Calculate the buoyant force exerted by the water on the sphere

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The buoyant force exerted by the water on the hollow plastic sphere is calculated to be 6860 N using the formula for buoyancy. The mass of the sphere is determined to be 622 kg by applying the relationship between buoyant force, mass, and tension in the cord. Once the cord breaks, the sphere rises, and the equilibrium condition shows that the mass of the sphere equals the buoyant force when at rest. The submerged volume is calculated, revealing that approximately 88.9% of the sphere's volume will be submerged when it reaches equilibrium. The calculations confirm the accuracy of the buoyant force and the submerged volume percentage.
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A hollow, plastic sphere is held below the surface of a freshwater lake by a cord anchored to the bottom of the lake. The sphere has a volume of 0.700 M^3 and the tension in the cord is 760 N.

Calculate the buoyant force exerted by the water on the sphere. Take the density of water to be 1000 kg/m^3 and the free fall acceleration to be 9.80 m/s^2.
**for this I set up Bouyant = density *Volume*gravity = 6860 N

What is the mass of the sphere? Take the density of water to be 1000kg/m^3 and the free fall acceleration to be 9.80m/s^2 .
**here I used Buoyant = mg+T and solved for m, so answer for m= 622 kg

The cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged? Express your answer as a percentage.
**For the sphere to be at rest, mg = B. There is no tension anymore, since the cord was broken. I express B as a function of the volume of the sphere that is still submerged:

mg = B --> rho*V*g = 0.7 how do I express it as a percentage? I tried 70%
 
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MAPgirl23 said:
A hollow, plastic sphere is held below the surface of a freshwater lake by a cord anchored to the bottom of the lake. The sphere has a volume of 0.700 M^3 and the tension in the cord is 760 N.

Calculate the buoyant force exerted by the water on the sphere. Take the density of water to be 1000 kg/m^3 and the free fall acceleration to be 9.80 m/s^2.
**for this I set up Bouyant = density *Volume*gravity = 6860 N

What is the mass of the sphere? Take the density of water to be 1000kg/m^3 and the free fall acceleration to be 9.80m/s^2 .
**here I used Buoyant = mg+T and solved for m, so answer for m= 622 kg

The cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged? Express your answer as a percentage.
**For the sphere to be at rest, mg = B. There is no tension anymore, since the cord was broken. I express B as a function of the volume of the sphere that is still submerged:

mg = B --> rho*V*g = 0.7 how do I express it as a percentage? I tried 70%

The first two look good. The mg on the left side of your last equation involves the mass of the sphere you already calculated. The buoyant force will be rho*V*g, but rho is the density of water and V is the volume of water displaced. This does not equal .700. Use the mg = B part to find B and use B = rho*V*g to find V. The ratio of that V (volume of displaced water) to the volume of the sphere is the fraction submerged.
 
622 kg * 9.8 = 6095.6 N

6095.6N / (1000 * 9.8) = 0.622

0.622 / 0.700 = 88.9% is this correct?
 
MAPgirl23 said:
622 kg * 9.8 = 6095.6 N

6095.6N / (1000 * 9.8) = 0.622

0.622 / 0.700 = 88.9% is this correct?

Looks good to me.
 
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