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A sphere is dropped in a can partially filled with water. What is the rise in the height of water if

they have equal diameters of 6in?​


Sagot :

Problem:

A sphere is dropped in a can partially filled with water. What is the rise in the height of water if they have equal diameters of 6in?​

Solution:

Diameter of sphere = 6 in

r of sphere = 6/2

r of sphere = 3 in

Volume of sphere

[tex]\[\begin{array}{l}V = \frac{4}{3}\pi {r^3}\\\\V = \frac{4}{3}\pi {(3)^3}\end{array}\][/tex]

[tex]\[V = \frac{4}{3}\pi (27)\][/tex]

V = 36π

A can is a cylindrical vessel

diameter of can = 6 in

radius of can = 6/2

radius of can = 3 in

Volume of can = πr²h

Volume of can = π(3)²h

Volume of can = π(9)h

to get the rise in height, equate the volume of both sphere and cylinder

Volume of sphere = Volume of can

36π = π(9)(h)

h = 4 in

Answer:

There is a rise of 4 in height after the sphere was dropped in a can

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