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A man of mass 80 kg and density 970 kg/m3 stands in a shallow pool with 31% of the volume of his body below water. Calculate the normal force the bottom of the pool exerts on his feet. (Take the positive direction to be up. Indicate the direction with the sign of your answer.)
Answers
Answered by
Damon
density of water = 1000
volume of man = 80 kg / 970kg/m^3 = .0825m^3
.31*.0825 = .0256 m^3
buoyant force up on man = .0256*1000*9.81 = 251 Newtons
weight of man = 80*9.81 = 785 Newtons
weight - buoyancy = 785 -251 = 534 N
volume of man = 80 kg / 970kg/m^3 = .0825m^3
.31*.0825 = .0256 m^3
buoyant force up on man = .0256*1000*9.81 = 251 Newtons
weight of man = 80*9.81 = 785 Newtons
weight - buoyancy = 785 -251 = 534 N
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