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6

From the Wikipedia article on sound: In physics, sound is a vibration that propagates as a typically audible mechanical wave of pressure and displacement. To fully understand how is air vibrating in an open pipe, you have to consider not only the acoustic pressure wave, $$\frac{\partial^2 p}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 p}{\partial ...


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You're wondering why pressure nodes form at an open end of a tube. The answer is, they don't! It's just a reasonably good approximation. Physically, consider the air molecules at the center of the tube. Since they're far away from the edges, there's no way for them to "know" exactly when the tube ends, so the sound wave must "leak out" slightly. The ...


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Assuming the deck doesn't bulge to much at at the sides (as the video suggests), the press exerts work $W$ on the deck acc.: $$W=\int_0^yFdy$$ With $F$ the force exerted by the press and $y$ the displacement. This work is converted to potential energy and stored in the deck. As suggested in the answer to this question, the high pressure is likely to ...


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If the parts are perfectly solid and they are at equilibrium, then the pressure is constant (ideally). $$q(x,y) = \frac{\int p(x,y)\,{\rm d}A}{\int \,{\rm d}A} =\frac{\text{applied force}}{\text{area}}$$ But if either the floor or the body are ever so slightly elastic then the pressure distribution is given by a non-hertzian contact which concentrates all ...


3

The effect of temperature is included in determining the density as a function of temperature and pressure via the equation of state. For example, in the case of an ideal gas in hydrostatic equilibrium, the density is related at each point in the gas locally to the temperature and pressure by $$\rho=\frac{pM}{RT}$$where M is the molar mass. So the ideal gas ...


2

Yes, it is independent in the case to which I believe you are referring, in which one has a container of liquid which is open to the atmosphere here on Earth, and the system is in static equilibrium. However these conditions are merely a special case in which the various effects of temperature cancel each other out perfectly. There are many possible ...


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Other answers have handled the oxygen issue quite well. In regards to the temperature, space itself has no temperature because it's a vacuum. Objects in space, however, do have a temperature. If a human is exposed, unprotected, to space near the sun (or any other star), the temperature change in their body could very well be terminal. Even near our ...


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The book answer looks wrong to me. I think $k$ can only appear in combination with $x$, since the tension $T=kx$ in the spring is relevant here but $x$ on its own is not. It could be relevant as a geometrical factor, but there is no length marked $x$ in the diagram. I agree with your answer - except that you should have $k$ in there with $x$. I do not ...


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In simple terms, the glass is being constantly shaken by weather. There isn't very much difference in mass between Oxygen and Nitrogen and the air at low altitudes is dense enough that molecules collide after a very short distance and so the atmosphere is very well mixed. At very high altitudes where the density is very low there is no weather, and little ...


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The other way around is more intuitive; if the pressure is lower on the right, the fluid would feel a net positive force in that direction and accelerate toward right. hence it will have higher velocity there. So, lower pressure will result in higher velocity. you can rephrase the above in a way that it sounds as what you may like but is not scientifically ...


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Since you have an equilibrium situation the net force on any portion of the water is zero. In region $B$ the horizontal forces acting on the water are either provided by the walls of the container in the left-hand diagram of by the water in regions $A$ and $C$ in the right-hand diagram. So as far as the water in region $B$ is concerned it makes no ...


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(b) is correct. Your equation applies only if all the fluid is the same fluid. Suppose that, as in your problem, E = gz, where z is the elevation above the base. So, the component of your equation in the z direction becomes $$\frac{dp}{dz}=-\rho g$$ Because points 3 and 4 involve the same fluid, the pressure at these two points is the same, say ...


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Gage pressure readings are taken assuming that $0$ gage pressure = $1$ atmosphere. By specifying a vacuum of $25~\textrm {kPa}$, that suggests that the gage pressure is $25~\textrm {kPa}$ lower than local atmospheric. When you want to find absolute pressure, then $P_{abs} = P_{atm} + P_{vac} = 70.7 - 25 = 45.7~\textrm {kPa}$.


1

Why do we not include the surface area of the container in the formula? Because it is not needed. Pressure $p$ is force $F$ per unit of surface area $A$: $$p=\frac{F}{A}$$ The pressure a gas exerts on the walls of a container is the collective force collisions of the gas molecules exert on the container walls, per unit of surface area. If we look at ...


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The Ideal Gas Law PV=nRT explains what happens. It can be written as PV/T=nR. For our purposes nR can be considered a constant, PV&T are all expressed in absolute units. Thus if the container volume is constant, P/T is a constant.


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Yes and No! The vessel is open to the surrounding atmosphere so that it will always be in equilibrium with it. However, if the opening is very small and the supply of high pressure gas of a high enough volume, then the pressure in the vessel would rise. This is analogous to a car tyre having a blowout or a slow puncture.


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In this case, it is allowed. As a thought experiment, you can replace the top bucket with the siphon hose by an S-shaped tube that has a gradual change in diameter, like this: _ / \ |www| | | \www| | | \ww| | | \w| | | \|__/ | | | With this type of analysis, it is more ...


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Many ways of looking at this: If there were any differences in pressure in horizontal direction, the pressure difference would mean a force that would induce motion of the fluid until the pressure difference would drop to 0. So in stationary state, pressure must be constant horizontally, and vertically, the difference in pressure between different heights ...


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In a liquid (or any fluid), the molecules are in random motion (best to say is vibration). So each molecule is vibrating and hence collides with each other. Likewise the molecules in contact with the container also collides with the container walls. Assuming perfect elastic collision, the collided molecules are pushed backwards as insisted by Newton's third ...


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The reason is because liquids don't have preferred directions. It's true that if you squeeze a solid by pushing down on it, it'll push back up on your hand but it won't push to the sides. If you model a solid as a cubic lattice of masses connected by springs, this makes sense, because only the vertical springs get compressed. A solid has enough order to ...


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The pressure at a particular depth, $h$, is $h\rho g$.Force on a rectangular strip of infinitesimal height $\text dh$ and width $w$ is $h\rho g\text dhw$ which is when integrated for the entire surface becomes $\frac{1}{2}\rho gh^2w$.Similarly for other sides.


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You cannot subtract a pressure from a pressure gradient because they have different dimensions. However suppose you express the pressure gradient as: $$ \frac{dP}{dx} = \frac{P_1 - P_2}{x_1 - x_2} $$ then you can subtract some constant pressure $P_0$ from both $P_1$ and $P_2$: $$\begin{align} P'_1 &= P_1 - P_0 \\ P'_2 &= P_2 - P_0 \end{align}$$ ...


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Bernouli does not explain wing lift. You can measure an older light plane with a "plank" wing, factor in the wing area, distance over the upper and lower surfaces, cruise speed, and air density, and come up with a total lift figure of about 25% of the aircraft weight. Bernouli equations were published in an aviation text decades ago and the error propagated ...



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