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The trick is that you're not just moving air, you're also increasing your volume. While the air is in the cylinder, it is at a very high pressure. A given amount of air takes up a rather small amount of space. When you push air into the BCD, you let it expand, decreasing it to a lower pressure (equal to the pressure of the water, actually). Now that same ...


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Using the ideal gas law: $P(h) = P_0 + \rho_w g h$, ${V_o P_0} = {V(h) P(h)}$ which gives $V(h) = {V_o P_0}{1 \over P_0 + \rho_w g h}$.


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As you go deeper, the pressure increases, decreasing the volume of the gas in the bubble. In water, the pressure is appoximately $14.7(1+\frac d{33})$ psi where $d$ is the depth is measured in feet. The volume goes as the inverse of the pressure.


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The complexity of the molecule, compared to H2 and O2, makes its thermodinamical properties completely different, up to the point that it is even in a different phase at room temperature.


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No, liquid air is not flammable. A flame requires an oxidant and something that can be oxidised. Liquid oxygen is a (very strong!) oxidant but neither nitrogen or argon are easily oxidised. The density of air is about 1.2 kg per cubic metre at room temperature. I don't know the density of liquid air, but the density of liquid nitrogen is around 800 kg/m$^3$ ...


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As Nogueira stated, the air is made up of more than one particle. If the idea you're using is that a force occurs between two objects, then you'd have to treat each air molecule as an individual object, and you'd have many, many forces, each between the skydiver (who you could treat as a single body) and an individual air molecule. Of course, this gets messy ...


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Yes. The air is a gas made by a lot of particles. The air resistence takes the energy from the skydiver and transform into heat. In other words, the molecules that made up the gas hits the skydiver, getting with some kinetic energy of the skydiver.


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The pressure of the tires would decrease because when the car is lifted, the force on the tires from the mass of the car decrease, causing the area of the tyre to increase. Since the formula for pressure is PRESSURE=FORCE/AREA, the bigger the area of the tyre, the less the pressure of the tyre.


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Well, I don't know if this is as specific as you asking, but the pressure in the tire has to balance the external forces acting on the tire. In this situation, there are two to consider: Air pressure pushing in on the tire and the amount the earth pushes up on the tire (3rd law equivalent of the weight of the car). When the car is up on a lift, the only ...


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They can sometimes have this shape as they stretch downward from a faucet, as seen in this video, however, surface tension usually pulls the droplet into a roughly-spherical package. The reasons that raindrops have "streaky" trajectories in photographs is not that the raindrops are distended by air resistance, but rather that the short exposure time of the ...


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The diagrammatic representation of rain is indeed incorrect. Raindrops do not resemble teardrops. This Youtube video by Minute Physics provides a really cool explaination of that: http://youtu.be/8lBvC7aFB40


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If you measure with sufficient accuracy, you will see a difference. Of course, any object you use to measure the air temperature might be exposed to sunlight and therefore experience significant difference in "apparent" temperature; but if we assume that you have a well shielded device that admits air but is insensitive to the effects of radiated heat (from ...


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The balloon will asymptotically approach the wind velocity. As it gets close, the force due to drag will decrease because the apparent wind will be small. To the extent it matches the wind velocity, flags will hang straight down. To the extent it is (barely) below wind velocity, the flags will point (just a little) in the direction of the wind. If the ...



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