University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 11 - Equilibrium and Elasticity - Problems - Exercises - Page 360: 11.34


$$\Delta V=-0.0527m^{3}$$ $$\rho =1.09\times10^{3}Kg/m^{3}$$

Work Step by Step

$\frac{1}{K}=B=\frac{-\Delta P}{\frac{\Delta V}{V_{0}}}\quad\rightarrow \qquad\Delta V=\frac{-\Delta P V_{0}}{B}$ $B=\frac{1}{K} =\frac{1}{45.8\times10^{-11}}=2.2\times10^{9}Pa$ $V=1 m^{3}\qquad \Delta P=1.6\times10^{8}Pa\qquad K=45.8\times10^{-11}$ $a) \quad \Delta V=\frac{-(1.6\times10^{8})(1m^{3})}{(2.2\times10^{9})}=-0.0527m^{3}$ $b)\quad \rho=\frac{m}{V_{F}}=\frac{1.03\times10^{3}Kg}{(1-0.0.527m^{3})}=1.09\times10^{3}Kg/m^{3}$
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