Question

A block of an unknown metal has a mass of 227.8 kg and a volume of 0.1189 cubic meters. If the object were a sphere, what would be its radius? (The value for pi ≈ 3.14159)

Answer 1

Step-by-step explanation:

The given data is as follows.

mass = 227.8 kg, volume = 0.1189
`m^(3)`

It is known that formula for volume of a sphere is
`(4)/(3) \pi r^(3)` and density is as follows.

Density =
`(mass)/(volume)` ........ (1)

=
`(227.8 kg)/(0.1189 m^(3))`

= 1915.895
`kg/m^(3)`

Hence, putting the given values into equation (1) as follows.

Density =
`(mass)/(volume)`

1915.895
`kg/m^(3)` =
`(227.8 kg)/((4)/(3) \pi r^(3))`

`8021.213 r^(3)` = 227.8

r = 0.305
`m^(3)`

Thus, we can conclude that radius of the sphere is 0.305
`m^(3)`.