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If you are referring to the animated TV series, remember this is a cartoon, you cannot expect much realism. In reality, divers open access to the water like this would require the air pressure above the water, to equal the external water pressure, you cannot have two different pressures meeting with nothing to separate them. Once in the air chamber above the ...


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The exact formula for the radius of curvature is $$ \frac{1}{R} = \frac{y''}{(1+y'^2)^{3/2}}. \tag{1}$$ Exactly at the minimum or maximum you have $y'=0$. Near the minimum/maximum you still have a very small $y'$, i.e. $|y'| \ll 1$. Therefore you can use the approximation $1+y'^2 \approx 1$. Then equation (1) from above can be simplified to $$ \frac{1}{R} \...


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The formula for the radius of curvature is $$ \kappa = \frac{y''}{(1+y'^2)^{3/2}} = \frac 1 R, $$


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I'm considering a local inertial frame, at a point $\mathcal{P}$ of $D$ dimensional spacetime, in which there's an electric field (any configuration). The calculations below are valid even in General Relativity, and for any electric field in spacetime of $D$ dimensions (I don't care about axes aligned to the field!). I'm using $c = 1$ and metric signature $...


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Here is an elementary argument for 3+1 dimensions, which I'll then generalize to $d+1$ dimensions, although I'm unsure if I got the latter right, and would appreciate comments. In the case of a purely electric (or purely magnetic) field, we expect the pressure and tension to depend only on the field at that point -- not on the field in some neighborhood, or ...


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First thing to note is the equation $$ \text{energy density} = |\text{tension along the $x$ direction}| $$ (in units with $c=1$) would be valid for any dimension, not just $4$. The reason is simple: for purely magnetic (or electric) field along the $x$ axis, the field strength tensor is invariant under boosts along the direction of the field. So the stress ...


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You have to look at the net forces on the ball and the paper individually, and not just the action-reaction pair. Gravity exerts a downward force of $mg$ on the ball and the paper momentarily exerts an equal and opposite force of $mg$ on the ball per Newton's third law. The net force on the ball is momentarily zero and the ball is momentarily stable. But ...


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Downward gravitational force applied on the sheet.Reaction force on the Ball in upward direction.But thin sheet can not sustain such heavy force and breakdown and losses its continuity and comedown by gravity .At that instant Metallic ball only experiences downward gravitational force so it's fall down.


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I am assuming the ball to be not properly infiltrated.Now because of this when you get the ball to the floor of pool,because of the pressure from the water above the ball the ball gets pressed and assumes the shape of a flat disk.Buoyancy force will affect a body only if the body has some fluid underneath it to push it upward,in all the other cases it will ...


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I think the role of pressure is to adjust itself immediately according to changes in the velocity field so that velocity is divergence-free at all times. In that sense, it does not have any thermodynamic meaning.


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For the first question, (e) is also a correct response. For the second one, you are meant to intuit that the pressure inside the beach ball is no greater than atmospheric pressure—otherwise the “thin” plastic would presumably “stretch”. I’d say that premise is a bit thin, and the reasoning a bit of a stretch, but otherwise the given answer is correct.


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The key distinction here is that the plastic is not stretched—this means that the pressure inside of the unstretched ball equilibrates with the external pressure. Assuming the air behaves like an ideal gas, Boyle’s Law indicates that $p_1 V_1 = p_2 V_2$, so the volume the air occupies decreases as the ball goes down thanks to the pressure increase from the ...


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Throwing water 100 meters in 1 second means the water has to have a speed of 100m/s. So the question becomes "how much pressure do I need to achieve those speeds?" What you're looking for is Bernoulli equation. They state that for any given incompressible flow (such as water out of a nozzle): $$\frac{v^2}{2}+gz+\frac{p}{\rho}=C$$ In this equation $v$ is ...


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You are correct in concluding that the pressure $p$ in the Navier-Stokes equations is not related to the equation of state. It is a purely mechanical quantity, defined as $p = -\sigma_{ii}/3$, where $\sigma_{ij}$ is the ($ij$-th component of) the stress tensor. The pressure that appears in the equation of state is a thermodynamic quantity and it is ...


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Why high voltage? To break the molecules of the gas into atoms and to remove the electrons from the outermost orbitals. Why low pressure? To allow the rays to move freely from one electrode to another and the possibility of collisions between rays and moleculess are minized.


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I know that $V_m=\frac{V}{n}=\frac{RT}{p_a}$ This is not correct. It should read $$V_m=\frac{ZRT}{p}$$ You can estimate the compressibility factor Z by using the corresponding states principle, which expresses Z as a function of reduced pressure, reduced temperature, and ascentric factor of the gas. Or you can determine Z from a non-ideal equation of ...


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Where does the other part of the force go for the funnel? It's supported by the horizontal component of the sloping walls of the funnel, not the bottom of the cylindrical portion of the funnel. See figures below. The bottom of the cylinder to the left supports the entire weight of the fluid above it. The weight of the fluid outside the cylindrical part of ...


