Question

A solid gold vase of mass 20.0 kg is suspended by a rope at sea bottom. What is the Buoyant 

Force of the seawater on the vase?

Answer 1

Solution.

$m=20.0kg;$

$\rho_g=19300kg/m^3;$

$\rho_w=1030kg/m^3;$

$F_B=\rho_wgV_g;$

$V_g=\dfrac{m}{\rho_g};$

$F_B=\dfrac{\rho_wgm}{\rho_g};$

$F_B=\dfrac{1030\sdot9.8\sdot20.0}{19300}=10.46N;$

Answer: $F_B=10.46N.$