A 49N (5kg) brick is placed into a pool of oil with a volume of 100L. The 5kg br

Question

A 49N (5kg) brick is placed into a pool of oil with a volume of 100L. The 5kg brick rests on the ground of the pool and is 26.51N. This raises the oil pool level to 102.5L.

1. The volume of oil displaced when the brick was placed in the pool is?

2.From balancing the forces acting on the brick, calculate the buoyant force on the brick. Note: You cannot use the definition of the buoyant force since you do not know the density of oil. Show your work for full credit including the equations used.

3.Calculate the density of the fluid (oil) from the definition of the buoyant force. Show your work for full credit including the equations used.

Explanation / Answer

1) The volume of oil displaced when the brick was placed in the pool = 102.5 - 100

= 2.5 L or 2.5*10^-3 m^3

2) Fnet = 0

N + B - Weight = 0

B = Weight - N

= 49 - 26.51

= 22.49 N

3) we know, Buoynat force, B = weight of the dispalced oil

B = rho*V*g

==> rho = B/(v*g)

= 22.49/(2.5*10^-3*9.8)

= 917.96 kg/m^3

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