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March 10

Cyclohexane smell

Why does cyclohexane smell like acetone? --70.244.234.128 (talk) 00:49, 10 March 2011 (UTC)[reply]

I don't think that they do. Cyclohexane has a sweet smell like benzene while acetone has a harsh smell. 75.41.110.200 (talk) 02:34, 10 March 2011 (UTC)[reply]

It has the same sort of "chemical" smell as acetone (and many other synthetic organic compounds, like paint, paint thinner, etc.). --70.244.234.128 (talk) 00:06, 11 March 2011 (UTC)[reply]

A "fuel savings device" question

I've read the articals here on fuel saving devices and understand that there is limited to no evidence of the value of the devices or the claims made by manufacturers. It appears that virtually all fuel saving devices are either additive based or impact the air flow and fuel mixture. However, I came across this (http://www.gfchips.com/fordf150.aspx) device that uses a computer chip to improve fuel economy, etc. Is there any validity to this company's approach (especially given the old adadage: if it sounds too good to be true, it probably is . . .")? 99.250.117.26 (talk) 01:14, 10 March 2011 (UTC)[reply]

I think if that chip really worked Ford would have done it already. Especially 60 more horsepower? They would love that. Even if this company thought of something Ford didn't, Ford has to be aware of it and would reverse engineer it. But one thing that really stood out is that they have a chip for every car in existence, and I'm sorry but there is no way in the world that is possible. Ariel. (talk) 02:14, 10 March 2011 (UTC)[reply]

It's a scam. See reviews here. --Stephan Schulz (talk) 10:21, 10 March 2011 (UTC)[reply]

I won't speak to this specific chip-model, which may be a scam. But electronic engine control is a real thing, and does improve fuel efficiency. Ford calls this EDIS and VCT; Honda calls it VTEC; and so on; these are special cases of an Engine Control Unit, a sophisticated computer/electronic box that "makes the engine go." The auto manufacturer usually "tunes" the car's ECU software, hardware, and the engine's mechanical systems for the best average performance. However, after-market "engine-chipping" is also a real thing - especially on the performance auto-enthusiast circuit (amateur racers and the like). You can tune for best performance for certain driving conditions. Here's a brief article Chip the engine which should give you an overview of the subject.

Will engine chipping give you 60 extra horsepower? Unlikely. Your engine has peak performance numbers; you can mechanically alter the engine to improve those specs, but not by a whole lot; and playing with your valve-timing isn't going to give you 50% more gas-mileage or horsepower. Custom engine-chipping will do things like moving your torque curve around so you get the car's peak performance at a different engine RPM; and it definitely can change your fuel efficiency (for better or for worse)! And as always, mucking with your car's ECU can be risky - if you screw up, you can permanently damage your car's mechanical systems, requiring costly repairs. Nimur (talk) 16:21, 10 March 2011 (UTC)[reply]

cdma2000

Hi there. Not really a _science_ question, but does anyone happen to know the theoretical ratio between the channel power (1.23 MHz bandwidth) and the RMS power for a cdma2000 SR1 signal? 80.254.147.84 (talk) 02:11, 10 March 2011 (UTC)[reply]

CDMA-2000 is pretty spectrum-filling, so I would "assume" for theoretical purposes that the spectrum is "flat" across the entire channel, and has a specified 3-dB frequency (or specified frequency rolloff rate). Since the signal is a digital QAM constellation with ~64 elements, it is actually pretty nasty when viewed as a pure frequency-spectrum; "theoretically" you should have 64 perfect frequency spikes; but as you of course know, the practical reality is that you have a fuzzy signal whose frequency-spread is determined by your QAM circuit's phase-noise specification. I doubt there's any use to construct a "thorough" theoretical model of the channel power - can you measure it? Otherwise, I'd report the channel power in watts per hertz (or mW/MHz, or other convenient unit), by "assuming" total RMS signal power divided by bandwidth; and spend some time spec`ing the out-of-band frequency rolloff to clean up my estimate. Two app-notes I found on the web search: Linear's AN99 LT5528 WCDMA ACPR, AltCPR and Noise Measurements discussing practical details for measurement; and Tektronix Digital Modulation technical series, Measuring ACPR of W-CDMA Signals with a Spectrum Analyzer. Don't underestimate the power of vendor app-notes - no matter how specific your need, you can probably search the major spectrum-analyzer manufacturers' public libraries, and if you've bought their equipment, you may be able to phone one of their engineers to discuss horrible engineering details with an experienced person. Nimur (talk) 16:46, 10 March 2011 (UTC)[reply]

