A Ping-Pong ball has a diameter of 1.99 cm and average density of 0.121 g/cm3. What force would be required to hold it completely submerged under water? The acceleration of gravity is 9.8 m/s 2. Answer in units of N.

0.035 N

Explanation:

Density = mass/volume

D = m/v

m = D * v ... Equation 1

Where D = density of the ping-pong, v = volume of the ping-pong, m = mass of the ping-pong

Note: The ping-pong is spherical in shape.

v = 4/3πr²

Where r = radius, π = pie

d = 1.99 cm, π = 3.14

v = 4/3 (1.99/2) ² (3.14)

v = 4.12 cm³

Also D = 0.121 g/cm³

Therefore,

m = 0.121 (4.12)

m = 0.499 g

W = mg

Where W = weight of the ping-pong

W = (0.499/1000) * 9.81

W = 0.005 N.

From Archimedes principle,

Upthrust = density of water * volume of water displaced * acceleration due to gravity.

U = D'vg/1000 ... Equation 2

Note: The volume of water displaced is equal to the volume of the ping-pong.

given: v = 4.12 cm³, g = 9.81 m/s², D' = 1 g/cm³

Substitute into equation 2

U = 1 (4.12) (9.81) / 1000

U = 0.04 N

The force required to hold the ball completely submerged under water = U-W

= 0.04-0.005 = 0.035 N