Question

What is the density (in kg/m3) of a woman who floats in freshwater with 4.92% of her volume above the surface

Answer 1

Answer:

The density of the woman is 950.8 kg/m³

Explanation:

Given;

fraction of the woman's volume above the surface = 4.92%

then, fraction of the woman's volume below the surface = 100 - 4.92% = 95.08%

the specific gravity of the woman $= \frac{95.08}{100 } = 0.9508$

The density of the woman is calculate as;

$Specific \ gravity \ of \ the \ woman = \frac{Density \ of \ the \ woman }{Density \ of \ fresh \ water }\\\\ Density \ of \ the \ woman = Specific \ gravity \ of \ the \ woman \ \times \ Density \ of \ fresh \ water$

Density of fresh water = 1000 kg/m³

Density of the woman = 0.9508 x 1000 kg/m³

Density of the woman = 950.8 kg/m³

Therefore, the density of the woman is 950.8 kg/m³