Thanks for the info. The problem is not so much measuring the signal parameters - I have a waveform that purports to be cdma2000, but which has a 99% OBW of 1.275 MHz (as against the CHP bandwidth of 1.23 MHz). I was wondering if that was correct, and, if not, how incorrect it is. You're probably right that a more specialist arena is a better place for this sort of question, though. 80.254.147.84 (talk) 16:45, 11 March 2011 (UTC)[reply]

entanglement

would entanglement support the conclusion that given two entangled particles at some large distance apart, if one of the particles were to somehow cease to exist that the other would simultaneously cease to exist? —Preceding unsigned comment added by 98.221.254.154 (talk) 05:03, 10 March 2011 (UTC)[reply]

No. Ariel. (talk) 08:09, 10 March 2011 (UTC)[reply]

Presumably you want a longer answer, and to start with particles can't just cease to exist - they always turn into something else. And even if they did it does not mean the other one will too. If anything it would be the reverse - one particles ceases to exist, the other ones doubles. (That doesn't actually happen, but it's more realistic than the reverse.) Ariel. (talk) 08:09, 10 March 2011 (UTC)[reply]

It's fairly misleading to comment on what physics would do if you forced it to violate itself. —Preceding unsigned comment added by 129.67.37.227 (talk) 14:12, 10 March 2011 (UTC)[reply]

I agree - Ariel's first response, while ultra-brief, was probably the appropriate one! Trying to trace out the consequences of a physical contradiction only leads to more bizarre contradictions. Particles can not cease to exist. If you come up with a realistic physical scenario, we can apply some established rules of quantum-mechanics to model what would happen. For example, if a particle decays, then we can model the trajectories of the decay products, and see how the quantum wave-functions would propagate to (and affect) the entangled partner particle. But when you ask a "what if...", all we can meaningfully say is "then our rules don't apply, so we can't answer that." Nimur (talk) 16:55, 10 March 2011 (UTC)[reply]

I thought one of them could be annihilated with antiparticle?? —Preceding unsigned comment added by 165.212.189.187 (talk) 18:32, 10 March 2011 (UTC)[reply]

You appear to misunderstand what annihilation means. Please see Annihilation - at least the introduction. -- kainaw ™ 18:36, 10 March 2011 (UTC)[reply]

Perhaps the OP is thinking of one particle falling into a black hole? I have a meagre understanding of QM but wouldn't then the entanglement be destroyed but not the particle? Blakk and ekka 18:46, 10 March 2011 (UTC)[reply]

No effect of entanglement is detectable with measurements on only one particle. So if you "throw the other particle away" in the sense that you never measure anything about it again, entanglement ceases to matter, though it's theoretically still there. Throwing a particle into a black hole is one way of getting rid of it, but there are easier ways. -- BenRG (talk) 00:52, 11 March 2011 (UTC)[reply]

"Since energy and momentum must be conserved, the particles are not actually made into nothing, but rather into new particles. Antiparticles have exactly opposite additive quantum numbers from particles, so the sums of all quantum numbers of the original pair are zero." OK, So one particle could be made into new particles while the other might not be? —Preceding unsigned comment added by 165.212.189.187 (talk) 19:06, 10 March 2011 (UTC)[reply]

If the two particles are entangled, then any operator that acts on the wavefunction of Particle 1 can cause the wavefunction of Particle 2 to change. That is not synonymous with "the decay of Particle 1 forces the decay of Particle 2." It means that the operator must modify both wave-functions. Tragically, in efforts to "simplify" descriptions of quantum-mechanical treatments, pop-science books do not emphasize the precise nature of this statement: both wave-functions may be affected is not the same as information was transferred, nor the particles are linked together, nor the action on particle 1 caused an action on particle 2. Two entangled particles form a composite quantum-mechanical system. Certain operators must act on the composite system. The composite system's final result will depend on the operator and the composite system's prior quantum state. The state of either particle is a subset of the composite system. Nimur (talk) 20:46, 10 March 2011 (UTC)[reply]

