures (called tertiary stage ovarian follicles) continue their development, but normally only one - the dominant follicle - completes this development and is released in ovulation. Sometimes two mature egg cells may be released in ovulation. Or sometimes no dominant follicle emerges, and the process starts over again with another group of tertiary follicles - this is why the follicular phase is variable in length. The response of immature egg cells to FSH and other hormones is not a simple on/off switch. Gandalf61 (talk) 10:48, 21 February 2010 (UTC)

You are not contradicting me here. I was getting to egg selections. Unlike male who produce billions and only the strongests get to achieve their destination, women complete this selection starting with a prederminated number of immature eggs. At the end, either the remainings aged too much or only bad ones in the first place. They can now predict the age of menopause based on the size of the ovaries, which assume the reserves. -RobertMel (talk) 16:17, 21 February 2010 (UTC)

Really ?? I can only find one study (Wallace & Kelsey, 2004) that suggests that ovarian volume is an accurate predictor of the onset of menopause in an individual woman (as opposed to a general statistical correlation), and a lot of commentary (here for example) that says that this method is based on faulty assumptions and has not been clinically proven. Gandalf61 (talk) 18:15, 21 February 2010 (UTC)

hmmm..., I should have been more careful, you are right not an individual woman. On the other hand, they also have a more recent replication. [7] The review by C.B. Lambalk et al. (2009) suggest it's not a view just shared by one or two researchers. -RobertMel (talk) 21:03, 21 February 2010 (UTC)

Are kale stems edible?

All my kale was infested by Whitefly, they only spared the stems. Are they edible, and would they need special treatment before eating? 95.115.163.171 (talk) 14:17, 19 February 2010 (UTC)

I'm pretty sure I've eaten kale stems. It's just a variety of cabbage. --Tango (talk) 14:26, 19 February 2010 (UTC)

Yes: raw or boiled. Bit woody either way if the plants are too old. A related question if anyone knows is whether Vitamin K is uniformed distributed between stalk and leaves for plants such as this, or broccoli. Anyone know? I happen to be very fond of the crudity made from the centre of a broccoli stalk and have to watch Vitamin K intake. --BozMo talk 18:14, 19 February 2010 (UTC)

Sorry to be picky, but I think it's crudité 86.4.186.107 (talk) 19:12, 19 February 2010 (UTC)

Crudity means crudeness. --Tango (talk) 20:59, 19 February 2010 (UTC)

If the stem is edible and spared by pests, is there any cultivar with thick and not-so-woody stems? 95.115.163.171 (talk) 22:46, 19 February 2010 (UTC)

Hmmm. I just read that Kale -like Broccoli, Cabbage, Brussels Sprouts, and Cauliflower- is derived from wild mustard by means of artificial selection. on a side note, why are you interested in maggot-infested stems? just go to the market and purchase some new Kale, no?Chrisbystereo (talk) 14:47, 20 February 2010 (UTC)

Who said they were maggot-infested? The OP probably doesn't want to waste the food they have grown if they don't have to. --Tango (talk) 15:48, 20 February 2010 (UTC)

That's right. Actually I (the OP) discarded leaves and stems already last year. I am not in a situation where I need to do "original research" on what might be edible. But 1.) I'm curious, 2.) times may change, and 3.) I don't want to deliberately waste food, for ethical reasons as well as for plain stinginess. 93.132.156.86 (talk) 17:54, 20 February 2010 (UTC)

H₂O a covalent compound

why ionic compounds dissolve in covalent compound water(H₂O)? how does a covalent compound sugar(C₆H₁₂O₆)dissolves in water and not a oil?is it necessary for all ionic compounds to be soluble in water. —Preceding unsigned comment added by Myownid420 (talk • contribs) 16:27, 19 February 2010 (UTC)

The bond in H₂O is a polar covalent bond which means it can be considered partially ionic. Read ionic bond#Ionic versus covalent bonds for a brief explanation. It is the polarity of the compound that matters when figuring out whether it is water soluble or not. Dauto (talk) 16:41, 19 February 2010 (UTC)

Rings around Earth

This video was very interesting however I wonder if earth would be different or possible for life to happen on earth if it had this. Does it matter what the rings would be made of or the size? --Reticuli88 (talk) 16:54, 19 February 2010 (UTC)

Not an answer but a further question: how do we go about building those way cool rings? Maybe even temporary ones, say, from water ice that will deorbit in a few decades -- how many kg of water? Launch cost in USD? (Yeah yeah, astronomers will complain about light pollution; we'll just ignore you.) 88.112.56.9 (talk) 17:21, 19 February 2010 (UTC)

To answer the cost question, a general estimate is that it costs $10,000/lb (roughly $20,000/kg) to get something into LEO. It would likely take many, many thousands or millions of tons of material in order to crate planetary rings. The more spectacular, the more material you probably need. So I estimate cost on a bare minimum guess of $2 trillion for a measly 100,000 tons of material. Keep in mind that is only 100,000 cubic meters of ice (approximately) so it would be spread pretty thin. Also, the people who have satellites might not like you if you did this. Googlemeister (talk) 17:40, 19 February 2010 (UTC)

How will the water de-orbit? --Reticuli88 (talk) 19:09, 19 February 2010 (UTC)

There's still plenty of atmospheric drag in low Earth orbit, and ice chunks have no fuel to counteract that. — Lomn 20:01, 19 February 2010 (UTC)

To answer the original question. The rings would have no ill effect whatsoever for life on earth. Dauto (talk) 20:51, 19 February 2010 (UTC)

I don't know about that - the shadow the rings cast across the planet could have some effect. It wouldn't prevent life from forming, but it might be a little different (in the appropriate region, anyway). --Tango (talk) 20:57, 19 February 2010 (UTC)

The amount of light blocked by the rings is negligible. Dauto (talk) 21:42, 19 February 2010 (UTC)

OK, our rings of saturn page says 5 to 12% gets blocked so not completely negligible but too small to cause any direct ill effect. Dauto (talk) 22:26, 19 February 2010 (UTC)

That's just the C ring (and is unreferenced), which is described as "faint". Presumably the darker rings block more. --Tango (talk) 22:44, 19 February 2010 (UTC)

Dauto, your claims in this thread are really sloppy. Can you provide references for your extravagant claims that there would be "no ill effect whatsoever" and "too small to cause any direct ill effect"? You're aware, aren't you, about possible tipping points, and that any weather system is so chaotic that you can't possibly predict the consequences of changes? 63.164.47.229 (talk) 23:12, 19 February 2010 (UTC)

Yeah, but who said anything about no changes? I said there would be no ill effect. Earth might become cooler by several degrees due to the rings and life would adapt the same it adapted to the last several glacial maxima. Note that the question is about whether life on earth would be possible at all. Dauto (talk) 23:40, 19 February 2010 (UTC)

That life would eventually adapt is a given - but if the rings formed quickly enough then there might not be time for that. Assuming Saturn's rings for a model (not necessarily a valid thing - but let's go with it), the rings block sunlight for half of the year and enhance it for the other half. That results in hotter summers and cooler winters. How much that blocking might be is tough to guess - because we don't have a solid description of how thick or how wide these rings are. Judging by the density of the shadow cast by the Saturnian ring system, it's certainly not negligable. We could certainly imagine more pronounced weather at the edges of the ring shadow where the temperature contrasts are strongest. The complexity of the changes caused by such rings are hard to unravel. So I'm sure Dauto's statement is too strong. This is a "Don't know" kind of a question.

It's worth mentioning that according to the Giant impact hypothesis the Earth did once have rings - shortly after the Mars-sized object sometimes known as "Theia" smacked into the young Earth there would have been some pretty impressive rings - much of which eventually formed the Moon - with the remainder falling back to Earth eventually. SteveBaker (talk) 00:50, 20 February 2010 (UTC)

I stand by what I said. I few rings around the earth wouldn't prevent life on earth. Life survived more extreme events in the past such as the K-T event. Dauto (talk) 01:11, 20 February 2010 (UTC)

But you went way further than that, and claimed there would be no ill effect whatsoever, which you could not possibly predict or know for sure. Being a little less sure of yourself in such answers would suit you. 63.164.47.229 (talk) 01:50, 20 February 2010 (UTC)

Rings in orbit really wouldn't hurt life on earth. They might make a few things a little different but still suitable for life. Hence no ill effect. Dauto (talk) 02:38, 20 February 2010 (UTC)

Butterfly effect. Too chaotic to predict what changes would occur, good or ill. Ks0stm ^(T•C•G) 03:08, 20 February 2010 (UTC)

The butterfly effect doesn't make it impossible to make predictions. For instance, I predict that six months from now it will be warmer in New York City than it is today. Dauto (talk) 03:17, 20 February 2010 (UTC)

Sure, you can predict that, but it is far from certain to be true. If you average the temperature over a month, it becomes a safer prediction. --Tango (talk) 13:11, 20 February 2010 (UTC)

Perhaps our problem is with the definition of "ill effect". Determining whether an effect is ill or not is very subjective. The only effect on life that is unquestionably ill is life being entirely wiped out, and I think we're all agreed that wouldn't happen. However, there are other effects that are possible which could be considered ill - some species going extinct, for example. --Tango (talk) 13:11, 20 February 2010 (UTC)

One bad effect is that it will make satellite launches very difficult as they will have to avoid the ring. Graeme Bartlett (talk) 21:09, 21 February 2010 (UTC)

You would have to avoid launching too close to the equator and make sure the orbit is either entirely inside or entirely outside the ring system. That would make launches more expensive (you can't exploit the Earth's rotation as much) and low inclination orbits (such as geostationary orbits) very expensive indeed. --Tango (talk) 00:25, 22 February 2010 (UTC)

(side issue) I wonder if 'Earth rings' would have any affect on the once widely held belief that the Earth was flat? Would it make people think more about such things, perhaps advance astronomy and science faster? (Though as mentioned earlier it will create light pollution, making observations more difficult)--220.101.28.25 (talk) 17:12, 22 February 2010 (UTC)

I question your premise. I don't think that belief was ever widely held. Most people wouldn't have thought about it at all and those that did realised from very early on that the Earth was round. --Tango (talk) 21:19, 22 February 2010 (UTC)

Circular, sinusoidal, and other interesting patterns in the heavens exist, a set of rings that ran from one horizon to the other would have probably just fueled the imagination of some Astrologers. The moon betrays the roundness of the earth on a regular basis and that didn't seem to convince a lot of people. --144.191.148.3 (talk) 21:58, 22 February 2010 (UTC)

Points taken. see also Myth of the Flat Earth—220.101.28.25 (talk) 07:39, 23 February 2010 (UTC)

Always radioactive?

