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# last physics problem need help!!?

A cube with an edge of 2.3 m is completely submerged in freshwater. The density of the cube is 4780 kg/m^3, and the density of freshwater is 1000 kg/m^3.

Weight of the cube-

buoyant force the water exerts on the cube-

calculate the apparent weight of the cube when submerged in freshwater-

### 2 Answers

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• volume is 2.3³ = 12.167 m³

4780 kg/m³ x 12.167 m³ = 58200 kg

weight is 58200 kg x 9.8 N/kg = 570000 N

mass of same volume of water =

1000 kg/m³ x 12.167 m³ = 12167 kg

diff is buoyancy = 45991 kg or 450700 N

apparent weight is 570000 – 450700 = 120000 N

• The buoyancy force is the weight of the displaced fluid (cube volume x water density x gravitational accleration), which is in the upwards direction. The weight of the cube itself is cube volume x cube density x gravitational acceleration, in the downwards direction. The difference is the apparent weight.

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