# What is the density of a bowling ball with a mass of 3.0 kg and a volume of 0.0050 m^3?

Dec 31, 2016

$6.0 \cdot {10}^{2} {\text{kg m}}^{- 3}$

#### Explanation:

In order to find the density of the bowling ball, $\rho$, you must determine the mass one unit of volume of this bowling ball would have.

Notice that the problem provides you with the volume of the bowling ball expressed cubic meters, ${\text{m}}^{3}$, which means that one unit of volume would be ${\text{1 m}}^{3}$.

Now, you know that a volume of ${\text{0.0050 m}}^{3}$ has a mass of $\text{3.0 kg}$. You can use this known proportion as a conversion factor to figure the mass of ${\text{1 m}}^{3}$

1 color(red)(cancel(color(black)("m"^3))) * "3.0 kg"/(0.0050 color(red)(cancel(color(black)("m"^3)))) = "600 kg"

So, you know that one unit of volume has a mass of $\text{600 kg}$, which means that the bowling ball's density is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\rho = 6.0 \cdot {10}^{2} {\text{kg m}}^{- 3}}}}$

The answer is rounded to two sig figs.