Are all chemical compounds of radioactive chemical elements radioactive? --88.76.18.70 (talk) 16:56, 19 February 2010 (UTC)

Simply put, chemical compounds concerns the electrons, radioactivity concerns the nucleus of the atom. So no matter the lenght of the chemical compound, as long as it contains radioactive elements, it is radioactive. -RobertMel (talk) 17:06, 19 February 2010 (UTC)

Maybe slightly off-topic, but it might be worth mentioning water vs. heavy water as an example of a compound whose chemical properties do depend appreciably on the nuclei of the constituent atoms. —Preceding unsigned comment added by 83.134.169.7 (talk) 10:01, 21 February 2010 (UTC)

Are there any radioactive chemical compounds which don't contain any radioactive atoms? --88.76.18.70 (talk) 18:08, 19 February 2010 (UTC)

No. -- Flyguy649 ^talk 18:10, 19 February 2010 (UTC)

I think the variations on what you are looking for in a single type of atom are called isotopes. ~AH1^(TCU) 23:41, 19 February 2010 (UTC)

February 20

star size limits

is there a limit for stars to grow before nuclear reaction is not enough to prevent them from collapse do to gravity —Preceding unsigned comment added by 161.184.96.146 (talk) 12:52, 20 February 2010 (UTC)

Yes. See Star#Mass and Eddington luminosity. I believe what the limit actually is depends on the metallicity of the star (which depends, among other things, on how old the universe was when the star was created). --Tango (talk) 13:06, 20 February 2010 (UTC)

For interest, last week's New Scientist magazine (dated 13 Feb 2010) had a cover-featured article [8] examining exactly this topic. 87.81.230.195 (talk) 16:30, 20 February 2010 (UTC)

Why are croc's more closely related to birds than lizards/snakes?

Is this because crocodiles and birds share a more recent common ancestor than the croc's and lizard/snakes do? Any help would be appreciated, thank you!!Chrisbystereo (talk) 14:32, 20 February 2010 (UTC)

Yes. See the nice tree at Archosaur#Phylogeny. --Mr.98 (talk) 15:07, 20 February 2010 (UTC)

Yes, that is a very common way of defining how closely related certain species are. That doesn't necessarily mean they have more genes in common, or more physical characteristics, etc.. The speed of evolution can vary widely, so one branch may have changed far more than another. --Tango (talk) 15:46, 20 February 2010 (UTC)

Well, it does mean that they have more genes in common (or rather greater similarity between homologous genes) -- that's the data that is used to work out the evolutionary history. Looie496 (talk) 17:45, 20 February 2010 (UTC)

They don't work out the evolutionary history in big jumps, though. On a small scale, what you say is correct, but when you are talking about branches that separated hundreds of millions of years ago it isn't so simple. Each species will have a lot of genes in common with other species that are very closely related to them, but when you are looking at distant relatives it is difficult to say. Consider this example. Species X has genes AAAAA and splits into species Y and Z with genes AAAAB and AAAAC. Species Y then splits into Y1 and Y2 with genes BAAAB and CAAAB. Y1 then stays as it is and Y2 evolves into Y2' with genes CBBBB. Y2' and Y1 are quite closely related (their common ancestor is Y), but only share 2/5 genes. Y1 and Z are more distant relatives (their common ancestor is X) but share 3/5 genes. --Tango (talk) 19:10, 20 February 2010 (UTC)

The last common ancestor of crocodiles and birds is more recent (in time) than the last common ancestor of crocodiles and lizards & snakes. That makes them more closely related in exactly the same way that brothers and sisters are more closely related than cousins. Birds are descended from dinosaurs - and dinosaurs and crocodiles are closely related. SteveBaker (talk) 17:48, 20 February 2010 (UTC)

Among other similarities, crocodilians have four-chambered hearts. DRosenbach ^{(Talk | Contribs)} 23:37, 21 February 2010 (UTC)

Decaying Copernicium

First off, chemistry is far from being my strongest subject. That said, when radioactive elements decay, they become other elements, yeah? So what does Copernicium become? If I'm totally off, please explain why in fairly simple terms. Thanks, Dismas|^(talk) 15:29, 20 February 2010 (UTC)

In the table at the top-right of each element page, it lists the stable isotopes, and for radioactive elements, has a "DP"="Decay product". So in this case, the most stable ones seem to become various isotopes of Darmstadtium (Ds). As for decay in general, there are a bunch of different ways radioactive elements can decay, some of which can modify its proton count (which makes it a different element and not just a different isotope)—see Radioactivity#Decay_modes_in_table_form. Anytime Z changes, you have a different element. The odd cases where Z goes up are from where other internal parts of atoms are converted into protons. --Mr.98 (talk) 15:52, 20 February 2010 (UTC)

132Sn is mentioned in the article text as well. (I would consider this a physics question, not chemistry) 75.41.110.200 (talk) 15:59, 20 February 2010 (UTC)

Though that's in the case of fission, not decay. (Which we can guess, immediately, by the fact that a fission product is going to be roughly 50% of the original, whereas a decay product is maybe 1 or 2 protons/neutrons different.) --Mr.98 (talk) 16:01, 20 February 2010 (UTC)

Spontaneous fission is a mode of nuclear decay. 75.41.110.200 (talk) 21:27, 20 February 2010 (UTC)

Erupting vanilla extract

While making waffles just now, I put the dry ingredients into one bowl and started mixing the wet in another. In the wet bowl, I put in milk and the eggs. Then when I put the vanilla extract in, it was roiling almost as if boiling. (so the combo was eggs, milk, van. extract) What causes this? Dismas|^(talk) 15:41, 20 February 2010 (UTC)

If it was bubbling when you poured it on to flour, I would suggest that percolation has the answers. Your flour and dry ingredients have pore spaces between the grains that are filled with air. As the fluid falls into those pore spaces, the air needs to bubble its way out. But it sounds like you've only mixed wet ingredients. Vanilla extract should really be pretty inert - it's mostly alcohol and water, with a trace amount of vanilla oil - but you could conceivably be seeing an acid-base reaction. I'm guessing your mixture is pretty viscous - so any trapped air (perhaps from whisking or mixing) might have formed bubbles that were slowly buoyantly rising to the surface. If the time constant for that buoyant rise is fairly slow, you might see air bubbling to the surface long after you stop mixing. Nimur (talk) 16:22, 20 February 2010 (UTC)

It's also possible that the vanilla extract is somehow acting as a surfactant and allowing the liquid to get more tightly into the pores of the flour, driving the air out. SteveBaker (talk) 17:40, 20 February 2010 (UTC)

Assuming that you only say liquid moving, with no visible gas bubbles and no liquid being forced higher than the normal surface of the liquid, the likely cause is mixing and surface tension effects. Vanilla extract usually contains a high proportion of alcohol, which has a lighter density than water, and certainly has a lighter density from the egg-milk mixture. When put together, they don't mix immediately, due to the density difference. As they slowly dissolve with each other, this sets up a number of concentration gradients, which can cause physical forces on the liquid. Additionally, and perhaps more importantly, water and water-alcohol mixtures have much different surface tensions. You can see this if you pour a small amount of alcohol into a very shallow layer of water. The water "tenses up" into a ball, and you get a "shimmery" movement at the water-alcohol interface as the two adjust to the surface tension differences. -- 174.21.247.23 (talk) 17:45, 20 February 2010 (UTC)

I'm sorry if I wasn't clear but the bowl only had the wet ingredients in it. Dismas|^(talk) 21:35, 20 February 2010 (UTC)

You've got a bowl with milk and eggs--are they beaten together or just poured in? Then you added vanilla extract. Dropped hard in one place (so that gravity makes it plunge towards the bottom) or gently sprinkled across the surface? When you say "roiling almost as if boiling", do you mean bubbling or just visible liquid mixing up-and-down? Alcohol is lighter than water, so if you drop alcohol to the bottom it will try to rise up again. DMacks (talk) 06:26, 21 February 2010 (UTC)

Just poured in. Dropped in one place. No bubbling, just a sort of eruption that didn't raise the surface of the liquid at that spot. It was fairly prolonged, on the order of 10-15 seconds. (on a side note, I just looked up 'roil' and apparently it's not a word though I've heard it used before to mean a rolling boil) After reading all the responses thus far, I'm betting that it's a surface tension thing. I wasn't aware that VE was mostly alcohol.

