Asked by Alex
Suppose that you place a square block of wood with dimensions 12.0 cm x 12.0 cm x 6.0 cm
into some water as shown below. The wood has a density of 600 kg/m3. What is the distance from the bottom horizontal face of the block to the surface of the
water? (In other words, how much of the block’s height will be underwater?)
into some water as shown below. The wood has a density of 600 kg/m3. What is the distance from the bottom horizontal face of the block to the surface of the
water? (In other words, how much of the block’s height will be underwater?)
Answers
Answered by
bobpursley
figure the mass of the wood:
.12*.12*.06*600kg== 0.000864(600)= = 0.5184 kg
so now, the volume of water displaced must be 518cm
12*12*h=518 solve for h in cm
.12*.12*.06*600kg== 0.000864(600)= = 0.5184 kg
so now, the volume of water displaced must be 518cm
12*12*h=518 solve for h in cm
Answered by
Henry
Vb = L*W*h = 12 * 12 * 6 = 864 cm^3. = Vol. of the block.
Db/Dw * h = 600/1000 * 6cm = 3.6 cm submerged.
Db/Dw * h = 600/1000 * 6cm = 3.6 cm submerged.
Answered by
Henry
NOTE: My vol. calculation was not required.
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