Answer
See answers.
Work Step by Step
a. The free-body diagram for the athlete has three forces: the athlete’s weight, the buoyancy force, and the upward force exerted by the scale. The net force is zero.
$$F_{buoyant}+F_{scale}-mg = 0$$
$$\rho_{water}Vg+m_{apparent}g-mg = 0$$
$$V=\frac{m-m_{apparent}}{\rho_{water}}=\frac{70.2kg-3.4kg}{1000\;kg/m^3}=6.68\times10^{-2}m^3$$
b. SG is the person’s density divided by the density of water.
$$SG=\frac{m/(V-V_R)}{\rho_{water}}=\frac{70.2kg/(6.68\times10^{-2}m^3-1.3\times10^{-3}m^3)}{ 1000\;kg/m^3}\approx 1.07$$
c. Use the given formula.
$$\frac{495}{SG}-450=\frac{495}{1.072}-450=11.9\;\%$$
