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.

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

Now calculate the density of the fluid (oil) from the definition of the buoyant force? Show your work for full credit.

Does this calculated value agree with the expected density of oil 920kg/m^3?

What forces act on the brick as it is lying on the bottom?

What is the volume of oil displaced when the brick is in the pool?

Please show all work for full credit

Explanation / Answer

Loss in Weight = Buoyant Force

Loss in Weight = 49 N - 26.51

= 22.49 N

V in * p * g = Buoyant Force

Vin = 102.5 - 100

= 2.5 L ( 2.5 * 10-3 m3)

2.5 * 10-3 * density * 9.8 = 22.49

Density = 918 Kg/m3

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