Can anyone help me in this physics problem?

Question

A 1.0 kg hollow ball with a radius of 0.10 m is filled with air and is released from rest at the bottom of a 2.0 m deep pool of water. How high above the water does the ball rise? Disregard friction and the ball’s motion when it is only partially submerged.

Answer 1

V = (4/3)πr^3
0.1 m = 10 cm
V = (4/3)π(1,000) cm^3 = 4,188.790 cm^3
The mass of that volume of water is 4.18879 kg, and the buoyant force on the ball is
9.8*(4.18879 - 1) = 31.250142 N while the ball is totally submerged. To simplify the problem, assume the force is constant until the ball is exactly halfway out of the water, and 0 thereafter. The force then acts through a distance of 1.9 m (2 m - 0.1 m), giving the ball a kinetic energy of
1.9*31.250142 = 59.3752698 J
the v^2 of the ball upon "leaving" the water is then
2*59.3752698 = 118.7505396 m^2/s^2
The height the center of the ball reaches above the water is then given by
118.7505396/(2*9.8) = 6.058701 meters

There's a boatload of simplifying assumptions in that answer, like ignoring the mass of the air in the ball, any expansion on rising, viscosity, and, finally, the really tough calculation of force on the ball as it pops out of the water.

Answer 2

The ball will displace 1 kg of water. You need to find the volume of the part of the sphere that holds 1 cubic metre(= 1kg) of water. The ball will sit at this elevation in the pool.

Make sure you get the right section... The volume of the sphere is about 4.2 m^3 - you need to ensure that it is the smaller portion of the sphere which is under water. :)

Answer 3

ill solve it here
f acting downward is 10N
upward f is v*d*g = 4/3(.001)*1000*10=13.33N
as 10 n=v
volume shid be equal to .001 i dont have calculator calculate it by assuming ball partilly submerged thats the ans

Answer 4

In practice , the ball will come up in the water and it will stop on the water surface. because it is not lighter than the air and it won't come up. it will stop at the surface.

Answer 5

the galaxies rotate at speeds inconsistent with their obvious mass is as a results of the fact we do see all of it. i'm touching on it as being the theoretical dark count. There are very solid proofs that shows that dark count exist. One the is the inconsistent velocity of and obvious mass. dark count makes up approximately seventy 5% to 80% of the situation interior the Universe...

Answer 6

well, the magnitude is obviously at an all time inflate, which means the volume would be equal to 2.0-1.0 over .10...which would solve ur answer, but dont forget to cyress the number u get, i would do it, but i dont have a calculator, its quite simple...