## Essential University Physics: Volume 1 (3rd Edition)

a) $49 \ kg$ b) $2.5 \times 10^3 \ kg$
We know the result from the last problem: $m_g = \frac{\rho_g M}{\rho_a-\rho_g}$ Thus, we find: a) $m = \frac{(280)(.18)}{1.22-.18}=49 \ kg$ b) $m = \frac{(280)(.9\rho_a)}{\rho_a-.9\rho_a}=2.5 \times 10^3 \ kg$