College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 9 - Problems - Page 367: 105


The mass of the lead is $8.68~kg$

Work Step by Step

We can find the volume of the wood: $V_w = (0.330~m)^3 = 0.03594~m^3$ We can find the mass of the wood: $m_w = \rho_w~V_w = (780~kg/m^3)(0.03594~m^3) = 28.03~kg$ If the water just covers the lead and wood, then the average density of the lead and wood must be equal to the density of water, which is $1000~kg/m^3$. Let the mass of the lead be $m_l$. Then the volume of the lead is $\frac{m_l}{\rho_l}$. We can find the mass of the lead $m_l$: $\frac{m_w+m_l}{V_w+V_l} = 1000~kg/m^3$ $\frac{m_w+m_l}{V_w+\frac{m_l}{\rho_l}} = 1000~kg/m^3$ $m_w+m_l = (1000~kg/m^3)(V_w+\frac{m_l}{\rho_l})$ $m_l = (1000~kg/m^3)(V_w+\frac{m_l}{11340~kg/m^3})-m_w$ $m_l-0.0882~m_l = (1000~kg/m^3)(0.03594~m^3)-28.03~kg$ $0.9118~m_l = 35.94~kg-28.03~kg$ $m_l = \frac{7.91~kg}{0.9118}$ $m_l = 8.68~kg$ The mass of the lead is $8.68~kg$.
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