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

Published by Pearson

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

#### Answer

The block's mass density is $667~kg/m^3$

#### Work Step by Step

We can find the volume $V_s$ of the part of the block that is submerged. $V_s = (0.020~m)(0.020~m)(0.040~m)$ $V_s = 1.6\times 10^{-5}~m^3$ The buoyant force on the block will be equal to the block's weight. The buoyant force is equal to the weight of the water which is displaced. Let $\rho_w$ be the density of the water. We can find the mass $M$ of the block. $M~g = F_B$ $M~g = \rho~V_s~g$ $M = \rho~V_s$ $M = (1000~kg/m^3)(1.6\times 10^{-5}~m^3)$ $M = 0.016~kg$ We can find the density $\rho_b$ of the block. $\rho_b = \frac{M}{V}$ $\rho_b = \frac{0.016~kg}{(0.020~m)(0.020~m)(0.060~m)}$ $\rho_b = 667~kg/m^3$ The block's mass density is $667~kg/m^3$.

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