Answer to Question #246132 in Physics for Faisal Yusuf

Question #246132
The apparent weight of a body holy emerged in the water is 32N and eat sweet in a is 96N Calculate the volume of the body. (Density of water =1000kgm-3,g=10ms-2
1
Expert's answer
2021-10-03T16:17:03-0400

The apparent weight of a body holy emerged in the water is 32N and eat sweet in a is 96N Calculate the volume of the body. (Density of water =1000kgm-3,g=10ms-2


The weight of the body in the water is the resultant of two forces: weight of the body in air Fg=96NF_g = 96N acting down and buoyant force FbF_b acting up:


W=FgFbW = F_g - F_b

The buoyant force is given as follows:

Fb=ρwVgF_b = \rho_w V g

where ρw=1000kg/m3\rho_w = 1000kg/m^3 is the density of water and VV is the volume of the body. Expressing VV and substituting W=32NW = 32N, obtain:


V=FgWρwgV=96N32N1000kg/m310N/kg=6.4×103m3V = \dfrac{F_g - W}{\rho_w g}\\ V = \dfrac{96N - 32N}{1000kg/m^3\cdot 10N/kg} = 6.4\times 10^{-3}m^3

Answer. 6.4×103m36.4\times 10^{-3}m^3.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment