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4

The pressure at any point is the total weight of everything above an area of a square metre. So for example the atmospheric pressure at Earth's surface, 101325 Pa, means the total weight of atmosphere above a square metre at the surface is 101325 N. So when you ask for the pressure at the centre of the Earth, the way to calculate this is to work out the ...


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Potential energy has absolutely nothing to do with stress-energy or pressure. The following reference is a good source about the origin of the pressure term in the stress-energy tensor: "Momentum due to pressure: A simple model" by Kannan Jagannathan in American Journal of Physics 77, 432 (2009);  http://dx.doi.org/10.1119/1.3081105 Potential energy ...


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The can is acting as a compact heat pipe. When you grasp the can, the liquified gas inside the can absorbs the heat from your hand and expands, carrying the heat away in convective currents that dissipate the heat to other parts of the surface that are at a lower temperature than your hand. Since it draws heat from your hand it feels cool to the touch.


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I can't speak to what sources you were reading, but the air under a hovercraft is higher than ambient air pressure, not lower. Wikipedia: Hovercraft use blowers to produce a large volume of air below the hull that is slightly above atmospheric pressure. The pressure difference between the higher pressure air below the hull and lower pressure ...


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The force would be proportional to the area, so the force on the hole would be 10^(-5) N


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You have to find out what kind of statistical ensemble you are dealing with. As soon as you know that, you can get the corresponding thermodynamical potential from the knowledge of the partition function. When you know the potential, you know everything! EDIT: Since you don't know to which ensemble this partition function $\mathcal{Z}$ belongs to you ...


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The required change in pressure for a change in melting point can be found from the phase diagram of water. The typical variations in atmospheric pressure are negligible, just as you neglect the additional water pressure experienced in the lower parts of the ice bath. I will leave it up to you to find the factor involved - it is the slope of the water/ice ...


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Static pressure in a compressible flow depends on the density but not the speed (not directly). Speed and geometry may affect the density. For isentropic flow (neglecting gravitational potential): $$ {p \over \rho^\gamma} = constant, \gamma = {c_p \over c_v} $$ which could be turned into this: $$ {p \over p_0} = ({1 \over 1+{(\gamma-1) \over ...


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Take an empirical approach; go to a oil-change place. Put the car up on a rack/lift (the kind that supports the frame, not the tires - allows tire rotation during oil change). Have them adjust tire pressure while car is up in the air; have them remeasure when the car is back on the ground (tell them why - give them "scientist for a day" certificate). Because ...


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If the temperature of the gas is kept constant during the compression then the bulk modulus of an ideal gas is just equal to the pressure. The definition of the bulk modulus is: $$ K = -V\frac{dP}{dV} \tag{1} $$ For an ideal gas $PV = RT$, so $P = RT/V$. If the temperature is constant this gives: $$ \frac{dP}{dV} = -\frac{RT}{V^2} \tag{2} $$ and ...


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It sounds like you are not asking about Bernoulli's principle, which describes energy conservation in a fluid, but about why fluids move faster in the thin section of a pipe. This is not Bernoulli's principle, it is just something someone might have mentioned when talking about Bernoulli's principle. Suppose the liquid moves the same speed everywhere in the ...


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Initially, I agreed with Olaf Chujko's answer to this question; however, on further reflection, I think the most accurate answer is 'it depends': Firstly, from the schematic that is given, when switch A is pressed, the cylinder volumes above the two cylinders will be vented to a reservoir at ambient pressure (trust me, I work in Oil & Gas and I look at ...


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The differential amount of force is given by: $$ dF = \Delta p* dA $$ You do need to add up the vertical components, which are given by: $$ dF_\downarrow(\theta) = \Delta p * dA * \sin(\theta) $$ This goes from $0 \leq \theta \leq \pi$. At each value of $\theta$ the differential area is the same: $$ dA = rd\theta * dx $$ This gives that: $$ ...


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How would you solve this for a single small hole? What happens if you now move that hole down by a small amount? What if you add up the contributions of all these holes? Congratulations, you just integrated the expression for the flow rate over the aperture.



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