And on another side note, I've known a few alcoholics in my lifetime and I'm aware that they don't keep mouthwashes with alcohol in their houses. Now I'm wondering if I would find VE in their cabinets. Dismas|^(talk) 13:20, 21 February 2010 (UTC)

You would have to be pretty desperate to want to drink undiluted Vanilla extract - it's amazingly strong stuff. SteveBaker (talk) 16:19, 21 February 2010 (UTC)

I've heard (from a not terribly reliable source) that sailors in the navy would drink some sort of alcohol based extract (was it almond extract?) to get around prohibitions on alcoholic beverages. If true, I wonder what the suppliers thought about the amount of almond extract being ordered. Buddy431 (talk) 18:03, 21 February 2010 (UTC)

I heard that story too - but I thought they were distilling the alcohol out of the extract rather than drinking it 'as-is'. SteveBaker (talk) 18:23, 21 February 2010 (UTC)

Cooper Sulfate use

What are the affects on wildlife such as ducks and geese if cooper sulfate is used in a small neighborhood pond for algae control? —Preceding unsigned comment added by 96.28.172.66 (talk) 16:52, 20 February 2010 (UTC)

Copper sulphate can be toxic to fish, aquatic plants and algae - which would obviously eliminate a food source for the ducks and geese - but a lot depends on the dosage. At low concentrations, it's used to treat swimming pools (so it's obviously not toxic to humans at those levels) but out article lists a large number of alarming toxic effects on humans (so it's obviously nasty stuff in higher concentrations). It's also used to treat skin diseases in goldfish - so even the known toxicity to fish can only occur at higher dosages. Sadly, our article doesn't discuss toxicity to birds. I strongly suggest that you discuss the dosage levels used in the pond with whoever is doing the treatment - and try to establish that it's being used in reasonable amounts. Follow the instructions on the packaging very carefully - with particular reference to using only the minimum amount needed to treat the algae. This can be tricky without knowing the volume of water in the pond - which can be really tough to estimate. SteveBaker (talk) 17:32, 20 February 2010 (UTC)

focal length of lens

does focal length of a lens depends on the refractive index of its surroundings? like if we put a lens in water, will its focal length change? i think f should remain constant becoz by 'lens maker formula' :-

         1/f = (n-1)(1/R1 + 1/R2) 

it seems that focal length depends only on refractive index of the material by which lens is formed and radius of curvature of both faces. but if we put a convex lens in say a denser medium(optically denser than the lens) it will become a diverging lens. will not it? so as we put it in another medium which is rarer than the first medium, divergence of light shall decrease. by this we can conclude that in air if the focal length of convex lens is x then in water f will be y such that y > x. this is my dilemma.thanx§§§§ --Myownid420 (talk) 17:00, 20 February 2010 (UTC)

Technically, the refractive index should be measured relative to the medium in which the lens is immersed. However, because air and vacuum have almost the same refractive index (1.0000 for vacuum, 1.0003 for air at standard temp & pressure), it is rare for people to state which medium they are measuring it in. So, if you read that the glass that some lens is made of has a refractive index of 1.5 - then it really doesn't matter whether that's in air or in vacuum. If you put that same lens into water, then you should really use the refractive index of the glass relative to that of water - but nobody lists those kinds of numbers. Hence, you have to divide the refractive index of the glass in air by the refractive index of water in air to get the refractive index of glass in water. That does indeed mean that the focal length of the lens will be different in air than in water. SteveBaker (talk) 17:21, 20 February 2010 (UTC)

To summarize, the n in that formula is always $n_{\text{lens}}/n_{\text{environment}}$; see Snell's law, which contains only that ratio and neither index separately. We often neglect the denominator because it's so close to 1 for air and vacuum. --Tardis (talk) 17:38, 20 February 2010 (UTC)

Ideas for activity

I'm a nurse working in the area of Injury Prevention. We're going to set up an Injury Prevention booth at a college health event which will be 2 hours long. The booth will be one of many others that deal with a health topic. I set up a bulletin board with brief info on preventing injuries. Now I am looking for some ideas on the types of interactive activities around the topic of Injury Prevention that might be suitable for college students who stop by at my booth. I've currently thought of having students answer 5~10 True or False questions, but I'm open to any other ideas. —Preceding unsigned comment added by 70.68.120.162 (talk) 20:49, 20 February 2010 (UTC)

Can you get hold of some fake injuries like those used in TV and film that you can stick on people? That might give people a better idea of the consequences of their carelessness. It would also be fun, which always helps at that kind of event. --Tango (talk) 20:54, 20 February 2010 (UTC)

You could make a big, hollow foam polystyrene cube - dress it up to make it look like something really heavy and put fake "Warning: Really Heavy" stickers on it - then ask people to demonstrate how they'd pick it up off the floor. You'd be able to instantly show them the postural mistakes they (almost inevitably) make while doing that - and explain how to avoid the resulting back injuries by doing it right. SteveBaker (talk) 21:11, 20 February 2010 (UTC)

Maybe a picture or a staging of a common workplace situation having the students point out where the risks for injuries are. (a trailing cable, floor that will be slippery when walking with wet shoes e.t.c) Just looking at a website about slips and falls: [9] there's plenty of situations that a college student may meet in his/her part-time job. EverGreg (talk) 21:20, 20 February 2010 (UTC)

That would be a lot of fun - take over a room someplace - see just how crazy dangerous you can make it - take photos and challenge people to count the number of preventable injury sources there are! Sit around with some buddies and brainstorm all of the ways you can make it dangerous. Just try not to get injured while doing it because that would be something that would get you a fast-track to the Darwin Awards. SteveBaker (talk) 03:34, 21 February 2010 (UTC)

In the UK, drawings of such a room are often used in Office Safety lectures mandatory in many companies (possibly at the behest of the Health and Safety Executive). Here's [10] one example, found from 20 seconds of googling, that might provide inspiration. (NB: those ringed in red on this version are not the only ones.) 87.81.230.195 (talk) 20:45, 21 February 2010 (UTC)

Bendroflumethiazide mechanism of action

Hi I understand that it is thought that some of the anti-hypertensive effects of Bendroflumethiazide may attributed to its inhibition of the enzyme carbonic anhydrase. (See the wiki article called thiazide) I'm not sure how the inhibiton of carbonic anhydrase would have an anti-hypertensive effect. How would this work? The article on carbonic anhydrases doesn't offer any suggestions. Thanks to anyone who can offer some insight. RichYPE (talk) 23:07, 20 February 2010 (UTC)

Well it affects bicarbonate levels in the blood and how well CO2 will dissolve in it. This in turn impacts blood acidity and H+ concentration. "Angiotensin II stimulates Na+/H+ exchangers located on the apical membranes (faces the tubular lumen) of cells in the proximal tubule and thick ascending limb of the loop of Henle in addition to Na+ channels in the collecting ducts. This will ultimately lead to increased sodium reabsorption" (Renin-angiotensin system). This is my guess ... John Riemann Soong (talk) 02:58, 21 February 2010 (UTC)

February 21

Green cones in the perception of colour

Hey, I was looking at this picture:

and it occured to me that we never really see green without quite a lot of red, and so we must be really sensitive to the difference between what our red and green cones are picking up. The purest green we'll ever see still looks like it'll maybe be 55% green to 45% red.

So I was wondering what it would be like to artificially stimulate only your green cones. This would essentially mean receiving RGB data in a ratio that you'd never experienced before, greener than the green. Would the experience of this be basically the same as seeing a normal bright green, or would one be able to tell that this is a colour they had never seen?

Thanks, Luke —Preceding unsigned comment added by 82.35.84.214 (talk) 01:00, 21 February 2010 (UTC)

I don't know what the result of that would be (or any way to do the experiment, actually), but it isn't easy to predict, because we don't "see" the raw output of the cones -- the neural activity is quite substantially transformed and recombined before it reaches the cerebral cortex, and then transformed even more inside the cortex. Along the way, the three-color cone activity pattern gets turned into an opponent process representation for which the primary hues are red, green, yellow, and blue. Looie496 (talk) 02:18, 21 February 2010 (UTC)

It's an interesting thought - what would happen if you got a green stimulus without any red stimulus at all? Certainly it's a bit of a guess because our brains are 'wired' to see things that way. Some colorblind people have no red sensors at all - so they are seeing what you're proposing. But the trouble is that their brains have formed since babyhood without that red sensor - so they don't know any different. The best answer is "We can't possibly guess without doing the experiment". If I had to guess, I think the result would be exactly what we see when we've been staring at a pure 625nm-ish red surface for a LONG time - then quickly glance at a green surface. Staring at the red surface will cause Neural adaptation which will make the brain ignore the "red" signal for a while. Then, when you look at a yellow or green surface, the red sensors are dumbed down and all you're registering is the green sensors. This is the basis of various Afterimage illusions. All I see when I do that is a really brilliant lime-green. But we have to be careful - we can't be 100% sure about the comparability of the result of this practical test with the impractical one you're thinking about. SteveBaker (talk) 02:46, 21 February 2010 (UTC)

Right. The color stimuli that correspond to impossible combinations of the three cone type activations are called Imaginary colors. Indeed, you can use the fact that human color adaptation is not instant, and have the L-cone gain reduced by strong input at 650-700 nm wavelength. Then, prompt preferential stimulation of the M-cones at around 550 nm should produce a "greener than green" color percept. However, I doubt the effect is going to be too strong, as V1 and higher visual areas have multiple negative feedback loops, making a highly unusual response next-to-impossible. Reducing this inhibition pharmacologically may actually produce an "impossible color" sensations via this mechanism; I think this was indeed reported as one of the LSD effects. Needless to say, any unauthorized use of controlled substances is VERY strongly discouraged. --Dr Dima (talk) 03:32, 21 February 2010 (UTC)

To be entirely accurate, LSD does not directly reduce inhibition. Still, it increases the excitability of the sensory areas of the cortex. --Dr Dima (talk) 03:40, 21 February 2010 (UTC)

