Ask Question
6 May, 04:50

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.

+1
Answers (1)
  1. 6 May, 05:20
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “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 ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers