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The pressure will decrease for two reasons. The pressure inside a bubble , $P_b$, is $P_b=P_{ext}+P_{ten}$, where $P_{ten}=2\gamma/R$ is the pressure added by the surface tension, $R$ is the radio of the bubble. As the bubble goes up: 1)the external pressure diminishes and 2) The surface tension diminishes too because as the internal pressure diminishes ...


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I believe the confusion is that you believe pressure will always increase as temperature increases. This is only the case in a closed environment such as inside the tire. In an environment such as the atmosphere which is, essentially, in an unconfined environment, the density will decrease with temperature as well. This does not happen inside of a closed ...


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When you open the tube, yes, air will be drawn into it. This is indirectly due to gravity, in the sense that what will drive the flow is a pressure gradient. The pressure in the tube is lower than the pressure outside (after you pump air out and let things settle down, this is true no matter where you open the hole), so air is drawn in. Gravity has a role in ...


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I once had the same problem understanding. But one day I realized pressure is just a measure of energy and Bernoulli's law is just another way of expressing the conservation of energy. The analogy: Total Pressure = Dynamic Pressure + Static Pressure >> Total Energy = Kinetic Energy + Potential Energy. In either case we assume no losses, or otherwise ...


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Since you know the force on the bottle (roughly 200N), you would have to get an approximate area over which this force is distributed. You could try by spreading some ink on either the bottle or the weight to estimate the contact area. While this is not completely correct (the wall of the bottle does redistribute the pressure on the outside to a larger area ...


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You are confusing pressure and force. Yes, the pressure is only a function of the water depth. However, the resulting force is pressure times area. If you have 2 feet of water in the tank, then the pressure at the bottom is .86 PSI. Since the pressure varies linearly and is 0 at the top, the average pressure on any vertical strip of wall is .43 PSI. Now ...


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What you observed is known as the Hydrostatic Paradox. You are confused that the force exerted onto the tank wall corresponds to the weight that is larger than the weight of water contained in the tank. You are probably familiar with levers: you can put a small weight onto the long arm of the lever and thereby exert a large force at the short arm. The gain ...


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The ideal shape for the water to just 'hang' is actually the cube that you've proposed. In this case all of the forces on the water are uniform across the interface. For one region to slip down, another needs to move up (as you've indicated in your diagram). But since every location is experiencing the same forces, there's no reason any spot to start ...


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Because what you are doing is a flow process, with mass inflow and no mass outflow, you need to use the thermodynamic equation: $dU_{cv}={m_{in}d}{H}_{in}-{m_{out}d}H_{out}+\delta Q-\delta W_{shaft}$ If you insulate your air cylinder well enough, $\delta Q = 0$. Assuming that your air cylinder does not deform, $\delta W = 0$. Since you are filling your ...


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The energy change a gas is expressed by: $$dE = TdS - pdV +\mu dN$$ Admittedly, the process of filling a cylinder is rather complicated because several variables in this equation vary, and the way you fill it will influence their behaviour, but you can see that the result will give a production of heat. When you fill the cylinder you are connecting it to ...


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Use this equation: $$\frac {P_1V_1}{T_1}=\frac {P_2V_2}{T_2}$$ Before, the pressure is 1 atm, the temperature is about 20°C and the volume is many cubic meters. After, the pressure is above 100 atm, the volume is about 1 cubit meter, and the temperature is significantly warmer, say 100°C. Filling in the equation: $$\frac {(1atm)V_1}{393K}=\frac ...


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I would explain it by simply using the first law of thermodynamics: $$Q-W=\Delta U$$ $Q$ is heat added to the system, $W$ is work done by the system, $U$ is the internal energy. Keep in mind that internal energy is closely bound to temperature, so a change $\Delta U$ also results in a temperature change (which is what we are talking about). If you look at ...


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The point where this pressure law brakes down is when the surface tension of the water starts to play a leading role. You can try this yourself with a narrow tube (I would say make it considerably smaller in inner diameter than a sixteenth of an inch). If you submerge the tube in water and you pull it out, the molecular forces between the water molecules and ...


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No. The answer is clearly no. This building is 800 meter high. Some comparison: Skydivers are falling more kilometers in free fall. They experience absolutely no damage from the pressure increase. Scuba divers moving fast upwardly or downwardly also don't get any wounds, although 10 meter deep water has the same pressure as there is between the sea level ...


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Here's a simple, non-mathematical, answer. Although the pressure at the surface does depend on the velocity of air molecules that's not the whole picture. It is more precise to say that it depends on the rate of collisions. The collision rate depends on the velocity of the molecules, i.e. the temperature. But it also depends on density of molecules. Higher ...


