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8

I take your question as Is there any substance with condensed (solid or liquid) equilibrium phase at zero pressure? No, because of statistical physics. Let's consider two things. (1) The potential energy of interaction between molecules. (2) The thermal energy distribution for molecules. The potential energy of interaction can generally be of any ...

3

The boiling point of liquids depends on temperature and pressure. If the pressure of the medium that the liquid presents increases, the boiling point of the liquid increases as well. Since perfect vacuum has no pressure, all liquids boil in a perfect vacuum. However, there is no such a thing as perfect vacuum(even space is %99.99 vacuum). If you are asking ...

0

Well - air does have "relative humidity" and this really affects the things that interact with it. For example - you will have a hard time cooling down by sweating when the relative humidity is very high, as the rate of evaporation that you can achieve (and therefore heat rejection) becomes quite low: this is why you end up "sweaty" on a hot muggy day. ...

0

Ion drag is associated with low frequency (RF) EM waveforms, and is what causes resistive heating, not dielectric heating. Dielectric heating (e.g. microwaves) relies on dipole rotation.

1

Interesting question. I suppose one should compare several scenarios Lie still Go forward - either straight, or hard to port, or hard to starboard Go in reverse The rate at which water enters the ship is (to first order) proportional to the pressure differential - lower the pressure and live longer. Maybe even long enough for the Carpathia to come and ...

2

The following picture (from http://hyperphysics.phy-astr.gsu.edu/hbase/waves/imgwav/circonwave.gif) gives you a better sense of how to reconcile your observation with "circular motion": As you can see - there is circular motion for particles at the surface: they don't have to go under water to do it though. Incidentally this also shows that in the trough ...

2

If you assume perfect efficiency, then the energy of dissociation of a liter of water is computed as follows: 1 liter of water, molar mass 18g, => 55.6 moles The energy needed is 237 kJ per mole (from your link - see under "thermodynamics"). 237 * 55.6 => 31.7 MJ of energy for a liter of water. In terms of power, this is 3.67 kW for one hour. This ...

1

We can do some calculations inserting numbers to actually see how is like the force thanks to the water dipole. Let's do the calculations classically. Obviously we need a non-homogeneous field. Therefore, I'll pick a point charge centered in the origin of the coordinates. It can be a uniformely charged sphere as well. Therefore, the eletric field is:  ...

1

There are two ways this can happen. One, the water becomes slightly polarized but net neutral charge. You end up with a dipole and a weak attraction. Two, the polarized water "breaks" so some charge is left behind. This charge can flow back through the main water pipe to ground and the container with the water becomes slightly charged. You could prove ...

3

Without repeating the mathematics of @alemi's answer, let's just think about one other thing: Buoyancy is a function of depth - that is, as you descend your lungs are compressed and your buoyancy decreases, then goes negative (depending on the fat content of your body - if you have enough fat, you remain positively buoyant at any depth since you do not need ...

7

I like Jerry Schirmer's answer, but I was worried that instead of modelling you swimming as fast as you can as a constant force, I thought it would be interesting to consider swimming as fast as you can as a constant power. This seems more logical to me as if you are really trying to go as fast as you can, you will be limited by how hard you work, how fast ...

2

OK, lets solve this in as simple an approximation as possible. I"m going to assume that the whole trip happens at the appropriate terminal velocity, and that acceleration times are very small. Furthermore, I"m going to model the resistant force${}^{1}$ of the water as $F = cv^{2}$, for some constant $c$. On the way down, you have the the resistant force ...

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