Question

What is the density of a woman who floats in fresh water with 2.25% of her volume above the surface? (This could be measured by placing her in a tank with marks on the side to measure how much water she displaces when floating and when held under water.) Round your answer to one decimal place.

ρ woman =____ kg/m3

What percent of her volume is above the surface when she floats in seawater (which as a density of 1025 kg/m3)? Round your answer to two decimal places.

Percentage above = ____%

Answer 1

Gravitational acceleration = g

Density of fresh water = _f = 1000 kg/m³

Density of seawater = _s = 1025 kg/m³

Density of the woman =

Volume of the woman = V

When the woman floats in fresh water 2.25% of her volume is above the surface.

Volume submerged in fresh water = V_f = 0.9775V

Vg = _fV_fg

V = _f(0.9775V)

= (1000)(0.9775)

= 977.5 kg/m³

Volume of the woman submerged in seawater = V_s

Vg = _sV_sg

V = _sV_s

(977.5)V = (1025)V_s

V_s = 0.9537V

Volume of the woman above the surface when she floats in seawater = V₁

V₁ = V - V_s

V₁ = V - 0.9537V

V₁ = 0.0463V

Percentage of her volume above the surface = P

P = (V₁/V) x 100

P = (0.0463V/V) x 100

P = 4.63%

a) Density of the woman = 977.5 kg/m³

b) Percent of her volume above the surface when she floats in seawater = 4.63%