The color you would see would not be green. Green is the color that is "55% green to 45% red". If you did only green cones you would have some other color, but not green. You have to remember our color is calibrated by the light we actually see, not by the labels we give the color sensors. Ariel. (talk) 04:33, 21 February 2010 (UTC)

Indeed. It is best not to refer to the different types of cones by colour, but by "short", "medium" and "long" (as they are labelled in that picture) to avoid confusion. --Tango (talk) 06:03, 21 February 2010 (UTC)

I think this has a straightforward answear. A situation where only the green cones are stimulated is closest to a situation where the green cones are stimulated the most and the blue and red cones are stimulated only a little and with equal intensity. From the figure it would seem that percentagewise, green is stimulated the most around 500nm, which corresponds to sea-green or blue-green according to a picture in the color vision article. EverGreg (talk) 13:15, 21 February 2010 (UTC)

Spectroscopy spectrum order

Okay, so I'm doing a lab right now on spectroscopy, and I'm given this equation: dsinθ=mλ. m is defined in the lab manual as "a positive integer equal to the order of the spectrum", but I am having trouble finding anything having to do with the order or how to know what value it may be.--十八 04:40, 21 February 2010 (UTC)

A light bulb of a flashlight seen through a transmissive grating, showing three diffracted orders. The order m = 0 corresponds to a direct transmission of light through the grating. In the first positive order (m = +1), colors with increasing wavelengths (from blue to red) are diffracted at increasing angles. Cuddlyable3 (talk) 23:14, 21 February 2010 (UTC)

We're missing a lot of context about what kind of spectroscopy you're doing, so I'll answer in general. The idea is that as you go along, the the paths taken by two waves vary in length, and therefore the phase difference between them changes. Say they start off in-phase; as the path-length difference increases, they get progressively more out-of-phase and then they wind up coming back into phase again, and then back out-of-phase, and then back into phase, and so on. The "order" talks about how many times that happens: at the lowest m they are very close (m=0 is the starting in-phase phase); at higher m, they have gone through many complete cycles and beyond. DMacks (talk) 04:53, 21 February 2010 (UTC)

I'm doing visible spectroscopy.--十八 06:16, 21 February 2010 (UTC)

Are you using a diffraction grating, and is our article helpful? TenOfAllTrades(talk) 15:00, 21 February 2010 (UTC)

how does magnet works in space?????????

hi,i am one of your member and i want to know , how permanent magnet works in space ? I will be waiting for your answer. —Preceding unsigned comment added by Debiprasadnayak (talk • contribs) 06:14, 21 February 2010 (UTC)

A magnetism works through the vacuum of outer space just like it works in air. Magnetism is like light; it is not transmitted via real particles or pressure the way sound is. DMacks (talk) 06:17, 21 February 2010 (UTC)

Light IS transmitted via real "particles" though... --antilived^{T | C | G} 11:02, 21 February 2010 (UTC)

No it's transmitted by waves. No, wait, it's particles after all! Actually it's neither, it's something else that sometimes appears particle-like, and sometimes wave-like, depending on how you measure it. 87.81.230.195 (talk) 20:26, 21 February 2010 (UTC)

Magnetism, the subject of the actual question, however, is transmitted by virtual particles, so DMack's is correct where it matters. --Tango (talk) 00:17, 22 February 2010 (UTC)

Coed Olympic Events

I just read the informative article on Olympic Sports but it does not cover the origin of the gender division, nor whether there have ever been any challenges to it. It occurred to me last night that there are more than a few Olympic events that could be fairly contested by both men & women together. In most shooting events strength plays only a very minor role, if not none at all (air pistols). I also suspect that gender would not imbalance a curling tournament, though perhaps the men could grind out a little extra bend to their throws?

Q1: Has there ever been a movement to unify some of the Olympic events?
Q2: In which events would gender impart very little or no advantage?

This question could have gone on Entertainment, but I find Q2 the more interesting and so I post it here. Thank you. 218.25.32.210 (talk) 07:15, 21 February 2010 (UTC)

Check out the equestrian events. B00P (talk) 08:25, 21 February 2010 (UTC)

Rightly or wrongly, the commentator on the US vs UK Women's Curling competition last night said something like: If this were the men's competition then this shot wouldn't have been a subject of debate - but it's a measure of how far Women's curling has come that they are able to even contemplate taking it" (after which the shot was taken and messed up). This suggests that the women's teams are getting closer to the men's game - but that they aren't there yet. I agree that there is nothing obvious in the biology of men and women to explain why women and men need to have separate games (unlike...say...weight-lifting). But the effect of merging the two events would undoubtedly to simply exclude women from Olympic curling. That would probably have a devastating effect on the number of young girls who see a female olympic curler and decide to take up the sport - and might well (effectively) keep women out of the competition forever. So I guess the Olympic committees prefer to keep male and female events separate until both genders are at about the same level. Is this "right"? Is it "fair"? Is it "sexist"? I don't know - it's like positive racial descrimination in college admissions - no matter what you do, there will be winners and losers. There are good arguments on both sides and no clear moral high-ground. If there were big protests amongst women curlers about being excluded from the men's game - then perhaps we'd be doing this differently. SteveBaker (talk) 16:12, 21 February 2010 (UTC)

Probably not what the OP means but it was a good play. Cuddlyable3 (talk) 17:00, 21 February 2010 (UTC)

I will note that there may be a preference for maintaining separate men's and women's events even where men and women are playing at comparable skill levels. This would be purely because it doubles the number of athletes who can participate, the number of events that can be televised, the number of medals to be won, and (potentially) the number of sponsorships and endorsement contracts available.

On the subject of curling, I will note that men (on average) are bigger and stronger than women. While this should make no difference for the majority of shots (which depend heavily on touch and finesse, rather than raw power), it will have an effect on a few of the high-power, high-risk shots (double peels, multiple raise takeouts, etc.) where a large amount of power needs to be finely controlled. I suspect that the biggest difference probably comes in sweeping, however. Bigger, heavier, stronger men are going to be able to put more weight on the brooms, and are going to be able to affect the course of a moving stone more effectively. TenOfAllTrades(talk) 18:07, 21 February 2010 (UTC)

There are still sex-related differences in areas like spatial cognition Sex-related differences in spatial cognition. Therefore even if a sport doesn't favour one sex on physical grounds, it may favour one sex through mental differences. Curling and shooting would both seem to require the kind of spatial cognition that would favour males. 82.132.248.88 (talk) 00:01, 22 February 2010 (UTC)

Litmus paper

how does Litmus paper or Universal indicator work. what chemicals are on the paper? are these harmful? and where can i buy the paper? —Preceding unsigned comment added by 67.246.254.35 (talk) 08:16, 21 February 2010 (UTC)

Have you tried reading the article on litmus paper? It is harmless (but obviously you wouldn't want to eat it). It is easy to buy on line.--Shantavira|^{feed me} 08:46, 21 February 2010 (UTC)

whats in it thou? —Preceding unsigned comment added by 67.246.254.35 (talk) 08:58, 21 February 2010 (UTC)

litmus is extracted from lichens(a symbiosis of algae and fungi). its really a solution of purple color and changes to red when in contact with acid . and turn to blue when in contact with base. if u just want a acid base indicator you can use turmeric it turns red with base( try it on soap water) and shows no difference for acid

most probably litmus when in contact with H⁺ ions give red color and when in contact with OH^- it gives blue.--Myownid420 (talk) 09:26, 21 February 2010 (UTC)

As Shantavira hinted at, what exactly from the article requires further clarification? Nil Einne (talk) 11:49, 21 February 2010 (UTC)

Here's a picture of Roccella tinctoria used to make Litmus paper. The article São Jorge Island under "Economy" mentions a possible source of this lichen. Cuddlyable3 (talk) 16:55, 21 February 2010 (UTC)

As the litmus test article indicates, the compound that changes color with a pH change is 7-hydroxyphenoxazone (pictured in article). It works like all other pH indicators - that is, it's a compound where the conjugate acid and the conjugate base have different colors (so it changes color around the pKa when the molecule gains or loses a proton). And the two have different colors because the presence or absence of a proton (hydrogen ion) changes the electronics of the conjugated system, perturbing the energy levels of the molecular orbitals, altering the energy difference between them, and thus the wavelength (color) of the photons the compound can absorb. -- 174.21.254.47 (talk) 18:37, 21 February 2010 (UTC)

field lines of magnet

lets assume a straight conductor studded in a square cardboard,which is parallel to the ground, If xA current flows through conductor than the concentric field lines formed are at distance of m cm from each other. now current yA flows though the conductor, Such that yA > xA, and dist between circles is n cm. so will n = m or n>m or n<m? in simple words if current is increased in a conductor the field lines get denser or distance between them remains same?--Myownid420 (talk) 09:41, 21 February 2010 (UTC)

First you need to know that the magnetic field lines fill the whole space so it is not meaningfull to talk about the distance between them. But conventionally when drawing them they're spaced inversely proportional to the magnetic field intensity. The magnetic field intensity around a wire drops with the distance from the wire so you really shouldn't draw equaly space concentric field lines. Dauto (talk) 15:00, 21 February 2010 (UTC)

Those "field lines" are not things that really exist. Think of them as the contour lines on a map of a mountain - when you visit the actual mountain, there are no lines to be seen! The spacing between the lines is something that the map-maker (or the magnetic-field-illustrator) chooses to make the illustration be informative. As Dauto points out, if the field lines were intended to represent fixed intervals of magnetic intensity (like the 100' foot contours on a map that are equally spaced in height) then they wouldn't be uniformly spaced in the first place. If they are uniformly spaced then they must represent varying 'step' in magnetic field intensity. SteveBaker (talk) 15:55, 21 February 2010 (UTC)

