A hollow, plastic sphere is held below the surface of a freshwater lake by a cor

Question

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 0.660 m3 and the tension in the cord is 1120 N .

Part A

Calculate the buoyant force exerted by the water on the sphere.

Express your answer with the appropriate units.

Part B

What is the mass of the sphere?

Express your answer with the appropriate units.

m =

Part C

The cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged?

Vsub/Vobj =

Explanation / Answer

Volume of sphere=v= 0.660 m^3

Density of water= d =1000 kg/m^3

Volume of water displaced by sphere=Volume of sphere=v= 0.660 m^3

Weight of water displaced by sphere=Ww =vdg=0.660*1000*g

Weight of water displaced by sphere=Ww = 660*9.8

Weight of water displaced by sphere=6474.6 N

The buoyant force exerted by the water on the sphere=Weight of water displaced by sphere =6474.6 N
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A) The buoyant force exerted by the water on the sphere is 6474.6 N
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Mass of sphere= m

Weight of sphere=Ws= mg

The tension in the cord =T= 1120 N.

The buoyant force =6474.6 N

For equilibrium of sphere

Weight of sphere+ tension in cord = buoyant force

Ws +1120 =6474.6

Ws = 6474.6 - 1120= 5354.6 N

Mass of sphere = m =Ws/g

Mass of sphere = m =5354.6 / 9.8

Mass of sphere =546.3877551 kg
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B) Mass of sphere is 546.3877551 kg
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Let volume inside water =Vi

Volume of sphere= v = 0.660 m^3

wt of displaced water = Vi*density of water*g=Vi*1000*g

Wt of sphere =546.3877551 *g

wt of displaced water = Wt of sphere

wt of displaced water = Vi*density of water*g

Vi*1000*g=546.3877551 *g

Vi=0.5463877551 m^3

volume of sphere =v =0.660 m^3

Vi / v = 0.5463877551 / 0.660 =0.827860235 or 82.7860235 %
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C) When the sphere comes to rest, 82.7860235 % of its volume is submerged

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