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Buoyant force and surface value

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Derive an expression for the buoyant force on a spherical balloon, submerged in water, as a function of the depth below the surface, the volume of the balloon at the surface, the pressure at the surface, and the density of the water. At what depth is the buoyant force half the surface value?

I know that the Buoyant force is equal to rho_water * g * V and that the pressure at the surface is P=P_0 + (rho * g * h). I'm not sure how to tie it together.

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Solution Preview

Because volume is a part of the sphere, where radius makes an angle Q(read as theta) with the vertical, given by

V' = 2*pi*(1 - cos(Q))*R^3/3

If theta < 90 degrees
V' = (2*pi*R^2/3)*(R - ...

Solution Summary

This solution is provided in approximately 158 words. It uses step by step equations to solve for the buoyant force and forces acting on the surface of the water. A diagram is attached in a .jpg file to enhance understanding.

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A 7.49 kg sphere of radius 5.25 cm is at a depth of 2.20 km in seawater that has an average density of 1025 kg/m3. What are the (a) gauge pressure, (b) total pressure, and (c) corresponding total force compressing the sphere's surface? What are (d) the magnitude of the buoyant force on the sphere and (e) the magnitude of the sphere's acceleration if it is free to move? Take atmospheric pressure to be 1.01 x 105 Pa.

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