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

Since a density of $7.8\times 10^3~kg/m^3$ is similar to the density of iron and steel, the sample is most likely made of iron or steel.
Let $M_a$ be the apparent mass of the sample. Let $\rho_w$ be the density of water. We can find the volume $V$ of the sample. $M_a~g = M~g-F_B$ $F_B = M~g-M_a~g$ $\rho_w~V~g = M~g-M_a~g$ $V = \frac{M-M_a}{\rho_w}$ $V = \frac{0.0635~kg-0.0554~kg}{1.00\times 10^3~kg/m^3}$ $V = 8.10\times 10^{-6}~m^3$ We can find the density $\rho_s$ of the sample as: $\rho_s = \frac{0.0635~kg}{8.10\times 10^{-6}~m^3}$ $\rho_s = 7.84\times 10^3~kg/m^3$ Since a density of $7.8\times 10^3~kg/m^3$ is similar to the density of iron and steel, the sample is most likely made of iron or steel.