## Physics: Principles with Applications (7th Edition)

The density of the rock is $2.99\times 10^3~kg/m^3$
Let $M_a$ be the apparent weight of the rock. Let $\rho_w$ be the density of water. We can find the volume $V$ of the rock. $M_a~g = M~g-F_B$ $F_B = M~g-M_a~g$ $\rho_w~V~g = M~g-M_a~g$ $V = \frac{M-M_a}{\rho_w}$ $V = \frac{9.28~kg-6.18~kg}{1.00\times 10^3~kg/m^3}$ $V = 3.10\times 10^{-3}~m^3$ We can find the density $\rho_r$ of the rock as: $\rho_r = \frac{9.28~kg}{3.10\times 10^{-3}~m^3}$ $\rho_r = 2.99\times 10^3~kg/m^3$ The density of the rock is $2.99\times 10^3~kg/m^3$.