Hello, i am havin trouble solving parts c and d of the problem, i already solved

Question

Hello, i am havin trouble solving parts c and d of the problem, i already solved parts a and b which were pretty easy and included the answers in case they are needed. How am i supposed to get the mass and density of the block with the information given? Couls someone help me.

A container has two layers: a layer of oil 10.21cm deep, and a layer of water 12.51cm deep. A cubical block of wood, 10.06cm on a side, floats at the interface between oil and water with its lower surface 2.42cm below the interface (the figure). The density of the oil is726kg/m3 . Keep in mind you must use the numbers given not the ones in the picture.

Part A) What is the gauge pressure at the upper face of the block? (183 Pa)

Part b) What is the gauge pressure at the lower face of the block? (964 Pa)

Part c) What is the mass of the block?

Part d) What is the density of the block?

Explanation / Answer

Fb = M*g

(P2*A) - (P1*A) = M*g

M = (P2-P1)*a/g = ((964-183)*0.1006^2)/(9.8)

M = 0.8065 kg

d = M /V =0.8065 / (0.1006^3) = 792.1857 kg /m^3

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