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Is an object that is neutral buoyant the equivalent of the same object in the microgravity (free fall) of the International Space Station? In some ways they are very similar, in other ways, not so much. In either case, the net weight is (approximately) $0$. This is the major similarity. Some differences would depend on the medium that you are buoyant in. ...


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One obvious difference is that the neutral buoyancy state of the ping pong ball on water is broken if it is moved vertically. If it is lifted and released it will fall back to the water surface; if it it pushed down and released it will rise back to the water surface. Whereas the ping pong ball in the ISS can be moved in any direction. Another difference is ...


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Another approach that does not require any calculus is to simply analyze a free body diagram of the object along with a prismatic region of fluid (height $H$ and cross-sectional area $A$) surrounding the submerged object. Envision the object being held in place by a string and let's assume that the object wants to float. Then, the force in the string will ...


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One way to compute the buoyant force is simply to compute the net force due to the pressure acting on the boundary of the object. Recognizing that the pressure has a gradient along the direction of gravity, $p = p_0 - \rho g z$, (where $\rho$ is the density of the fluid the object is submerged in) then allows you to convert the surface integral into a ...


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The total buoyant force on the object comes out of integrating all of the forces per unit area acting on the surface of the object—in this case, due only to pressure. Because all the forces per unit area add up—the physics of buoyancy is linear—you can think of this physical system as two separate blocks immersed in each fluid, bridged by an ...


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At room temperature and 1 atmosphere pressure air weighs 1.3 g/l. The pressure will increase the density by a factor of 1099, making it denser than water. Obviously the ideal gas law no longer applies and also the air can no longer be gaseous but actually is solid. Solid air is heavier than water at 1 atmosphere but the density of water at that depth has ...


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Due to the immense pressure, the balloon will shrink in size. Once let loose, the balloon will start accelerating by quite a bit (somewhat both 2 and 3) and will eventually attain terminal velocity at a point. Then onwards, the balloon will rise at the same pace.


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A Simpler Solution to the Mathematical Problem When the original ice cube of volume $a_0^3$ has melted completely it shrinks to a volume $ka_0^3$ of water, where $k$ is the ratio of densities of ice to water. This water fills the cylinder to a final depth of $h=\frac{ka_0^3}{A}$ where $A$ is the area of its base. Now imagine that we run time backwards ...


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When an object is floating, which means a part of it is in water and another part in air. Part of the object does not have to be partially in the air in order for the object to be "floating". All that is required is the upward buoyant force acting on the object equals the downward weight of the object. That occurs when the weight of the volume of water ...


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1.)When an object floats, is the volume of the object submerged and displaced water equal? Yes, the submerged volume of the object equals the volume of the water displaced. V(object)p(density of the object)g=Buoyant force=V(displaced liquid)p(density of the liquid)g $$V_{o}ρ_{o}g=V_{l}ρ_{l}g$$ This equation applies if the object floats (is in ...


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Archimedes' Principle would be the answer to both of your questions and fix the apparent paradox (or dilemma as the OP calls it). Archimedes' Principle says that The upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces. The other key ...


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When an object floats, not all of the object is submerged. There is some portion that remains above the surface. This means that the total volume of the object is greater than the volume of the object that is submerged. Since, as you said, the volume of the object that is submerged is equal to the volume of displaced water, this means that the total volume ...


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This answer will overlap with previous ones. When you cut the thread a mass of water equal to the volume will fall down, so the indicated weight will drop by this amount. If the balloon can escape to the surface without friction, the scale wil continue showing this weight until the water hits the remaining water. It will have accelerated so the scale will ...


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