Buoyant force physics? I have been studying for 7 hours straight and still don't understand it
[From: ] [author: ] [Date: 11-04-24] [Hit: ]
2 !bThe weight needed on top of the block will weigh the same as 8-5.2=2.8m^3 of water2.8m^3*1000kg/m^3 = 2,800kg of water which weighs = 27,......
Notice that 650/1000=0.65 and 0.65*8=5.2 !
b The weight needed on top of the block will weigh the same as 8-5.2=2.8m^3 of water
2.8m^3*1000kg/m^3 = 2,800kg of water which weighs = 27,440N
So 27,440N/9.8 = 2800kg of lead whixh about 0.25m^3 of lead
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ok here's the question
"a cube of wood having and edge dimension of .2 m and a density of 650 kg/m^3 floats on water.
a) what is the distance from the horizontal top surface of the cube to the water level
b) what mass of lead should be placed on the cube so that the top of the cube will be just level with the water surface?"
Don't get too complicated.
From the cube side length, (20cm. x 20 x 20) = 8,000cc. wood volume. Leave that for now.
Its density is 0.65 x that of water, so 0.65 of its 20cm. side length will be submerged, and 0.35 will be above water.
a) (.35 x 20cm) = 7cm. above the water.
Now. back to the lead bit.
The block, fully submerged, will displace 8,000cc. of water, the mass of which is 8,000g.
The block, in air, has a mass of (.65 of 8,000) = 5,200g.
Mass of lead required = (8,000 - 5,200) = 2,800g., or 2.8kg.
Water density is 1,000kg/m^3, or 1g/cc.
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