In: Physics

# To determine an athlete's body fat, she is weighed first in air and then again while she's completely underwater.

To determine an athlete's body fat, she is weighed first in air and then again while she's completely underwater. It is found that she weighs 690 N when weighed in air and 48.0 N when weighed underwater. What is her average density?

## Solutions

##### Expert Solution

Concepts and reason

The concepts required to solve this problem are density and weight. Firstly, find the mass of athlete in air and in water. Then, find the mass of the water displayed by subtracting the masses of athlete in air and water. Then, find the volume of the water displaced by the expression of volume. Finally, find the density of the athlete by using the expression of density.

Fundamentals

The density of a body is given by the following expression:

$$\rho=\frac{m}{V}$$

Here, $$\mathrm{m}$$ is the mass of the body and $$\mathrm{V}$$ is the volume of the body. The weight of a body is given by the following expression:

$$W=m g$$

Here, $$\mathrm{m}$$ is the mass of the body and $$\mathrm{g}$$ is the acceleration due to gravity.

The weight of a body is given by the following expression:

$$W=m g$$

Rearrange the above expression for $$\mathrm{m}$$. $$m=\frac{W}{g}$$

Substitute $$9.8 \mathrm{~m} / \mathrm{s}^{2}$$ for $$\mathrm{g}$$ and $$690 \mathrm{~N}$$ for $$\mathrm{W}$$ in the above expression to find the mass of athlete in air.

$$m_{\mathrm{a}}=\frac{690 \mathrm{~N}}{9.8 \mathrm{~m} / \mathrm{s}^{2}}$$

$$=70.408 \mathrm{~kg}$$

Substitute $$9.8 \mathrm{~m} / \mathrm{s}^{2}$$ for $$\mathrm{g}$$ and $$48 \mathrm{~N}$$ for $$\mathrm{W}$$ in expression $$m=\frac{W}{g}$$ to find the mass of athlete in water.

$$m_{\mathrm{w}}=\frac{48 \mathrm{~N}}{9.8 \mathrm{~m} / \mathrm{s}^{2}}$$

$$=4.898 \mathrm{~kg}$$

The water is displaced because the athlete is in the water. The change in mas of the athlete is equal to the mass of the water displaced. The mass of the water displaced is as follows:

$$m=m_{\mathrm{a}}-m_{\mathrm{w}}$$

Substitute $$70.408 \mathrm{~kg}$$ for $$m_{\mathrm{a}}$$ and $$4.898 \mathrm{~kg}$$ for $$m_{\mathrm{w}}$$ in the above expression.

$$m=(70.408 \mathrm{~kg})-(4.898 \mathrm{~kg})$$

$$=65.51 \mathrm{~kg}$$

When the athlete is completely underwater, it displaces some amount of water. The mass of the water displaced is equal to the difference between the masses of athlete in air and in water.

The density of a body is given by the following expression:

$$\rho=\frac{m}{V}$$

Rearrange the above expression for V. $$V=\frac{m}{\rho}$$

The volume of water displaced is as follows:

$$V_{\mathrm{w}}=\frac{m}{\rho_{\mathrm{w}}}$$

Substitute $$65.51 \mathrm{~kg}$$ for $$m$$ and $$1000 \mathrm{~kg} / \mathrm{m}^{3}$$ for $$\rho$$ the above expression

$$V_{\mathrm{w}}=\frac{65.51 \mathrm{~kg}}{1000 \mathrm{~kg} / \mathrm{m}^{3}}$$

$$=0.06551 \mathrm{~m}^{3}$$

The volume of the athlete is equal to the volume of the water displaced. The density of a body is given by the following expression:

$$\rho=\frac{m}{V}$$

Substitute $$70.408 \mathrm{~kg}$$ for $$\mathrm{m}$$ and $$0.06551 \mathrm{~m}^{3}$$ for $$\rho$$ in the above expression.

$$\rho=\frac{70.408 \mathrm{~kg}}{0.06551 \mathrm{~m}^{3}}$$

$$=1075 \mathrm{~kg} / \mathrm{m}^{3}$$

The average density of athlete is $$1075 \mathrm{~kg} / \mathrm{m}^{3} .$$

The average density of a body is equal to the total mass of the body divided by the total volume of the body.

The average density of athlete is $$1075 \mathrm{~kg} / \mathrm{m}^{3}$$.