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 0.650m^3 and the tension in the cord is 670Newtons. Part A Calculate the buoyant force exerted by the water on the sphere.(In Newtons) Part B What is the mass of the sphere?(In kilograms) 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?(%) Will give the first and best rating to the most detailed answer. Please use formulas before and after and show ALL WORK. Thank you.

Explanation / Answer

A) Density of water = 1000 kg/m3

Buoyant force = density of water*volume*sphere*g = 1000*0.65*9.81 = 6376.5 N

B) T + W = 6376.5

670 + W = 6376.5

W = 5706.5 N

m = W/g = 5706.5/9.81 = 581.7 kg

C) Let fraction of volume submerged be x.

Buoyant force B' = density of water*volume submerged*g = 1000*(0.65*x)*9.81 = 6376.5*x

Putting B' = W we get

6376.5*x = 5706.5

So, x = 0.8949 = 89.49 %

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