The title is about a magnet but the question is about the magnetic field of a straight conductor. The cardboard and the ground seem to be unnecessary parts of the question. The magnetic field lines are indeed concentric with the wire but as Dauto pointed out the lines are just a way of showing the direction of the field which actually fills the whole space (it is a Vector field). As one moves away from the wire the distance between the lines should increase because the field gets weaker. To the question: Increasing the current by the factor y/a increases the magnetic field at a given point by the same factor so the line spacing gets less, n<m. You may like the article Field line but note that its top diagram shows electric field lines not magnetic ones. Cuddlyable3 (talk) 16:38, 21 February 2010 (UTC)

Michael Faraday was a 19th century genius and an early experimenter in electricity and magnetism. He had little or no mathematical ability. He used "lines of magnetic force" to create theories of electromagnetism. James Clerk Maxwell was another 19th century genius who theorized about electricity and magnetism, and who was a gifted mathematician. Maxwell considered at length Faraday's writings and then said that their views of electromagnetic fields agreed 100%. "Field lines" do not exist. The field exists everywhere, and not just in discrete "lines." But they are a useful aid to gaining a physical understanding of magnetism and electromagnetism. In the crude model of "field lines," greater current would result in more densely spaced field lines, which would correspond to a stronger magnetic field. If you consider a permanent magnet in the form of a horseshoe or bar, with a north and south pole, the lines of force leave the north pole and end at the south pole. They are densely clustered at the poles, and some take a short path between the poles of the horseshoe, while others branch out to cover the entire plane, getting farther apart as you move away from the poles. The strength of the magnetic field, as seen in its effect on a current carrying wire or one pole of a long bar magnet, varies with the density of the fictitious "field lines." When iron filings are sprinkled over a pane of glass or a sheet of cardboard over a bar or horseshoe magnet or an electromagnet, the filings tend to align like the fictitious "lines of force," but there is a bit of hand waving in the demo. Each iron filing becomes by induction a little magnet itself, and they attract each other. Tapping on the glass or cardboard causes the "visible lines" to shift closer to the magnet's poles, so the "lines" are not discrete in space at all. "Electric field lines" can be modelled in a shallow tray of water between a DC + and - terminal just like the magnetic "field lines", with a voltmeter probe tracing out equal potential contours, but the electric field exists everywhere in the tray, and not just on the "field lines." More voltage between the + and - terminals in the tray would create more "one volt contours" just as more current in a wire creates denser "magnetic field lines." Edison (talk) 05:16, 22 February 2010 (UTC)

No Edison, you are describing equipotential contours not electric field lines. Electric field lines such as these show the direction of force on an isolated positive charged Test particle. Mathematically electric field lines are the Gradient of the scalar electric potential field and they lie at right angles to your measured contours. BTW you won't measure anything in pure water. Cuddlyable3 (talk) 19:24, 22 February 2010 (UTC)

Thanks for the correction. This lab writeup describes the procedure, and in fact calls for "deionized water", which would not work in your understanding. Tap water should also work fine. They use low voltage AC. Edison (talk) 20:20, 22 February 2010 (UTC)

Stars ejected from the Milky Way

Are there any known stars that are no longer part of the Milky Way but are just floating away in extragalactic space? TheFutureAwaits (talk) 17:54, 21 February 2010 (UTC)

This section of our Milky Way article describes some globular clusters that are a long way outside of the galactic disk which may be leaving the galaxy. Also, it is thought that the Canis Major Dwarf Galaxy is close enough to the Milky Way to be ripped apart by it - and it would be surprising if some of the Milky Way's stars were not being pulled out in the opposite direction. But at these great distances, we aren't really looking so much at individual stars as larger groups like globular clusters. So I doubt we could point to a particular "known star" and say "Ooops! We lost that one!". Also, this stuff happens on truly spectacular time-scales - so it's not like we can see it happening. SteveBaker (talk) 18:17, 21 February 2010 (UTC)

See hypervelocity star. According to that article (or rather section), 10 are currently known. Looie496 (talk) 18:23, 21 February 2010 (UTC)

There are nine stars in Category:Extragalactic stars. TenOfAllTrades(talk) 18:39, 21 February 2010 (UTC) ...though most of those aren't free-flyers, but are in the Large Magellanic Cloud.... TenOfAllTrades(talk) 18:43, 21 February 2010 (UTC)

nutrition labels

How would a nutrition label for the average human body (male and female) be broken down and for the normal distribution of (male and female) humans? No, I am not planning to join a cannibal cabal but just want to know what nutrients the human body already contains so I can print a human nutrition label on a t-shirt to give people who look at me in that sort of way a printed answer to their question. 71.100.3.6 (talk) 19:44, 21 February 2010 (UTC)

The nutritional content of the edible parts of a human would not be significantly different than what you would find on the nutritional label for other types of meat (say pork, for example). So you can just find one of those and copy it - though some things (like fat content) are going to vary from one individual to the next. There would be no significant difference between a male and female body, though serving size may vary depending on the consumer's predilections. —Preceding unsigned comment added by 72.94.164.21 (talk) 21:25, 21 February 2010 (UTC)

Females have a higher fraction of fat on average (some tables), but you could personalize the label by bioelectrical impedance analysis. Icek (talk) 21:43, 21 February 2010 (UTC)

Wikipedia has an article on Cannibalism which does not give a nutritional breakdown. It would not be a good idea to wear the OP's T-shirt in North Korea where some people reportedly do look at one in that sort of way. Cuddlyable3 (talk) 23:06, 21 February 2010 (UTC)

Actually I do not sympathize with cannibals since it is very easy to have a complete and balanced diet with all of the calories you need from items you might not think of. All that is required is that you know the nutrients the items have, such as a nutrient label provides. While such things as snails and slugs might not be on your normal breakfast menu, as with the information provided by any nutrition label, if you had one for them a meal could be prepared to fulfill your every nutritional need simply through using the right proportion. Instructables: Meal in a Cup 71.100.3.6 (talk) 23:31, 21 February 2010 (UTC)

Here is an example of data from one such source... Mollusks and Snails as Shellfish 71.100.5.197 (talk) 03:41, 22 February 2010 (UTC)

research methods

A society has a very high gini coefficient, a high level of social inequality. The same society has a very high level of HIV infection. The hypothesis could be that social inequality 'causes' or exacerbates HIV infection, or conversely that high HIV infection leads to social inequality. We're trying to avoid a fundamental methodological mistake...is it possible that both can be true at the same time, or does scientific method assume that only one of these hypothesis can actually be 'true'? —Preceding unsigned comment added by 83.98.238.113 (talk) 20:02, 21 February 2010 (UTC)

What of the possibility that neither are true? 87.81.230.195 (talk) 20:17, 21 February 2010 (UTC)

I have not considered this specific case in detail, but our article on Positive feedback may be relevant. You may also wish to consider that the causes and effects in this situation (or indeed any situation) may not be constant over time.131.111.185.68 (talk) 20:28, 21 February 2010 (UTC)

They can certainly both be true—a feedback situation would mean that each exacerbates the other dynamically. Or, put another way, HIV exacerbates social inequality, and social inequality exacerbates HIV transmission. There is no reason to expect a complex socio/epidemiological issue to be monocausal, and nothing about the scientific method suggests it has to just be one way or the other a priori. It's also possible that neither are true, that the correlations you see do not imply any causation. (Classic example: in the summer, crime rates and ice cream sales both increase. This is not because ice cream causes crime or the other way around, it is because both have some causation due to outside temperatures.) --Mr.98 (talk) 20:31, 21 February 2010 (UTC)

A lot depends on how good the correlation is. In the case of crime versus ice cream - I doubt the correlation is as exact as (for example) lung cancer and smoking. Probably the crime wave starts a month before the ice cream sales spike - and outside of summer there may be other crime and ice-cream 'spikes' that are completely uncorrelated. But when the correlation is exceedingly close, it's more likely that there is a causative link than when it's merely a sloppy fit. So it is perhaps better to say that "correlation cannot prove causation" - but it can certainly provide a sufficiently strong suspicion as to justify a presumption of causation until such time as better information may be found. SteveBaker (talk) 20:51, 21 February 2010 (UTC)

Even in cases of very tight fits there are always the possibilities of more complicated dynamics, though. Anyway, I wasn't trying to say that correlation and causation have nothing to do with each other, just that without really studying something closely, seeing a correlation does not tell you what the causation is. In the case given here relating to social inequality and HIV transmission, both could be by-products of corrupt government, for example, one that lacks both services in public health/education and supports a regime of inequality. In such a case you might expect extremely tight correlation between public health issues and economic ones, but it doesn't necessarily mean that the public health situation is causing the economic situation or vice versa (or, at least, that such an answer is an adequate understanding). (I'm not saying that is the case here, but one would, I assume, want to make sure one is not missing out on a larger causative factor when trying to draw conclusions about variables that might otherwise be linked.) --Mr.98 (talk) 21:44, 21 February 2010 (UTC)

The measurements being compared may have different time scales. HIV was first clinically reported in 1981 and has spread rapidly to reach the latest reliable figures that the OP has, perhaps for a year like 2007. I don't know what the timeline of the data being used to calculate Gini coefficient of incomes dispersion but only a change in consistently calculated gini since 1981 can support the "HIV infection leads to social inequality" hypothesis. The converse hypothesis "social inequality exacerbates HIV infection" (I think it is obvious that it does not cause HIV) is best tested by comparing the rate of HIV infections in two (preferably more) different societies of different long-term ginis. Cuddlyable3 (talk) 22:41, 21 February 2010 (UTC)

Agree 100% that a question like this must have historical data incorporated into a convincingly causal conclusion. HIV is not transhistorical (and neither probably is social inequality, but the latter is probably generally more constant). --Mr.98 (talk) 04:32, 22 February 2010 (UTC)

Our article on correlation does not imply causation may provide some insight into the risks of drawing these sorts of parallels. (Is there a third factor – perhaps poor access to antiretroviral therapy – which is associated with both HIV transmission and social inequality?) TenOfAllTrades(talk) 23:04, 21 February 2010 (UTC)

Well, the fundamental methodological mistake is to think that it is possible to conclude anything at all from post-hoc analysis of a single data point.

