A cubical block of wood, 10.5 cm on a side floats at the interface between oil (

Question

A cubical block of wood, 10.5 cm on a side floats at the interface between oil (depth d = 10.5 cm) and water with its lower surface 1.80 cm below the interface. The density of the oil is 790 kg/m3.

(a) What is the gauge pressure at the upper face of the block?
Pa

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

(c) What is the mass of the block?
kg

What is the density of the block?
kg/m3

A cubical block of wood, 10.5 cm on a side floats at the interface between oil (depth d = 10.5 cm) and water with its lower surface 1.80 cm below the interface. The density of the oil is 790 kg/m3. (a) What is the gauge pressure at the upper face of the block? Pa (b) What is the gauge pressure at the lower face of the block? Pa (c) What is the mass of the block? kg What is the density of the block? kg/m3

Explanation / Answer

A. gauge pressure at upper block

Pup - P0 = Poil*g*h = 790*9.8*0.018 = 139.35 Pa

B. gauge pressure at lower block

Pdown - P0 = Poil*g*hoil + Pwater*g*hwater = 790*9.8*0.105 + 1000 * 9.8* 0.018 = 989.31 Pa

C. Ftotal = 0

Pdown*A - Pup*A - mg = 0

m = Pdown*A - Pup*A/g = (989.31 - 139.35) *10.5*10.5/9.8 = 0.956 kg

d = 0.956/0.105^3 = 825.82

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