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Why does the air pressure at the surface of the earth (resulting from collisions of molecules on the surface of the earth which has to do with the velocity of the particles) exactly equal the weight of the entire air column above it (which just has to do with the number and mass of the molecules in the air column)? That's not exactly true. Deviations ...


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Suppose the pressure at the Earth's surface is $P$. Consider an air column of cross-sectional area $A$. The upward force on the column is $F_{\text{up}}=PA$. Denote the weight of the column as $W$. By definition of "weight", the downward force on the column is $F_{\text{down}}=W$. Suppose the pressure is too low, such that ...


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You have a mistake in the line $$ΔV= (.0314m^2)(.002m)$$ Note that 2 cm = 0.02 m, not 0.002 m...


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How do slight changes in these properties result in a large change in pressure, microscopically? Slight change of volume is not so easy to accomplish for solids - it takes a great force to achieve it. Considerable external force applied by different body (wall) needs to be maintained. The pressure is a measure of this force per unit area and since the ...


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When a ball bounces lower each time, it loses kinetic energy. This is a result of internal friction - the kinetic energy is converted to internal heating. When two atoms bounce off each other inelastically, where does the energy go? I can think of two mechanisms. The first is heating. In that case, one of the atoms must get the energy, because that's how ...


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Sorry for the enormous delay, I was caught with more work than I thought, then, here it is. @DanielSank, relativity is not necessary, it would help with what you said, since it would pinpoint exactly you are calling momentum in your system. My answer would be an extension of DanielSank comment. When there is the conservation of a continuous quantity, the ...


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The simplest possible answer is that in a closed system the lowest entropy state is one where the temperature is (statistically) uniform.


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As the comments to the question have stated, in real gasses ( contrasted to ideal gasses which just bounce around elastically) there exist both elastic and inelastic scatterings controlled by quantum mechanical interactions. Photons are generated leading to what we call Black Body radiation and an isolated gas volume will lose energy according to the ...


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Yes you are right. The 10kg piston acts as a force over the area of the piston, increasing the pressure and decreasing the volume of the gas inside. When the set up is tilted, the force no longer acts on the gas, but sideways, so the pressure equalizes.


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The rotation is part of the key to the storm itself. Primarily the pressure and temperature differences are what causes these systems to take the shape and forms that they do. Once a tropical depression starts to form you can already see rotation in the moisture around the low pressure zone, even through it typically looks nothing like a hurricane. Not ...


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I answer my own question and give a good thanks to DavidPh, who has not really gave the answer, but in fact, it was impossible for him to give it. Here is "why": I'm French, so I've many fire hydrant data but from France. And when applying them to the formulas, the result was wrong... In fact, the problem is not the formula but the way we measure the ...


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Like any object moving through a fluid, a high-speed train distorts the air as it moves through it. Broadly speaking, there are three main regions of flow structure around a high-speed train: the upstream distortion, boundary layer and wake. These can be collectively referred to as the slipstream. The effects of the slipstream on a static observer (e.g. a ...


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Therefore could one say that the dust actually has a negative pressure p? In mechanics of solids, negative pressure (positive tension) means the internal forces resist expansion of the body due to external forces. If you have dust (rarified set of particles) in a syringe acting on each other with non-negligible gravitational forces, the gravitational ...


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The discrepancy is that the pressure as measured by the Pitot tube is not just the kinetic energy term of the pressure, but instead is a combinaiton of static pressure and the kinetic energy term. See if pages 16-34 of the following reference are helpful, though not metric: http://www.southsaltlakecity.com/uploads/documents/%5E_Fire_Flow_Calculations.pdf ...


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No, there is no contribution to the pressure from the gravitational attraction between the particles. To see this you need to appreciate that the pressure is an ensemble property, and look at the stress-energy tensor for a single point particle. This is: $$ T^{\alpha\beta}({\bf x},t) = \gamma m v^\alpha v^\beta \delta\left( x - x_p(t) \right) $$ where $v$ ...


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The front of the train compresses air which can blow you away, while at the back of the train air rushes back in after the train has displaced it. This backdraft is especially troublesome in closed areas such as subways, where a train exits a small tunnel near a platform and the displaced air rushes back into the vacated tunnel. Next time you see a big truck ...


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First and foremost, a primer over what "buoyancy" is is needed. Pressure decreases with altitude. The atmospheric pressure at the top of the balloon is a tiny bit less than the atmospheric pressure at the bottom of the balloon. This pressure difference results in a tiny net upward force on the balloon. The balloon rises if this tiny net upward force exceeds ...



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