Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 9 - Static Equilibrium; Elasticity and Fracture - Problems - Page 256: 44

Answer

The new volume is $997.5cm^3$

Work Step by Step

We can solve for the change in volume using the following formula: $\frac{\Delta V}{V_0}=-\frac{1}{B}\Delta P$ $\frac{\Delta V}{1000cm^3(1m/100cm)^3}=-\frac{1}{B_{alcohol}}(2.6*10^6N/m^2-1.0*10^5N/m^2)$ $\Delta V=-\frac{1}{B_{alcohol}}(2.6*10^6N/m^2-1.0*10^5N/m^2)*1000cm^3(1m/100cm)^3$ $\Delta V=-\frac{1}{1.0*10^9N/m^2}(2.6*10^6N/m^2-1.0*10^5N/m^2)*1000cm^3(1m/100cm)^3$ $\Delta V=-2.5*10^{-6}m^3$ or $-2.5cm^3$ Then; $V=V_0+\Delta V$ $V=1000cm^3+-2.5cm^3=997.5cm^3$ The new volume is $997.5cm^3$.
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