1. A cylindrical disk of wood weighing 45.0N and having a diameter of 30.0 cm floats on a cylinder of oil of density 0.850 g/cm3. The cylinder of oil is 75.0 cm deep and has the same diameter of the wood. (a) What is the gauge pressure at the top of the oil column? Suppose now someone puts a weight of 83.0N on the wooden disk and no oil seeps through. (b) what is the change in pressure at the top of the oil [3 pts] and at the bottom of the oil column?

asked by Francis

7 years ago

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2 answers

∆p=p-p_o=p_g W/A=W/(πr^2 )
p_g=(45 N)/(π(.150m)^2 )=636 Pa
∆p=(83 N)/(π(.150〖m)〗^2 )=1170 Pa

answered by Dafne Perea

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answered by Anonymous

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Question

1. A cylindrical disk of wood weighing 45.0N and having a diameter of 30.0 cm floats on a cylinder of oil of density 0.850 g/cm3. The cylinder of oil is 75.0 cm deep and has the same diameter of the wood. (a) What is the gauge pressure at the top of the oil column? Suppose now someone puts a weight of 83.0N on the wooden disk and no oil seeps through. (b) what is the change in pressure at the top of the oil [3 pts] and at the bottom of the oil column?

2 answers

∆p=p-p_o=p_g W/A=W/(πr^2 )
p_g=(45 N)/(π(.150m)^2 )=636 Pa
∆p=(83 N)/(π(.150〖m)〗^2 )=1170 Pa

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Dafne Perea · Accepted Answer

∆p=p-p_o=p_g W/A=W/(πr^2 ) p_g=(45 N)/(π(.150m)^2 )=636 Pa ∆p=(83 N)/(π(.150〖m)〗^2 )=1170 Pa

Anonymous · Answer

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