February 22

Minimum Temperature For Human Survival

What is the smallest temperature that a human can survive in indefinitely? Which is to say, suppose a human (with no clothes or shelter) is provided enough food and water to live, and that he/she is in a container of air at some constant temperature and with no wind. Further assume that the amount of air present is such that oxygen is not a concern and the temperature of the air is not influenced by the radiated heat from the person. What is the minimum temperature that would let that person survive for as long as the food and water lasts? Certainly, if the air were about 98 Farenheit, the person would be able to maintain appropriate body temperature forever. But there must exist some temperature where the loss of energy to the surrounding air exceeds the creation of energy by metabolism (or rather, the rate of conversion of food into thermal energy) such that the person is unable to sustain an internal temperature sufficient to survive. Does this temperature depend on the amount of calories provided per unit time, or is there some maximal level of calories per time that the human body can process? Assume that the food and water are provided at the outside temperature -- i.e. no heated food or water is provided. Also, would the provision of clothes and/or shelter decrease this minimum temperature, or simply prolong the period of time before hypothermia sets in? 71.70.143.134 (talk)

The article for thermoregulation states that "In experiments on cats performed by Sutherland Simpson and Percy T. Herring, the animals were unable to survive when rectal temperature fell below 16°C [~60 Farenheit]." This is not necessarily the answer to my question in that humans are exothermic, and so can maintain an internal temperature greater than the exterior given enough calories/time.

I suspect the temperature is surprisingly high. Clearly a person running in circles could maintain a sufficiently warm core temperature in very low air temperature, but the problem is sleep. Eventually that person is going to have to lay down, cease moving (save for shivering I suppose), and go to sleep - at which point their body temperature will plunge. I have no references, so I shall offer no predictions. 218.25.32.210 (talk) 01:11, 22 February 2010 (UTC)

It depends partly on whether the person has lived in the cold for a while. I've read news accounts of people in Russia taking cold showers and fishermen who use bare hands to catch fish from ice holes in -50 degree weather. Imagine Reason (talk) 03:41, 22 February 2010 (UTC)

The OP asked about the minimum environmental temperature for long term survival of a naked human with adequate food and water. This is not a situation that would frequently occur outside medical experiments or torture, since humans have been smart enough to wear clothing in cold environments for tens of thousands of years. Death would be attributed to "exposure." Wind would be a very important variable in determining the drop of core body temperature. Core body temperature likely determine survival of the individual, but temperature of fingers and toes would determine loss of digits from frostbite. A recent TV show had some "Survivor man" replicating the experience of people who nearly died when stranded during a snowstorm, In an experiment, he sat in a cold room, perhaps 23 degrees Fahrenheit (-5C), and the temperature of his core and his fingers was monitored. His awareness and manual dexterity decreased until the experiment was stopped when he seemed to be failing physiologically. Turning on fans to increase "wind" was devastating. In any realistic scenario, the person's clothing, however thin, would act as a windbreak and provide as insulation a dead air layer next to the body. Experimental data I read once (years ago, no reference citation possible) said that when an Australian aborigine and a European were asked to lie down at night on cold ground with minimal clothing, the European's body tried to maintain normal temperature in all body parts and he shivered miserable, with a severe and steady drop in core temperature, while the aborigine's body let the peripheral parts drop in temperature and defended core temperature, letting him sleep peacefully if not comfortably, implying greatly improved survival prospects under the specified conditions (which were likely too warm for frostbite). There is probably some research from military research departments on survival in cold temperatures, but it is not so likely they assumed no clothing, which would be more like some of the torture interrogation methods used by the Bush administration for suspected terrorists. In the U.S., I know of an incident wherein some young Girl Scouts survived overnight with clothing and sleeping bags under windy conditions when it got down to 13 degrees Fahrenheit (-10.5 C). Inadequate clothing is a more likely condition to be experienced. With plenty of food, shivering would exercise muscles and help maintain core temperature in the OP's scenario, but falling asleep would eventually occur and that is when the "death from exposure" would likely occur. (I recall a joke from an early TV show wherein someone tells of an unfortunate photographer who got locked in his darkroom in a building where the heat failed during a cold snap, and who ultimately died of exposure). Edison (talk) 04:57, 22 February 2010 (UTC)

It's unclear what relevance any of this has to my question. Certainly the question is hypothetical and I'm not pushing for the creation of experiments to test any offered hypotheses. Instead, I'm asking if there are any arguments from first principles which might allow one to state x temperature is such that the rate of heat loss is greater than the rate of heat creation by metabolism. Further, frostbite seems to be extraneous to the question, since it can only occur at less than freezing, and the minimal survival temperature is certainly greater than that. I understand why clothing allows us to survive extremely low temperatures, but again, I'm less interested in feats of extreme survival than in feats of prolonged survival.

This is not a direct answer but we are talking about Hypothermia, so the article Hypothermia is a good place to start. --220.101.28.25 (talk) 09:59, 22 February 2010 (UTC)

And I have just noticed that you have already looked at that article (sorry!) this is not an easy question to answer! 220.101.28.25 (talk) 11:07, 22 February 2010 (UTC)

Is wearing a layer of your food considered cheating? I suspect lard might be used for insulation. Googlemeister (talk) 16:52, 22 February 2010 (UTC)

Okay, I think I figured out an answer. This claims that the basal metabolic rate for a human is 100 watts (although it's based on a dead link). So we can lose up to 100 watts to the surrounding environment, and we can calculate the two main ways to loose heat: through radiation and conduction. Radiation is governed by the Stefan-Boltzmann law, $P=\sigma \cdot A\cdot T^{4}$, where σ is the Stefan–Boltzmann constant and A is the radiating surface area. Since the surrounding air is also radiating heat into the person, we end up with $P=\sigma \cdot A\cdot (T_{h}^{4}-T_{a}^{4})$, where T_h is the temperature of the human, and T_a is the temperature of the surrounding air. Conduction is governed by Fourier's law, $P=-kA{\frac {\Delta T}{\Delta x}}$ where A is the surface area, k is the thermal conductivity of the material (.025 for air), $\Delta T$ is the temperature difference, and $\Delta x$ is the distance. I'll pick 5 cm as about the width of the layer of warm air sitting on the skin. So the total energy loss is simply the sum of these two: $P=\sigma \cdot A\cdot (T_{h}^{4}-T_{a}^{4})+-kA{\frac {\Delta T}{\Delta x}}$. Solving this for P = 100 watts, the smallest temperature we can survive indefinitely is about 302 kelvin, or about 84 fahrenheit. This seems very high, but I suppose we did evolve on the savannah. The question is, can someone raise his or her basal metabolic rate while sleeping? Certainly someone can keep warm by aerobic exercise, but I don't know if shivering while asleep has a significant impact. 71.70.143.134 (talk)

Your $T_{h}$ isn't the standard 37 °C; in a cold environment, the skin temperature will be significantly lower even when the core temperature is maintained. Also, you can reduce the effective A by folding the body (basically, go for the fetal position). But you also lose heat through the ground (we haven't said what it's made of) and via convection (which serves to reduce the $\Delta x$; I don't know if your 5 cm is a good guess including that effect or not). Finally, that 100 W is for comfortable humans; though I don't know by how much, shivering and other mechanisms will certainly raise that power when it's needed. --Tardis (talk) 19:59, 22 February 2010 (UTC)

Speed skating

I'm watching the Winter Olympics, and the speed skaters don't run but sway from side to side. I've never skated, so I don't know the mechanics. Doesn't the side-to-side movements make the skates cut through ice that much harder? Thanks. Imagine Reason (talk) 00:13, 22 February 2010 (UTC)

To push yourself forward on ice skates you have to turn one skate sideways and push against it, you then lift that skate up, move it forwards, put it down and turn the other skate sideways and repeat. That is pretty much the only way to propel yourself (at any significant speed, anyway). A running motion wouldn't work since the skates would just slide on the ice and you would go nowhere (unless you put spikes of some kind on the toes that can dig in, which you do see on some skates, but it's not a very effective method for going quickly). --Tango (talk) 00:20, 22 February 2010 (UTC)

That is not how I would describe it. Speed skaters do not move their feet forward to backwards very much or angle the blade "side ways", except for the very first few steps off the start line. Most of the power is driven from moving the legs from the inside to the outside in a sideways motion. Admittedly the blade is angled, but I would not call it "side ways," it is still very much acute to the direction of travel. This is what creates the long and powerful "power stroke" rather then if you move your legs forward and backwards with the blades sideways. Vespine (talk) 01:03, 22 February 2010 (UTC)

The key is that skates will slide frictionlessly along the line of the blades. So a straight walking or running motion, like you would do on dry land, would only produce a cartoonish running in place effect until you eventually fell on your butt a few seconds later. Just about everyone does this their first time on ice skates. APL (talk) 04:26, 22 February 2010 (UTC)

Nuclear isomers

What exactly is a nuclear isomer? The article is way too technical for me to understand, as are most of the articles linked in the intro. I'm going to guess that it's (1) the most stable isotope of an element, or (2) the isotope of an element with the longest half-life, or (3) perhaps most stable = longest half-life, and therefore both 1 and 2. Nyttend (talk) 02:03, 22 February 2010 (UTC)

Neither of those. More like you know how you can expose glow-in-the-dark toys to light and then they give off a low glow for hours. Metastable compounds are charged with gamma rays and reradiate them slowly. By definition if a frequency of gamma radiation reradiates quickly it isn't "metastable". Absorbing or emiting a gamma ray changes the energy in an atom but doesn't change the isotope - gamma rays are pure energy unlike alpha and beta radiation. 75.41.110.200 (talk) 02:45, 22 February 2010 (UTC)

Just to be sure no-one gets the wrong idea here, let's be clear that the reference to glow-in-the-dark toys is an analogy. They actually work by luminescence or phosphorescence, which involve transitions between electron energy states, not nuclear energy states. Glow-in-the-dark toys do not involve nuclear isomers and do not absorb or emit gamma rays ! Gandalf61 (talk) 09:07, 22 February 2010 (UTC)

Here is a very vulgar attempt at popularization, which others are encouraged to improve on and/or correct. The nucleus of an atom contains a certain amount of energy in it just holding it together, like a coiled spring. In most atoms this is the minimum amount of energy required to do so. A nuclear isomer is an atom whose nucleus contains more than the minimum amount of energy. In shifting from this more energetic state to the minimum state, it can radiate a gamma ray or an electron. This does not change its overall proton or neutron count, so it remains the same element, and the same isotope, despite having undergone a form of radioactive decay.

