Answer
(a) $1.04\times 10^3Kg/m^3$
(b) $0.0745m^3$
(c) $26N$
Work Step by Step
(a) We know that
$x_f=\frac{4950Kg/m^3}{\rho_p}-4.50$
This can simplify as:
$\rho_p=\frac{4950Kg/m^3}{x_f+4.50}$
$\rho_p=\frac{4950Kg/m^3}{0.281+4.50}=1.04\times 10^3Kg/m^3$
(b) We can find the required volume as
$V_p=\frac{1}{\rho_p}(\frac{W}{g})$
We plug in the known values to obtain:
$V_p=\frac{756}{(1035)(9.81)}$
$V_p=0.0745m^3$
(c) We can find the apparent weight of the person as
$W_{ap}=W-\rho_w V_p g$
We plug in the known values to obtain:
$W_{ap}=756-(1000)(0.0744)(9.81)$
$W=26N$