"Any operator that acts on the wavefunction of Particle 1 can cause the wavefunction of Particle 2 to change" is incorrect. There's only one wavefunction, which describes the entire system. If the particles aren't entangled then you can factor the wavefunction into a product of separate wave functions for each particle, but if they are entangled then you can't. That's one way of defining entanglement. -- BenRG (talk) 00:52, 11 March 2011 (UTC)[reply]

Yes or no, can one decay and not the other? —Preceding unsigned comment added by 98.221.254.154 (talk) 00:41, 11 March 2011 (UTC)[reply]

Yes, one can decay and not the other. For example, one could be an electron (stable) and the other a neutron (unstable). They don't have to be particles of the same type. But "decay" doesn't mean "cease to exist". Decaying doesn't destroy entanglement. -- BenRG (talk) 00:54, 11 March 2011 (UTC)[reply]

Sorry, so there could be more than one "particle" in an entangled "pair"?98.221.254.154 (talk) 01:47, 11 March 2011 (UTC)[reply]

Yes, any number of particles can be entangled. -- BenRG (talk) 03:26, 11 March 2011 (UTC)[reply]

Is every particle in our galaxy entangled with every other one? —Preceding unsigned comment added by 165.212.189.187 (talk) 14:41, 11 March 2011 (UTC)[reply]

Yes, I suppose so. (Slightly.) -- BenRG (talk) 09:02, 13 March 2011 (UTC)[reply]

Myasthenia Gravis

What is the right way of presentation "Myasthenic crisis" or "Myasthenia Crisis"? aniketnik 10:06, 10 March 2011 (UTC) — Preceding unsigned comment added by Aniketnik (talk • contribs)

To describe the occurrence of symptoms in a person with myasthenia gravis, you would call it a "myasthenic crisis." --- Medical geneticist (talk) 11:01, 10 March 2011 (UTC)[reply]

Feeling cold in the morning and body weight

Hello. When I wake up in the morning, I have a cold sensation that usually lasts an hour or more. When I mentioned about this to a friend, he said he has experienced the same, and it's related to having a metabolism that effieciently converts extra energy into heat, and that this is probably why we're both thin while we eat a lot and don't get much exercise. Could this be the case? 212.68.15.66 (talk) 10:09, 10 March 2011 (UTC)[reply]

If I were you I'd see a doctor. We're not allowed to give medical advice. --TammyMoet (talk) 20:02, 10 March 2011 (UTC)[reply]

I'm not ill, why would I? I've always felt chilly in the morning, but in fact I've been very healthy, I hardly ever catch any disease. What I'm asking is whether the "super metabolism" could explain the chilly feeling and my apparent "immunity" of gaining too much weight. 212.68.15.66 (talk) 07:50, 11 March 2011 (UTC)[reply]

OP it is quite possible for you to be quite ill without being aware of it! My original advice still stands. --TammyMoet (talk) 13:56, 11 March 2011 (UTC)[reply]

In my opinion you're overreacting, as there is nothing that suggest that the OP is ill in any way. I've always felt a bit cold when I wake up, and I too have been very healthy. Time for some LOGIC: The OP mentions his/her friend has experienced the same, and I also feel cold in the mornings usually for half an hour. Therefore, it's probably rather common, and if feeling cold in the mornings was related to a common medical condition, we would probably know about. On the other hand, it might be a symptom of a very slowly advancing terminal disease... but my money's on Looie496's suggestion on reluctance to leave the warm bed behind. Zakhalesh (talk) 20:46, 11 March 2011 (UTC)[reply]

I can think of two conditions which match the symptoms described by the OP, but our prohibition on diagnosing mean I can't even mention them. Both conditions are common in the Western world, and can remain undiagnosed for many years, silently causing many problems which, if diagnosis had been made earlier, would not have occured. --TammyMoet (talk) 21:23, 11 March 2011 (UTC)[reply]

Since interpreting everything as a sign of a disease is itself a disease, you'd better get yourself to a doctor right away, Tammy. :-) StuRat (talk) 00:45, 12 March 2011 (UTC)[reply]

_{I'm actually speaking to mine on Monday because I suffer from one of the conditions I was writing of... --TammyMoet (talk) 10:04, 12 March 2011 (UTC)}[reply]

I don't know what "a cold sensation" means. Nearly everybody experiences a circadian rhythm of body temperature in which the temperature drops during the night and rises around daybreak. Whether there is anything unusual about your sensations is unclear without a better description. Lots of people are averse to getting out from under warm covers on a chilly morning. (I don't consider this medical advice.) Looie496 (talk) 23:02, 10 March 2011 (UTC)[reply]

A few points:

1) If you turn the temperature down at night (or have an automatic thermostat that does so), it may actually be colder. Even if the temp is turned back up by the time you get out of bed, it still may take some time for the house to get up to the new temperature.