Now, the question of what we mean by the energy that holds a nucleus together (binding energy), and how it can be that some nuclei have different energy than others, and so forth, I think this requires a somewhat deeper quantum mechanical account to make sense of, but if you think of it as just "energy" and don't worry about where it comes from, it is much simpler for these purposes.

It probably should not be confused with different isotopes of an element (same element, different mass, often radioactive and unstable), which is what seems to have thrown you off. ^180m
₇₃Ta
and ¹⁸⁰
₇₃Ta
are the same isotopes of the same element, but they have different energies in their nuclei (the metastable, m one has more), and are thus isomers. --Mr.98 (talk) 04:24, 22 February 2010 (UTC)

One thing to add is that m represents a discrete amount of energy. If an atom absorbs x amount of energy it enters a metastable (somewhat, relatively stable) state which we designate as m. If it absorbs a different amount, say 1/2x or 2x, it decays almost instantly (not even slightly stable!) For some isotopes there will be a large amount of energy y which yield a different somewhat stable stable which we call m2 with a different half-life than m. The amount of energy represented by x and y and the ratio of x/y are all dependent on the particular isotope. Rmhermen (talk) 17:40, 22 February 2010 (UTC)

So different isomers have different amounts of binding energy, and I can always count on atoms of the same isomer to have about the same amount of binding energy? I like the "don't worry about where it comes from" idea — I already knew that there was such a thing, but I don't know anything more than it exists and basically what it does, and I really don't care about it other than that :-) Finally, do I understand you rightly to say that different isotopes would have different isomers, since you'd need different amounts of binding energy to hold together different numbers of neutrons? Nyttend (talk) 21:13, 22 February 2010 (UTC)

Let me have a go at it: consider each nucleon (proton and neutron) has a specific energy associated with it, which contributes to the total nuclear energy. Something like vibrating or rotating in place. The most stable state for a nucleus is to have each one at its lowest energy (the "ground state") possible for that nucleus (bound collection of nucleons). A nuclear isomer has a nucleon at a higher energy state than normal in that nucleus. DMacks (talk) 21:24, 22 February 2010 (UTC)

Birthday

Is there any statistical bias towards a particular day or month of a year when most of the people are born? —Preceding unsigned comment added by Amrahs (talk • contribs) 04:04, 22 February 2010 (UTC)

Footnotes 2 and 3 of the Birthday problem article argue that the answer in both cases are "yes" (months depend on seasons when people conceive, days depends on hospital schedules), but they are not terribly well-sourced (there is a link to this page which has some graphs of the month distribution for 1978-1987, but that's it). Perhaps if people dig up better sources on this they can fix up that citation... --Mr.98 (talk) 04:28, 22 February 2010 (UTC)

There used to be a claim that more people are born in August/September due to the holiday season. While that may have been the case 50+ years ago, it hasn't been so in recent times. The reason that I've heard is simple birth control. People can have all the holiday sex they like and not increase births 9 months later. -- kainaw™ 18:15, 22 February 2010 (UTC)

Salt in the Great Lakes

If all of the Great Lakes are connected to the the ocean, then why don't they have a massive amount of salt in them? JackSlice^Talk _Adds 05:21, 22 February 2010 (UTC)

They are all fresh water lakes, significantly above sea level. All are fed by freshwater rivers/streams. They are drained via Lake Ontario by the St. Lawrence River.-- Flyguy649 ^talk 05:25, 22 February 2010 (UTC)

Thank you very much. JackSlice^Talk _Adds 05:30, 22 February 2010 (UTC)

Great Salt Lake and other terminal lakes accumulate salts because they have no outlet streams (their water leaves mainly through evaporation leaving behind their dissolved mineral load.) Rmhermen (talk) 17:23, 22 February 2010 (UTC)

When the water and other fluids in comets evaporate, do they become hollow asteroids?

Friends and I were trying to understand the implications of Comet#Debate over comet composition and the question came up about whether a comet with a near-sun orbit would eventually gas out entirely and become the rock and dust crust alone. Would it then be a hollow asteroid?

Are partially evaporated comets likely to be partially hollow?

Googling on "hollow asteroid" and "hollow comet" are almost entirely science fiction and space game hits. Thank you! 99.191.75.124 (talk) 06:27, 22 February 2010 (UTC)

Far as I've understood the literature, we think of comets as "dirty snowballs". That is, a mix of ice and rock. So from that the comets would become loose piles of rubble rather than hollow asteroids. Remember that for the rock in the comet to fuse into an compact asteroid with holes, you'd need a certain temperature. That heat would presumable cause the ice to melt away.

That said, I think that what evaporates from the "dirty snowball" is dirty water. In which case the comet gets smaller and smaller and evaporates completely, perhaps with the loose rubble disintegrating. Googling melting or disintegrating comets might turn up some good sources. EverGreg (talk) 08:53, 22 February 2010 (UTC)

Comet tail has a nice illustration of the fact that the evaporated water and the dust form separate tails (either in front or behind the comet) - so it's pretty clear that the dust is being transported from the surface as well as the ice. The answer probably comes down to whether the comet is made up of large rocks or very small pieces - but I think it's clear that the answer is always either a loose rubble pile or there is nothing whatever left at the end. SteveBaker (talk) 13:45, 22 February 2010 (UTC)

Insulin secretion in healthy adults per 24 hours

How much insulin, in international units (IU), is secreted during a 24 hour period in healty (non diabetic, non insulin resistant) adults?
(And how do one measure it?)
--Seren-dipper (talk) 06:58, 22 February 2010 (UTC)

Insulin dosage for a type 1 diabetic is roughly 1 IU/kilogram body weight/day. It varies between individuals and for the same individual between different days due to physical activity, infections etc. but it gives an indication of the body's need for insulin and hence insulin production.Sjö (talk) 12:30, 22 February 2010 (UTC)

That was useful, thank you! (Do you have a reference. So I can back up the allegation?)
Hmm, an indication maybe yes, but then again maybe not. I can imagine that, to some degree, a kind of insulin resistance might be present in all diabetics. Thus, the total, per 24 hour, insulin secreted could be significantly less for a healthy non-insulin-dependent, non-diabetic, compared to the 24 hour need for a diabetic.
So my question still stands.
--Seren-dipper (talk) 20:09, 22 February 2010 (UTC)

The secretion of insulin is a response mechanism, largely (though not solely) dependent on blood-glucose levels. As blood-glucose levels are largely dependent on what a person has (or hasn't) eaten, insulin production will vary fairly widely from hour to hour and day to day. See [here] for more information. A carbohydrate-rich diet will cause fairly high levels of insulin production in a non-diabetic, for example. Bielle (talk) 20:24, 22 February 2010 (UTC)

Uranus change in Orbit

How would the strucuture of Uranus change if it migrated into the habitable zone and stayed there for a couple hundred million years? TheFutureAwaits (talk) 09:34, 22 February 2010 (UTC)

That would depend largely on whether the inner planets are still there or not. If they are, the interactions with them would be highly significant. Ignoring that, the increased temperature and solar wind would strip the planet's atmosphere of some of its hydrogen and helium (I'm not sure how much of it) and the weather would probably get more violent (since there is more energy available). I'm not sure what else would change. --Tango (talk) 16:26, 22 February 2010 (UTC)

I guess more what I'm asking is would water form on it's surface? Does it have the necessary components for life but it's just too cold to work for the moment? You can assume the other planets don't interact with it. TheFutureAwaits (talk) 17:54, 22 February 2010 (UTC)

Did you look at Uranus? From the article: "water clouds are hypothesised to lie in the pressure range of 50 to 100 bar (5 to 10 MPa)". Also from the article: "The abundances of less volatile compounds such as ammonia, water and hydrogen sulfide in the deep atmosphere are poorly known. However they are probably also higher than solar values." Also also from the article: "The abundance ratio of water is around 7 × 10–9." -- kainaw™ 18:01, 22 February 2010 (UTC)

Only Uranus' outer atmosphere is cold, its "surface" is frozen due to pressure rather than cold. If enough of the atmosphere is stripped off the pressure might get low enough for liquid water to form, I don't really know. --Tango (talk) 18:10, 22 February 2010 (UTC)

Human Survival to Exposure

How long could a person survive (with and without injury) to a sudden exposure to each of the planets' surfaces? TheFutureAwaits (talk) 09:35, 22 February 2010 (UTC)

