Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 14 - Fluids and Elasticity - Exercises and Problems: 23

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. Therefore; $M_a~g = Mg-F_B$ $F_B = Mg-M_a~g$ $\rho_w~V~g = Mg-M_a~g$ $V = \frac{Mg-M_a~g}{\rho_w~g}$ $V = \frac{(28.4~N)-(17.0~N)}{(1000~kg/m^3)(9.80~m/s^2)}$ $V = 1.16\times 10^{-3}~m^3$ We can find the mass $M$ of the statue as: $M = \frac{weight}{g}$ $M = \frac{28.4~N}{9.80~m/s^2}$ $M = 2.90~kg$ We can find the density of the statue as: $\rho = \frac{M}{V}$ $\rho = \frac{2.90~kg}{1.16\times 10^{-3}~m^3}$ $\rho = 2.5\times 10^3~kg/m^3$ The density of the statue is $2.5\times 10^3~kg/m^3$.
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