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Elden [556K]
2 years ago
10

An 80 N bowling ball has an apparent weight of 60 N when completely submerged in water. What is the average density of the bowli

ng ball
Physics
1 answer:
Allisa [31]2 years ago
3 0

Answer:

V * ρB = WB        volume of ball * density of ball = weight of ball

V * ρW = Ww        volume of ball * density of water = buoyant force

ρB / ρW = WB / Ww = 80 / 20 = 4    water provides 20 N of buoyant force

ρB = 4 ρW = 4 gm/cm^3

ρW = 1000 kg / m^3 = 1 gm / cm^3

(1 gm/cm^3) = .001 kg / .000001 m^3 = 1000 kg/^3

ρW = 1 gm / cm^3

multiply by 9.8 or 980 to get weight densities

In this case

4 * 1000 kg/m^3 * 9.8 m/s^2 = 39200 N /m^3      weight density of ball

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#2

As it is given here

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#3

As it is given here

initial it starts from rest

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#4

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