What force would be required to hold it completely submerged under water ?

Question

A ping pong ball has a diameter of 4.78 cm and average density of .046 g/cm^2. The volume of a ball is 4/3(pie)R^3, where R is the radius of the ball. what force would be required to hold it completely submerged under water ? answer in units of N.

Answer 1

The volume of the ball is:
4/3(pi)*2.39^3
=57.185 cm^3
That much water would mass that many grams.
Multiply # by 0.046 for mass of the ball:
57.185 * 0.046
=2.63 g.
displaced water minus mass of ball is:
=54.55 g.
convert to newtons:
9.6 * 0.05455
=0.52368 newtons.
Edit: I just changed the semantics from weight to mass.

The volume of the ball is:
4/3(pi)*2.39^3
=57.18502458373722202431958932589cm^3
That much water would mass that many grams.
Multiply 0.046 by number for mass of the ball:
0.046*57.18502458373722202431958932589
=2.630511130851912213118701108991g.
displaced water minus mass of ball is:
=54.554513452885309811200888216899g.
convert to newtons:
9.6 * 0.054554513452885309811200888216899
=0.52372332914769897418752852688223 newtons.
This time I did it with the full precision of the MS calculator.

Answer 2

V = 4/3 ? r³ & r = ½ × 5.seventy 5 cm ? V = ninety 9.5 cm³ Ball's mass = ? × V = M = 0.0397 g/cm³ × ninety 9.5 cm³ = 3.ninety 5 g Water's ? = a million.0 g/cm³ Displaced water mass = a million.0 g/cm³ × 796.3 cm³ = 796.3 g Mass difference = 796.3 g ? 3.ninety 5 g = 792.3 g = 0.792 kg Buoyant unfavourable weight (in gadgets of tension) = 0.792 kg × 9.8 m/s² = 7.8 N