Gabriel L.

asked • 10/30/20

The volleyball is played with a ball of about 270 g and 21 cm in diameter.

In one game, the ball ended up in a pool with so much strength that it almost reached the bottom of it and then went up to the surface.

Calculate:


a) The force exerted by the pool water on the ball.

b) The acceleration with which the ball rose to the surface.

c) Assuming that the density of the ball was uniform, what part of it stood out from the water?

1 Expert Answer

By:

Daniel B. answered • 10/31/20

Tutor
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A retired computer professional to teach math, physics

Anthony T.

tutor
(4849 cm3 . 1g/cm3 - 270g) 9.8 m/s2, shouldn't the acceleration be 980 cm/s2 to be consistent with the other units?
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10/31/20

Daniel B.

It is not necessary. Notice that the two cm3 cancel each other, so there is no problem. And keeping g as m/s2 then makes the conversion to N easier. But this is a very good question, because in general we need to make sure that all units are compatible. My preferred approach is to carry the units together with all the numbers; this reduces the chances of making a mistake and makes any necessary conversion automatic.
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10/31/20

Gabriel L.

I didn't understand why the Thrust Force has to be subtracted by the weight of the ball to determine the Resulting Force. In the a) question.
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11/01/20

Daniel B.

There are two forces acting on the ball: 1) Gravity, mg, acting downward, which is the same whether the ball is in water or not. 2) Thrust force, Vdg, acting upward as long as the ball is submerged. Those two forces act in opposite direction, and therefore they are subtracted. If F = Vdg - mg is negative then gravity wins and the object sinks. If F is positive (as in our case), then the upward force, Vdg, is winning, pushing the object out of the water. As it starts emerging from water, the submerged volume V starts decreasing, decreasing the force F. When the submerged volume is reduced so much that the force F becomes 0, then the object floats.
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11/01/20

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