The density of a 6 foot tall man that weighs 840. Newtons fluctuates from a valu

Question

The density of a 6 foot tall man that weighs 840. Newtons fluctuates from a value slightly less than the density of water when the man completely inhales to a value slightly more than the density of water when the man completely exhales.

Note: Assume that the water pressure does not affect the volume of the man.

A)The man completely exhales and jumps into the deep end of a pool with a depth of 9 feet. The man sinks to the bottom of a pool and lands on a scale; at equilibrium, the scale reads 44.5 Newtons.

1)Calculate the buoyant force of the water acting on the man with his breath completely exhaled.
2)Calculate the volume of the man with his breath completely exhaled.

Explanation / Answer

1)buoyant force = 840 - 44.5 = 795.5 N

2) thus v*d*g = 795.5

=> 1000*9.8*v = 795.5

=> v = 0.081173469387755 m^3 (exhaled)

thus density = mass/volume = 840/(9.8*0.081173469387755) = 1055.939660591 kg/m^3

difference in density = 1055.939660591 - 1000 = 55.939660591

thus inhaled density = 1000 - defference = 1000-55.939660591 = 944.060339409 kg/m^3

thus volume (inhaled) = mass/density = 840/(9.8*944.060339409) = 0.0907932280768669 m^3 answer

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