## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

# Chapter 15 - Fluids and Elasticity - Exercises and Problems: 48

#### Answer

The density of the statue is $2.5\times 10^3~kg/m^3$.

#### Work Step by Step

The apparent weight of the statue when it is submerged is the actual weight minus the buoyant force on the statue. The buoyant force is equal to the weight of the water that is displaced. Let $\rho_w$ be the density of water. We can find the volume of the statue as: $M_a~g = Mg-F_B$ $F_B = Mg-M_a~g$ $\rho_w~V~g = Mg-M_a~g$ $V = \frac{Mg-M_a~g}{\rho_w~g}$ $V = \frac{(28.4~N)-(17.0~N)}{(1000~kg/m^3)(9.80~m/s^2)}$ $V = 1.16\times 10^{-3}~m^3$ We then find the mass $M$ of the statue; $M = \frac{weight}{g}$ $M = \frac{28.4~N}{9.80~m/s^2}$ $M = 2.90~kg$ We then find the density of the statue; $\rho = \frac{M}{V}$ $\rho = \frac{2.90~kg}{1.16\times 10^{-3}~m^3}$ $\rho = 2.5\times 10^3~kg/m^3$ The density of the statue is $2.5\times 10^3~kg/m^3$.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.