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27 April, 21:58

A heavy metal sphere

with radius 10 cm is dropped into a rightcircular cylinder with

base radius of 10 cm. If the originalcylinder has water in it that

is 20 cm high, how high is the waterafter the sphere is placed in

it?

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Answers (1)
  1. 28 April, 00:42
    0
    Answer: the height of the water after the sphere is placed in

    it is 33.33 cm

    Step-by-step explanation:

    The cylinder is called a right circular cylinder because its height make a right angle with its base. The formula for determining the volume of the cylinder is expressed as

    Volume = πr^2h

    Where

    π is a constant whose value is 3.14

    r represents the radius of the cylinder.

    h represents the height of the cylinder.

    From the information given,

    r = 10 cm

    h = height of water in the cylinder = 20 cm

    Volume of water in the cylinder before the sphere was placed in it would be

    V = 3.14 * 10^2 * 20 = 6280 cm^3

    The formula for determining the volume of the sphere is expressed as

    Volume = 4/3 πr^3

    V = 4/3 * 3.14 * 10^3 = 4186.67cm^3

    Total volume of the sphere and the cylinder = 6280 + 4186.67 = 10466.67 cm^2

    To determine the new height of the water,

    10466.67 = 3.14 * 10^2 * h

    h = 10466.67/314 = 33.33 cm
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