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New answers tagged fluid-statics

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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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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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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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There is a buoyant force pushing upwards on the submerged object, which is equal to the weight of the water that is displaced. If the submerged object is suspended from a string, and doesn't touch the bottom of the beaker, the weight of the beaker will increase by the magnitude of the buoyant force, because an upwards buoyant force on the object requires an ...

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First note that LBM doesn't actually work for in-compressible fluids unless you replace the density in your D2Q5 vectors with pressure (or energy etc.. something that doesn't relate to the number of particles per unit volume/area), as density is constant in in-compressible fluids (and I'll be from this point onward considering your LBM implementation in ...

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You have a 1000cc volume, 720cc displaces water, it tells you what the water weighs. Now 280cc displaces oil, it tells you what the oil weighs. Find the weight of both displacements and add them together for the total weight. You should be able to work it with this. Also you put .78 in your equation for water volume, it should be .72

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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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In general it is not the same. If the interface is not flat, the difference in pressures (Laplace pressure) can arise from non-zero surface energy of interface. Or if there is preffered curvature of the interface (that can arise from highly assymetric molecules), additional terms in pressure difference can be present. However, if the interface is flat and ...

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No. If the object has a compressibility much less than water (nothing is perfectly incompressible) it would only sink part way. At a depth of 10,000 ft, the water would be about 1.4% denser, assuming temperature remained constant.

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The answer to both your questions in the last paragraph is yes. You understood it correctly. The first sentence is not carefully worded, instead of "buoyant force is greater" is should say something like "buoyant force [on the fully submerged body] is greater".

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The molecules at the liquid to air interface feel this downward pressure due to cohesion with other liquid molecules. The molecules below the surface feel no net force (unless external force is applied) because cohesion with other molecules is pulling them equally in all directions. They are in motion, as you stated, but they are not accelerating. ...

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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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It would only make a difference if the water could run from one end to the other, such as a single straight tube with air and water in it. If the water was contained as it appears to be, where the see saw's center of mass did not change, it would work the same as a solid.

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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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Keeping in mind that the buoyant force depends on the difference between the water pressure below and above the rod, then the buoyant force acts uniformly along the rod and the effective force can be taken at the middle of the submerged section. With the force equation: buoyant force + tension = weight, and and a torque equation (choose an axis), you can ...

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In a gas, the pressure on a surface is associated with the momentum change of rebounding molecules. Extra pressure requires a higher density or temperature in the gas. In a liquid you must add the contact forces between the molecules. The total pressure results from a combination of both, and the upward component must be adequate to support the weight of ...

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Why don't we consider the force by the movement of liquid molecules due to their thermal energy? This does not change the pressure. Essentially, increased thermal energy will make there be fewer collisions that are individually more energetic. You can think of a ball bouncing on a floor. Regardless of the energy of the ball the average force is the same ...

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Consider a small cube (of size $\Delta x$) of fluid at equilibrium. If you draw a free body diagram you see that there are 7 forces acting on the box of fluid: one pressure force acting on each face and the weight. Since it is in equilibrium the forces on the horizontal faces are all equal and therefore the pressure does not change horizontally. However, to ...

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Since they represent different tubes, h and h are probably the same height, otherwise they would be labeled differently. Water pressure increases with depth from the surface, not volume. So with water of the same purity and temperature, the pressure at the same depth from the surface will be equal in both tubes.

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What you have neglected to consider is the effect of the walls of the container which exert forces on the water. Look at the sequence below where on the left there is just water and then as you progress to the right the walls of the container are added nd the hole closed which does not change the pressure at point $P$ and finally the water outside the ...

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The water pressure exerted by the fluid already includes the weight of the water. In fact, the pressure is caused by the fluid's weight. Therefore, the force exerted by the water onto the piston is just the water pressure at the piston multiplied by the area of the piston. The water pressure is given by p=\rho g\Delta h=\frac{mg\Delta h}{A\Delta h}=\...

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Quote from Here: "The Capillary number describes the ratio of viscous to capillary forces, while the Bond number indicates the ratio of gravitational to capillary forces". Additional links 1 2

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The annular region of the cylinder lid exerts a downward force on the fluid below also. So the total force on the piston is equal to the pressure at the piston times the total area of the piston.

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I assume the question is to determine the force required by the piston. Notice that if the piston at the very bottom is in equilibrium, the magnitude of the force from the water on the piston must equal the force exerted by the piston. So we just need to find the force of the water on the piston. At the very top of the tube the pressure is atmospheric ...

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Short answer Wikipedia: Pressure  (symbol $p$ or $P$) is the force applied perpendicular to the surface of an object per unit area So direction of the force caused by pressure is determined by surface orientation. If pressures at A and B were different, horizontal pressure force would be pushing fluid from higher pressure to lower one to equalise the ...

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