2) Activity warms you up, and a bit of exercise in the morning might do the trick.

3) It might be warmer in your bedroom, particularly under the covers, due to your body heat overnight, and walking into another room may cool you down. This seems to especially be a problem with bare feet on uncarpeted floors, as in the bathroom or kitchen. A cold toilet seat will also wake you right up. StuRat (talk) 00:50, 12 March 2011 (UTC)[reply]

It may just be a side-effect of wearing light clothing. I like to keep my bedroom cool. When I get up in the morning at this time of year, from under the warm bedclothes, I feel cold until I put on some warm clothing. You may just be not wearing warm enough clothing. Your body has to use extra energy to keep itself warm, hence you do not get fat. Feeling cold in the morning is a small price to pay for being perpetually slim - lucky you. 92.15.11.100 (talk) 13:23, 14 March 2011 (UTC)[reply]

Will Global Warming cause the Moon recede faster or slower from the Earth?

Global Warming will cause sea level rise. This will have some effect on the tidal friction the Moon experiences, causing the rate at which it recedes from the Earth to change a bit. Will this rate increase or decrease? Count Iblis (talk) 15:02, 10 March 2011 (UTC)[reply]

It will recede faster. With more liquid water instead of solid, the tidal friction will increase, so the moon's recession will occur faster. For partial support of that statement, see the second paragraph of Tidal acceleration#Angular momentum and energy. Red Act (talk) 15:24, 10 March 2011 (UTC)[reply]

On the other hand, by melting ice into water some of the mass that is now near the poles will move closer to the equator changing earth's Moment of inertia slowing down its rotation and therefore increasing the time between a high tide and the following low tide reducing the rate of the motion of the tides around the planet and decreasing the pace with which the moon is moving away from the earth :). Dauto (talk) 01:45, 11 March 2011 (UTC)[reply]

And note that any effect would be so slight it would be difficult or impossible to even measure. StuRat (talk) 00:39, 12 March 2011 (UTC)[reply]

Hmmm, wouldn't the change in total lengths of coast lines be a factor here, as this ultimately provides for the friction that moves the tidal bulge ahead of the Moon? It seems to me that this could change quite dramatically if e.g. all the ice on Antarctica were to melt, but I'm still not sure if this woild increase or decrease the recession rate.Count Iblis (talk) 15:28, 12 March 2011 (UTC)[reply]

Chemical catalog

In a chemical catalog item description, what is the meaning of something like (N/10)? For example, "Hydrocholric acid solution, 0.1N (N/10)". I am looking specifically at Cole Parmer and I've been unable to find any explanation in their information. ike9898 (talk) 16:56, 10 March 2011 (UTC)[reply]

N means 'normality'; it's a way to express concentration. See Concentration#Normality for details. TenOfAllTrades(talk) 17:53, 10 March 2011 (UTC)[reply]

Well, I understand that, but is "0.1N (N/10)" any but just completely redundant? ike9898 (talk) 19:30, 10 March 2011 (UTC)[reply]

Why would redundancy be bad? The catalog wants to sell to people who use different methods of expressing concentration, and it's in their interest to make the concentration instantly understandable to both sets of people. - Nunh-huh 20:42, 12 March 2011 (UTC)[reply]

Solution of NaOH at 40% m/m

I need to prepare a 100ml solution of NaOH at 40% m/m (in water). As NaOH absorbs the humidity in air, I have consulted an old Handbook, section sodium hydroxyde, concentration properties of, in varying concentrations but I am still unsure as to how much NaOH (in grams) I need. Please help.

Thank you very much. --192.197.51.41 (talk) 18:06, 10 March 2011 (UTC)[reply]

First, the preparation of weight/weight solutions is tackled by our RefDesk's foremost competitor here.

I should add that I'm a bit surprised to see "m/m" or "mass/mass" coming up. It seems from searches that this is in vogue for recent homework problems. On the one hand, yes, the number on the balance is in grams of mass; on the other, what the balance measures is weight.