Some planets don't necessarily have what you'd call a "surface", just a gradient of gradually thicker and thicker gases. You'd have to define surface for those. Certainly, for the bigger gas giants, any human would be instantly crushed by the pressure if they were deep enough to be on a solid surface. I don't think Uranus or Neptune have such a surface, though I'm not certain. Vimescarrot (talk) 10:53, 22 February 2010 (UTC)

(edit conflict) Assuming you mean "exposure without any sort of environment suit or breathing apparatus", then

• Mercury - instant death due to heat (light side) or cold (dark side). Pick just the right intermediate spot near the terminator and it may be survivable for, say, 30 seconds before the near-vacuum kills you.
• Venus - instant death due to heat (hotter than Mercury) and pressure (>90 atmospheres).
• Mars - maybe survivable for a minute or two (but don't try this at home, folks !).
• Jupiter, Saturn, Uranus, Neptune - as per Vimescarrot, depends on how you define "surface" for a gas giant, but probably instant death due to pressure/heat.
• Pluto (let's call it a planet for the purposes of this question) - instant death due to cold - max surface temperature colder than liquid nitrogen. Gandalf61 (talk) 11:08, 22 February 2010 (UTC)

So if I wanted to impress my friends on Mars I could hop out of the airlock, run around for 30 seconds, come back in and have a good laugh? TheFutureAwaits (talk) 11:31, 22 February 2010 (UTC)

No, you would likely stagger out of the airlock, being careful not to hold your breath to avoid rupturing your lungs, collapse in agony due to decompression sickness, lose conciousness in 10-20 seconds, and hope that someone drags you back inside and repressurises the airlock before you finally die from hypoxia - see our article on space exposure. Gandalf61 (talk) 11:58, 22 February 2010 (UTC)

Impress them with your silliness perhaps! The pressure gradient is not all that high, but I wonder if Decompression sickness might still be possible? Pressure_suit#Exposure_to_space_without_a_spacesuit may be of interest, as it suggests that very tight fitting 'clothing' makes a big difference. See also Space_activity_suit--220.101.28.25 (talk) 12:10, 22 February 2010 (UTC)

(EC) You included pluto but forgot the third planet from the sun??? Nil Einne (talk) 12:11, 22 February 2010 (UTC)

A sudden exposure to the surface of this planet - depending on the height you'd been dropped from - would result in more or less instant death!--TammyMoet (talk) 12:42, 22 February 2010 (UTC)

Depending on where you arrived, a sudden exposure to the surface of this planet could easily result in death by drowning (in minutes or hours depending on how well you swim) or hypothermia (also in minutes or hours, but depending more on the place). In other places you might die by dehydration (in a few days) or possibly by starvation (in a few weeks) or being eaten (after a variable length of time). --Anonymous, edited 20:23 UTC, February 22, 2010.

Hmm, so are there ANY feats of daring do I could complete on Mars to impress the crowds?TheFutureAwaits (talk) 12:46, 22 February 2010 (UTC)

Just getting there would impress the heck out of me! SteveBaker (talk) 13:39, 22 February 2010 (UTC)

Obligatory xkcd link. Land on Deimos instead, then use a bike and a ramp to launch yourself to Mars. (I get the feeling this wouldn't work, but I've no idea why not...) Vimescarrot (talk) 14:26, 22 February 2010 (UTC)

You could get yourself clear of Deimos, but Deimos is pretty solidly in Mars orbit. You'd have a hard time getting enough of a velocity change to leave Mars orbit (instead, you'd just orbit roughly in parallel with Deimos). — Lomn 15:06, 22 February 2010 (UTC)

Indeed. Deimos's average orbital speed is 1.35 km/s - I can't see a bike getting up to those speeds. You would just gradually drift away from Deimos while staying in roughly the same orbit. --Tango (talk) 16:29, 22 February 2010 (UTC)

Could you survive on Mars if you had a breathing apparatus (Say like a diving mask) on a warmer Mars day? I think it does sometimes get up to 20C in places. Googlemeister (talk) 19:33, 22 February 2010 (UTC)

No. The air pressure is far too low. The air you're breathing has to be roughly the same pressure as the air pressing on the outside of your body. (Otherwise you won't be strong enough to exhale and besides you run the risk of puncturing your lungs when you inhale.) APL (talk) 19:55, 22 February 2010 (UTC)

How is it different then a diver at 10m below the surface? Googlemeister (talk) 21:16, 22 February 2010 (UTC)

I don't think that "Instant Death from cold" is the right answer for Pluto or the Dark Side of Mercury. It may be unbearably cold, but in total vacuum and wearing good shoes, there'd be nothing to conduct your body heat away. I think you'd get your full thirty seconds of agonizing depressurization death. APL (talk) 19:59, 22 February 2010 (UTC)

Pipes

What is the difference between schedule 40 and schedule80 pipes? —Preceding unsigned comment added by Rohanpradhan21 (talk • contribs) 11:40, 22 February 2010 (UTC)

Nominal Pipe Size has a table showing the differences - but essentially, the higher the schedule number, the thicker the walls of the pipe. Hence, for example, for a 2" (DN 50mm) pipe the schedule 40 has 0.154 in (3.912 mm) thick walls while the schedule 80 has 0.218 in (5.537 mm). This allows schedule 80 to support higher pressures and to generally be more robust. SteveBaker (talk) 13:36, 22 February 2010 (UTC)

Chemical differences between E150a, E150b, E150c, and E150d.

Why do they have different E numbers please? 84.13.16.216 (talk) 12:46, 22 February 2010 (UTC)

Because there are differences in the way they are prepared - and hence in chemical composition. SteveBaker (talk) 13:31, 22 February 2010 (UTC)

See Caramel color#Classification for more detail. hydnjo (talk) 15:09, 22 February 2010 (UTC)

It was that article that prompted the question! Do they have a different chemical formula? Are there different chemicals mixed in with them as a result of how they are made? 89.243.87.3 (talk) 19:00, 22 February 2010 (UTC)

L-DOPA is less water-soluble than dopamine?

OK, why does adding a charged group like a COOH group on a water-soluble compound like dopamine end up reducing the solubility? I understand the whole molecule might become zwitterionic, but they carry two formal positive charges which would seem to increase the water solubility. Is it solvent effects and things like water-cage effects? I mean surely the extra charge would make it less soluble in things like ether. John Riemann Soong (talk) 17:37, 22 February 2010 (UTC)

Human Survival to Exposure (on Earth)

Induced by the question above I would like to ask a question that troubles my mind every now and then: how did ice age people survive? Could we nowadays average people at least survive the cold? 95.112.177.38 (talk) 18:07, 22 February 2010 (UTC)

Ice age people survived by use of clothing, Shelter and fire. Dauto (talk) 18:18, 22 February 2010 (UTC)

People didn't live on top of glaciers, and the globe was not covered by ice. You might want to start with Paleolithic#Paleogeography_and_climate. BrainyBabe (talk) 18:33, 22 February 2010 (UTC)

The ice age Earth was only 3 C (5 F) colder than today on average, and that was primarily due to changes at high latitudes. If one lived in the tropics it would be little different than today and you'd still worry about surviving the heat rather than the cold. Dragons flight (talk) 18:36, 22 February 2010 (UTC)

Looking at the map in the Inuit article, this culture increased their habitation of Greenland to cover most of the coast between AD 1300 and AD 1500. Presumably they had something resembling Stone Age technology, and presumably they were not "physically" different form people in sedentary cultures like modern-day Western countries. It follows that they survived in the same way as modern people do - by technology and adaption. That said, mortality in the Ice Age in the cold climates (and elsewhere) is likely to have been much, much higher than today, so for parts of the population, the answer to your question is "they didn't". Jørgen (talk) 20:07, 22 February 2010 (UTC)

That picture from the temperature record article seems to indicate a temperature difference of a about 8 C. Dauto (talk) 20:23, 22 February 2010 (UTC)

Thai (?) translation request

Hi Refdeskers, I got this t-shirt at a thrift store and am wondering what the logo says. I assume it's in Thai, but my ignorance in this area is both deep and broad. Thanks. --Sean 19:02, 22 February 2010 (UTC)

I'm afraid you forgot to post a link to the logo. I'd also recommend you post this to the Language Desk instead of the Science Desk, unless the Thai logo appears to be a chemical or nuclear decay formula. Comet Tuttle (talk) 19:23, 22 February 2010 (UTC)

Egg Whites

Egg Whites

I've been trying to find out about the protein that contains egg whites in it, i did a research and came to find out that by consuming to many raw egg whites can lead to Biotin Deficiency. Egg whites from what a read here at Wikipedia contains high leves of AVIDIN, a protein that binds the vitamin biotin strongly. Now if we cook the egg whites by normal boiling in water, does the protein in the egg whites has the same effects in the human body campare as when consumed raw?71.52.59.50 (talk) 21:01, 22 February 2010 (UTC)

Supervised and Unsupervised Learning

Sorry to post this here. I posted it at the computing reference desk, but it's been ignored. We should relabel the computing reference desk as the "My internet and/or home computer isn't working problem desk". I was reading the supervised learning article. At the end of the overview section it says that the Gaussian Mixture Model is one of the most commonly used classifiers. But the Gaussian Mixture Model article says that a mixture model can be regarded as a type of unsupervised learning. Can something be regarded as both supervised and unsupervised learning (as these links seem to show), or are they mutually exclusive (as I was beginning to believe)? •• Fly by Night (talk) 21:42, 22 February 2010 (UTC)

You should try being more patient and less rude. The Computing Desk is filled with some really smart dudes & dudettes. Duration before receiving an answer to your question depends largely on difficulty x GMT time it was asked. 61.189.63.173 (talk) 21:58, 22 February 2010 (UTC)