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The average density of the material in intergalactic space is approximately 2.5 × 10–27 kg/m3. what is the volume of a lead sample, ρ = 11 300 kg/m3, that has the same mass as 8.0 × 1024 m3 of intergalactic space?
a.1.0 × 10–6 m3
b.2.2 × 10–5 m3
c.5.4 × 10–6 m3
d.3.6 × 10–5 m3
e.1.8 × 10–6 m3

Respuesta :

Answer:

e. [tex]1.8\times 10^{-6}m^3[/tex]

Step-by-step explanation:

It is given that,

The density of intergalactic space material is [tex]2.5 \times 10^{-27}[/tex] kg per cubic meter.

And the volume of intergalactic space material is [tex]8.0\times 10^{24} m^3[/tex]

So the mass of that much intergalactic space material is,

[tex]m= \rho\times v[/tex], where 'm' is the mass, and 'v' is the volume.

Putting the values we get,

[tex]m=2.5 \times 10^{-27}\times 8.0 \times 10^{24}[/tex]

[tex]m=20\times 10^{(-27+24)}=20\times 10^{-3} kg[/tex]

It is also given that the mass of lead is the same as the mass of the intergalactic space material. Therefore, the mass of lead is [tex]20 \times 10^{-3}=2 \times 10^{-2}kg[/tex]

So the volume of lead is,

[tex]v=\frac{m}{\rho} =\frac{2 \times 10^{-2}}{11300} = 0.00000177 m^3[/tex]

[tex]v=1.77 \times 10^{-6}=1.8\times 10^{-6}m^3[/tex]

So the volume of lead is [tex]v=1.8\times 10^{-6}m^3[/tex].

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