Physics
6.23) In a measurement of a man's density he is found to have a mass of 65.0 kg in air and an apparent mass of 1.50 kg when completely submerged in water with his lungs empty.
a. What mass of water does he displace?
b. What volume of water does he displace?
c. Calculate his density.
d. If his lungs capacity is 2.0 liters, is he able to float with his lungs filled with air without treading water?
Note: Guys help me, provide me an answer with solution/formula. Thanks!!!?
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Answers
a. Mass of water displaced = Mass in air - Apparent mass in water = 65.0 kg - 1.50 kg = 63.5 kg
b. Volume of water displaced = Mass of water displaced/Density of water = 63.5 kg/1.000 kg/m^3 = 63.5 m^3
c. Density = Mass/Volume = 65.0 kg/63.5 m^3 = 1.02 kg/m^3
d. In order for the man to float in water, the apparent density of his body must be less than the density of the water he is in, which is 1.000 kg/m^3. With his lungs filled with air, the total volume of his body increases by 2.0 liters. Therefore, his apparent density would be:
Density = Mass/Volume = 65.0 kg/(63.5 m^3 + 2.0 L) = 0.972 kg/m^3
Since this is less than the density of water, the man would be able to float.