Now the real question you have is what the purity of the NaOH is. You could run various tests on it, or try to purify it e.g. by baking it at high temperature to drive out the water, but the traditional chem lab way to handle the problem is to titrate the sodium hydroxide solution after you've made it, against a known standard. In this way you can adjust the molarity to be exactly right, and therefore the m/m value. (Though usually people titrate to get an exact molarity rather than an m/m value) Wnt (talk) 20:53, 10 March 2011 (UTC)[reply]

Thanks, I got it figured out. --192.197.51.41 (talk) 22:20, 10 March 2011 (UTC)[reply]

Does copier or printer ink contain metal?

I was wondering whether photocopier ink or laser-printer ink contains any metallic substances or metal compounds?

Thanks. 71.252.113.85 (talk) 20:37, 10 March 2011 (UTC)[reply]

It depends on the ink. Our article toner may help. Most toner is pure carbon (say, graphite powder), or something similar (like carbon black), but may also include a plasticizer, a binder, one or more liquid solvents, and so on. In general, these are synthetic polymers. Colored ink may contain pigment, which may include synthetic polymers, or metallic pigments. Nimur (talk) 20:51, 10 March 2011 (UTC)[reply]

Is it possible for the polymers in black copier or printer ink to be magnetic? 71.252.113.85 (talk) 03:03, 11 March 2011 (UTC)[reply]

Definitely possible to formulate a magnetic ink--see MICR. DMacks (talk) 18:37, 12 March 2011 (UTC)[reply]

Xerography in photocopiers often use the metalloid selenium, which is not likely found in the ink. ~A H 1^(TCU) 02:28, 13 March 2011 (UTC)[reply]

Probability cloud of a free electron?

Would the probability cloud of a free electron look the same as an electron in the ground state bound to a hydrogen atom? ScienceApe (talk) 20:40, 10 March 2011 (UTC)[reply]

No, not as far as I can tell. Until you get a better answer, consider these related questions: why would you think that the electron_cloud would look the same in both cases? Compared to a 'free electron' (presumably with no other particles around), an electron bound in hydrogen has at least a few additional forces acting on it, especially the electric field generated by the proton. Shouldn't this difference in forces change the shape of the probability cloud (and perhaps radically so)? SemanticMantis (talk) 22:40, 10 March 2011 (UTC)[reply]

No --Gr8xoz (talk) 22:31, 10 March 2011 (UTC)[reply]

Without a nucleus to bind the cloud, the autorepulsion will cause the cloud formation to disolve. Plasmic Physics (talk) 22:41, 10 March 2011 (UTC)[reply]

Here's the problem. You can't formulate a wavefunction for an electron in free space. Let's trace out how we do it for a simplified quantum mechanical treatment of an electron in an Hydrogen atom: (see our article on Hydrogen atom, particularly the solution of the wave function.

First, define a coordinate system by assuming the proton to be heavy and therefore immobile. (The hydrogen proton is located at the origin).
Now, define the Coulomb potential - for practical purposes, this is $1/r$ , where r is the radius from the origin. (How can you define the coordinate-system of the free electron, if there's no reference origin? You have to pick an origin at random, which isn't a problem, but then there's zero potential energy gradient, and so the electron has a probability of being anywhere relative to your undefined origin).
Solve the Schrodinger equation using the Coulomb potential and normalize using physical constraints (probability that the electron exists somewhere must equal 1).

So, in free space - you don't have anything to define your wave-function, and the electron can be anywhere. The only thing you can do is measure an electron - determine its current state (its momentum and its position, subject to the uncertainty principle), and then you can predict where it will be next (which is, of course, anywhere, since it is unbound). When you measure it again, at some other position/momentum, the principle of indistinguishability means you can not know whether you've measured the same or a different electron! So, there's no way to predict the position of an electron in an unbound system (no defined coulomb potential). It has an equal probability of being anywhere; and anything you measure has a likelihood of being a different electron. Nimur (talk) 23:29, 10 March 2011 (UTC)[reply]

This is completely wrong. See Wave packet. Truthforitsownsake (talk) 02:17, 11 March 2011 (UTC)[reply]

