To keep a beach ball with a volume of 0.050 m3 completely submerged underwater, a force equal to the buoyant force must be applied. The buoyant force, calculated using Archimedes' principle, is 490 N, derived from the density of water (1000 kg/m3) and the gravitational acceleration (9.8 m/s2). Therefore, to counteract this buoyant force, an additional downward force of 490 N is required to keep the beach ball submerged.
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This is not correct. Look up Archimedes principle.jsalapide said:I learned that for an object that is completely submerged in a liquid, the density of the object is equal to the density of the liquid.
For all practical purposes, the density of the air filled beach ball can be neglected...that is, assume it has no weight.I used the formula p=m/v to get the mass of the ball. I set p as the density of water which is 1000 kg/m^3.
The buoyancy force is 490N upward. What downward force must be applied to keep the ball completely underwater?The answer I got was 50 kg. Then I multiply it to 9.8 m/s^2 to get the buoyant force.
My answer was 490 N...
Am I correct?