A Ping-Pong ball with mass 2.5 g is attached by a thread to the bottom of a beaker. When the beaker is filled with water so that the ball is

Question

A Ping-Pong ball with mass 2.5 g is attached by a thread to the bottom of a beaker. When the beaker is filled with water so that the ball is totally submerged, the tension in the thread is 0.029 N . 2.5 g Determine the diameter of the ball. The density of water is 1000 kg/m3 and the acceleration of gravity is 9.81 m/s 2 .

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Trúc Chi 2 months 2021-07-22T18:25:01+00:00 1 Answers 5 views 0

Answers ( )

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    2021-07-22T18:26:48+00:00

    Answer:

    0.022m or 2.2cm

    Expxlanation:

    Step 1:

    Data obtained from the question. This includes:

    Mass (m) = 2.5g = 2.5/1000 = 2.5×10^-3Kg

    Tension (T) = 0.029 N

    Density (ρ) = 1000 kg/m3

    Acceleration due to gravity (g) = 9.81 m/s2

    Diameter (d) =?

    Step 2:

    Finding an expression to calculate the diameter of the ball. This is illustrated below:

    Tension = weight displaced – weight of the ball

    Weight displaced = Mass of water x acceleration due to gravity

    Mass of water = Density x volume

    Mass of water = ρxV

    Weight displaced = ρxVxg = ρVg

    Weight of the ball = Mass of the ball x acceleration due to gravity

    Weight of the ball = mg

    Therefore,

    Tension = weight displaced – weight of the ball

    T = ρVg – mg

    Make V the subject of the formula

    T = ρVg – mg

    T + mg = ρVg

    Divide both side by ρg

    V = ( T + mg) /ρg. (1)

    Recall that the ball is spherical in shape and the Volume of a sphere is given by

    V = 4/3πr^3

    Radius (r) = diameter (d) /2

    V = 4/3π(d/2)^3

    V = 4/3πd^3/8

    V = πd^3 /6

    Substituting the value of V into equation 1, we have

    V = ( T + mg) /ρg

    πd^3 /6 = ( T + mg) /ρg.

    Making d the subject of the formula, we have:

    πd^3 /6 = (T + mg) /ρg.

    d^3 = 6(T + mg) /πρg.

    Taking the cube root of both sides

    d = [6(T + mg) /πρg]^1/3

    Step 3:

    Determination of the diameter of the ball. This is illustrated below:

    T = 0.029 N

    m = 2.5×10^-3Kg

    g = 9.81 m/s2

    ρ = 1000 kg/m3

    d =?

    d = [6(T + mg) /πρg]^1/3

    d = [6(0.029 + 2.5×10^-3×9.81)/ πx1000x9.81]^1/3

    d = 0.022m

    Therefore, the diameter of the ball is 0.022m or 2.2cm

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