Question #b30cd

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Question #b30cd

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Answer 1 · 2016-04-13T23:55:51

I used the definition of density:

Explanation:

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

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

$ρ = \frac{m}{V}$ $rho=m/V$

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

1) $V_{1} = 2 m^{3}$ $V_1=2m^3$ of bricks:
$m_{1} = ρ_{bricks} \cdot V_{1} = 2000 \cdot 2 = 4000 k g$ $m_1=rho_("bricks")*V_1=2000*2=4000kg$ Lowest mass of bricks

2) $V_{2} = 4 m^{3}$ $V_2=4m^3$ of bricks:
$m_{2} = ρ_{bricks} \cdot V_{2} = 2000 \cdot 4 = 8000 k g$ $m_2=rho_("bricks")*V_2=2000*4=8000kg$

3) $V_{3} = 6000 m^{3}$ $V_3=6000m^3$ of air:
$m_{3} = ρ_{air} \cdot V_{3} = 1 \cdot 6000 = 6000 k g$ $m_3=rho_("air")*V_3=1*6000=6000kg$ lowest mass of air

4) $V_{4} = 10000 m^{3}$ $V_4=10000m^3$ of air:
$m_{4} = ρ_{air} \cdot V_{4} = 1 \cdot 10000 = 10000 k g$ $m_4=rho_("air")*V_4=1*10000=10000kg$

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Question #b30cd

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