InterviewSolution
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InterviewSolution
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Air is a fluid and our bodies displace air. And so a buoyant force is acting on each of us. Estimate the magnitude of the buoyant force on a 75 kg person due to air displaces. Is the actual weght of the person more or less than the scale reading? |
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Answer» Volume of person, `V_p=(m)/(rho_p)=(75kg)/(1000(kg)/(m^3))=0.075m^3` Buoyant force, `F_B=rho_agV_p=(1.29(kg)/(m^3))(9.8(m)/(s^2))(0.075m^3)=0.95N=1N` (as estimated density of person, `rho_p=1000(kg)/(m^3)`,density of air, `rho_a=1.29(kg)/(m^3)`) The weight of the person is `1N(=0.1kg)` more than the scale reading. |
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