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It presses on the funnel. Imagine that the lower cylinder of the funnel extended all the way to the top. So you had water in the central cylinder pressing on the floor, and you had a separate conical area with the center cut out, full of water. You could drain the center cylinder and the rest of the water would still be there, pressing down on the funnel. ...


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Not that I'm aware of, no. The interface between the air pocket and the surface of the water should be approximately the same pressure. If they were not at the same pressure, then once would be pushing against the other with a greater force, and the interface would move until the forces equalized (therefore the pressures approximately equalize). Adding ...


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You can think of the equation as meaning how hard the molecule hits the wall, multiplied by how frequently it hits it. The time taken to travel to the other side and back is just the gap between successive hits.


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Shouldn't we put instead "the actual time of the collision/contact" between the molecule and the wall for $\Delta t$? I believe the kinetic theory of gases assumes the gas molecules consist of hard spheres whose collisions with the walls are perfectly elastic. That being the case, the assumption may be that the amount of time the molecule is in contact ...


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We will consider the case of real transport with a given flow rate $J=\rho u$. We assume that all compressors are identical, capable of maintaining a given air flow. Specifications: the flow velocity is limited by the condition $1\le u\le 10$ m/s. It is necessary to determine how many compressors are needed and how to optimally position them when ...


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In thermodynamics it makes little sense to speak about the real transient force that a molecule exerts on molecules of the wall. Instead we are more interested in pressure, i.e. the force averaged in space and time. The average force over the time $\Delta t$ is the same as the change of the momentum divided by the time interval. $$ f_{avg} = \frac1{\Delta t}...


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Is there more or less atmospheric pressure pressing down onto water having a larger atmosphere than of a smaller atmosphere? If you mean by "larger atmosphere" you mean the height of the atmosphere, then yes there will be more atmospheric pressure down on the water if the height of the atmosphere is greater. You know now that I rethink of this I now ...


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As shown in the figure, the red arrow is the direction of the boat, the blue arrow is the direction of the water, and the green line is the plane of the drainage outlet. Because the drain plane is so inclined, the water tends to leave along the normal direction (the black arrow is the normal direction of the drain plane), so low pressure will occur in the ...


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when it is being deformed in the hot press, the atoms in its lattice are being pushed past one another, and the solid metal flows like stiff taffy. The slag on the outside is brittle, and when it exfoliates, a fresh new red-hot iron surface is exposed to air. It violently oxidizes, producing flames and sparks. Each time the ingot is deformed more, the ...


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Ice VII occurs above 3 GPa. Pressure at depth $z$ is $p =\rho g z$, so the critical depth for $\rho=1050$ kg/m$^3$ is $z=290,951.4$ m. So we need a 291 km deep ocean to get high pressure ice. Not very likely on Earth.


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Buoyancy does not really change with depth (water temperature differences can slightly change it's density). Once a sealed container weighed more than the water it displaced (negative buoyancy), it would begin to sink and go to the bottom. If it weighed less than the amount of water it displaced (positive buoyancy) it will float. So to decide if your ...


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If your sample of air were in a container open at the bottom with extra weights attached, sinking would require that the total weight of the air, container, and weights must exceed the weight of the water displaced by all three. Once submerged, the volume of the air would start to decrease and the system would continue to sink. If the air were in a sturdy ...


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Vacuum is not the same as zero gravity. So if bubbles rise in a kettle they will rise in a vacuum chamber. The pressure will eventually be the saturation pressure of water at the temperature of the environment (about 0.02 bar at room temperature). Yes, the vapor can be condensed again, by reducing the volume.


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We can approximate this as a process where the existing volume, say $V_0$, is reduced to $V_1$, where: $\Delta V=V_0-V_1$ is the volume of water you want to add. First we compress $V_0$ to $V_1$, to make room for water to be added. During the compression pressure will increase from $p_0$ to $p_1$ and the compression requires work as per your question. ...


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There is a simple intuitive explanation for the factor of two difference. The thrust has two contributions, as follows. 1) The air pressure in the rocket imposes a force down the nozzle of F=PA exactly as you surmised in your second method. 2) The exhaust water at the instant it leaves the nozzle is still pressurised, and imposes a back-pressure on the ...


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The pressure drop from tank pressure to atmospheric pressure does not occur instantly at the nozzle but rather is spread out according to the area of the flow channel. This reduced pressure results in additional thrust that was not accounted for in your second solution. Here is one way to calculate this additional missing thrust: Bernoulli's equation (your ...


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Often these rockets are launched off a vertical section of pipe or rod that extends up inside the bottle, acting as a piston until the rocket has moved far enough to clear the end of the pipe. During this phase, your second method is correct: the thrust is simply the nozzle area times the pressure. Why should the thrust increase when the rocket separates ...