Well, in actuality, I never apply quantum-mechanics when I'm describing free electrons. Most of the time, I can treat my plasmas as a superposition of an electron gas and an ion gas, and solve using classical or relativistic electrodynamics. By the very nature of the involved length, time, and energy scales, quantum-mechanical properties are less relevant to free electron clouds, in practice. I'm not surprised that my attempt to describe a quantum mechanical system for an electron gas sounds awkward. I respectfully disagree that my explanation is "completely wrong" - I think it's completely correct, although it's surely "stretching the limits of the conventional application of quantum mechanics." Nimur (talk) 14:54, 11 March 2011 (UTC)[reply]

The hydrogen atom orbitals are a complete basis for smooth electron wave functions that approach zero at infinity, so they don't tell you anything about the electron by themselves. An electron that's nowhere near the nucleus can still be described by a sum of these solutions. The only thing that's special about them is that they're energy eigenstates. If you remove the nucleus, leaving the background potential zero everywhere, then it's no longer possible to write down a basis of energy eigenstates, but I think that's just a pathological corner case. You can still work in some other basis like a wave-packet basis, or you can write down energy eigenstates if you add, say, a single nucleus a billion light years away. -- BenRG (talk) 02:35, 11 March 2011 (UTC)[reply]

magnetic earth

will we ever know for sure what is inside of the earth or if it is hollow will we ever know for sure what causes the earth's magnetic field — Preceding unsigned comment added by Lufc88 (talk • contribs) 21:41, 10 March 2011 (UTC)[reply]

I think we're pretty sure about the structure of the Earth. Is there any reason you don't believe the scientific explanations presented in that article? Geology and geophysics are evidence-based sciences, and there is a lot of evidence to back up the current theories. Have a read at those articles for details on the experimental toolkits available to help scientists gather evidence about Earth's interior. Nimur (talk) 21:46, 10 March 2011 (UTC)[reply]

We will never know anything about the world for sure, we could just be hallucinating or in a computer simulation. We have a rather good model for the structure of the earth, we may never know some finer details but as long as we can trust the measurements the overall structure are known, it is not hollow and could not be hollow since no known material would be strong enough to keep it from collapsing. --Gr8xoz (talk) 22:41, 10 March 2011 (UTC)[reply]

For why we know the Earth isn't hollow, see Hollow Earth#Contrary evidence. Also, the interior of the Earth has been mapped in detail using seismic tomography, without any evidence of any hollow space; see Seismology#Seismic waves.

As per Earth's magnetic field#Field characteristics, the currently accepted theory is that the Earth's magnetic field is primarily due to electric currents in the (liquid) outer core, similar to an electromagnet. This is called the dynamo theory, although a numeric model of the Earth's magnetic field consistent with the dynamo theory has not yet been achieved. Red Act (talk) 22:45, 10 March 2011 (UTC)[reply]

how can we prove that it is the dynamo effect any ideas — Preceding unsigned comment added by Lufc88 (talk • contribs) 22:49, 11 March 2011 (UTC)[reply]

The main way is by comparing Earth with other (terrestrial) planets and moons which lack a magnetic field. Venus doesn't appear to have one, and Mars appears to have a relatively weak one, as does our Moon. So, if these lack the molten iron outer core of the Earth, that's fairly good evidence that this is the reason. The Moon shows little evidence of recent volcanism, so that supports it being solid, not liquid. Mars shows signs of massive ancient volcanoes, but less activity more recently. More study is needed there. Venus is difficult to study, due to it's thick cloud cover and corrosive atmosphere. StuRat (talk) 00:31, 12 March 2011 (UTC)[reply]

@@ Line 99: / Line 99: @@
 :3) It might be warmer in your bedroom, particularly under the covers, due to your body heat overnight, and walking into another room may cool you down.  This seems to especially be a problem with bare feet on uncarpeted floors, as in the bathroom or kitchen.  A cold toilet seat will also wake you right up.  [[User:StuRat|StuRat]] ([[User talk:StuRat|talk]]) 00:50, 12 March 2011 (UTC)
+:It may just be a side-effect of wearing light clothing. I like to keep my bedroom cool. When I get up in the morning at this time of year, from under the warm bedclothes, I feel cold until I put on some warm clothing. You may just be not wearing warm enough clothing. Your body has to use extra energy to keep itself warm, hence you do not get fat. Feeling cold in the morning is a small price to pay for being perpetually slim - lucky you. [[Special:Contributions/92.15.11.100|92.15.11.100]] ([[User talk:92.15.11.100|talk]]) 13:23, 14 March 2011 (UTC)
 == Will Global Warming cause the Moon recede faster or slower from the Earth? ==