Answer to Question #216344 in Molecular Physics | Thermodynamics for Dullah

Question #216344

11. 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? 


1
Expert's answer
2021-07-12T12:17:29-0400

The Buoyant Force is given as follows:


F=ρwgVvF = \rho_w g V_v

where ρw=1024kg/m2\rho_w =1024kg/m^2 is the density of sea water, g=9.8N/kgg = 9.8N/kg is the gravitational acceleration, and VvV_v is the volume of the vase.

By definition, the volume is:


Vv=mρgV_v = \dfrac{m}{\rho_g}

where ρg=19300kg/m3\rho_g = 19300kg/m^3 is the density of gold, and m=20kgm = 20kg is the mass of the vase.

Finally, obtain:


F=ρwgmρgF=10249.8201930010.4NF = \rho_w g\dfrac{m}{\rho_g}\\ F = 1024\cdot 9.8\cdot \dfrac{20}{19300}\approx 10.4N

Answer. 10.4N.


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