# Question #b30cd

Apr 13, 2016

I used the definition of density:

#### Explanation:

I think you mean density of air not water (water density should be approx. $1000 \frac{k g}{m} ^ 3$).

You can use the relationship between density, $\rho$, volume, $V$, and mass $m$:

$\rho = \frac{m}{V}$

in your case (for the 4 options) you have:

1) ${V}_{1} = 2 {m}^{3}$ of bricks:
${m}_{1} = {\rho}_{\text{bricks}} \cdot {V}_{1} = 2000 \cdot 2 = 4000 k g$ Lowest mass of bricks

2) ${V}_{2} = 4 {m}^{3}$ of bricks:
${m}_{2} = {\rho}_{\text{bricks}} \cdot {V}_{2} = 2000 \cdot 4 = 8000 k g$

3) ${V}_{3} = 6000 {m}^{3}$ of air:
${m}_{3} = {\rho}_{\text{air}} \cdot {V}_{3} = 1 \cdot 6000 = 6000 k g$ lowest mass of air

4) ${V}_{4} = 10000 {m}^{3}$ of air:
${m}_{4} = {\rho}_{\text{air}} \cdot {V}_{4} = 1 \cdot 10000 = 10000 k g$