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The first method is correct. In the second you have assumed that the pressure at the nozzle is still $P$ despite the water exiting with some velocity. i.e. You have neglected the dynamic pressure. You need to use Bernoulli's principle $$ P + \frac{\rho v^2}{2} + \rho h g = {\rm constant}$$ Your first method assumes that the top surface of the water is ...


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As you have neglected atmospheric pressure, let's assume that you are doing the experiment in a vacuum(I know that there is a spherical cow analogy somewhere :) ). You are assuming that the pressure is exerted by air inside the ballon at the surface. The air will exert the pressure in the empty part as shown by your arrows as well as on the horizontal ...


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If you consider a gas then it has a pressure because the molecules of the gas are whizzing around rebounding off the walls of whatever container it's in. For an ideal gas the pressure is proportional to the temperature, and the temperature is proportional to the internal energy, so the pressure is proportional to the internal energy i.e. the kinetic energy ...


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Figure the surface area of a piston (pi times radius squared), then divide the weight on top of it by the square inches of it's surface, this will give the pounds per square inch (PSI) of pressure exerted on the water. Do this on each piston and compare the PSI of each. The piston with greater PSI will push down and lift the piston with lighter PSI. If ...


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Would the buoyant force increase if I inflated a balloon that's inside a closed chamber hypothetically submerged underwater? And vise versa would the chamber's buoyancy decrease if deflated? First, I am assuming (from the diagram) that the two chambers are fastened to the bottom of the tank and that the chamber walls are rigid. The upward buoyant force ...


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I think one can get the expression for $P$ without that for $E$. Let $p_o$ be the pressure due to the fluid outside and $p_i$ be the pressure due to the fluid in the bubble. Since the bubble is initially spherical, \begin{equation}\tag{e1}\label{e1} p_o - p_i = \frac{2\alpha}{R}. \end{equation} When the bubble is deformed its radius is given by $r = R + \...


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As Ched Miller points out ideal gas law (https://en.m.wikipedia.org/wiki/Ideal_gas_law) states that PV=nRT, with P = pressure, V = volume, n = number of moles, R = constant, T = temperature. Since V, R and T are the same in all your half boxes, you are left with P = k n, k being a constant. In other words, pressure will be linearly proportional with the ...


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The atmospheric pressure drops about 11.3 pascals per meter the first 1000 meters above sea level. The pressure at sea level is 101,325 pascals. The fractional difference between the bottom and top of a cubic meter of air is therefore only 0.00011152 Hope this helps


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The ideal gas law assumes constant pressure and temperature throughout the volume of the gas. In case these quantities were vary in the volume, you can use a local form of the law, for instance: $$P=\rho \dfrac{R}{M}T$$ where $P$, $\rho$ and $T$ are the local pressure, density and temperature, respectively, and $M$ is the molar mass of the gas. As you said, ...


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The author based his tank pressure reference point on the bottom of the tank, which in my opinion is mathematically correct, but more difficult to interpret from a physical standpoint. Another way to formulate the Bernoulli equation for this problem is to only consider the height of the liquid above the hole, which gives: $P + \rho g h = P_0 + \frac{1}{2} \...


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${\rho}gy_1$ and ${\rho}gy_2$ denote the potential energy of the fluid. This is important because the two points are not at the same level. Point 2 is at a higher potential than point 1(that is why fluid is flowing from 2 to 1). Also,yes the external pressure at both point 1 and point 2 will be equal to the atmospheric pressure and they will cancel out.


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Check your sources for the exact terms they are using for kg. It is possible they are referring to kilogram-force as opposed to kilogram-mass. According to Wikipedia kilogram-force is a gravitational metric unit of force. It is equal to the magnitude of the force exerted on one kilogram of mass in a 9.80665 $\frac{m}{s^2}$ gravitational field. Hope this ...


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Mathematically pressure is defined as force per unit area: $$ p = \frac{F}{A} $$, if the only force which is acting on area is weight then we can extend expression according to second Newton law as: $$ p = g\frac{m}{A}$$, so because $g$ is constant we can consider a pressure as a mass per unit area as well.


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Technically, pressure is better defined as force per unit area, without any specific units in mind. But your confusion is justified. Pressure should be measured in units of force per area, in metric, we typically use Pascals ($Pa$) which are given by $\frac N{m^2}$. $\frac {kg}{m^2}$ is not a measure of pressure, so your reasoning seems solid to me. ...


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A balloon is comprised of surface patches that expand with hydrostatic pressure and taken on a certain shape. If the material shape and properties are known, the direct problem of finding out what shape it takes at a given pressure and thus the volume is straight forward. The inverse problem; however, isn't. In this case one would normally model the volume ...


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most bourdon tubes are of circular cross-section, or nearly so. However, for measuring low pressures in the range of 1 to 5 psi, the tube will not only have a thin wall but it will be flattened into a rectangular cross-section, with semicircular sides, and with the broad axis at right angles to the length of the curved tube.


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