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 m3b.2.2 × 10–5 m3c.5.4 × 10–6 m3d.3.6 × 10–5 m3e.1.8 × 10–6 m3

e.

Step-by-step explanation:

It is given that,

The density of intergalactic space material is kg per cubic meter.

And the volume of intergalactic space material is

So the mass of that much intergalactic space material is,

, where 'm' is the mass, and 'v' is the volume.

Putting the values we get,

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

So the volume of lead is,

So the volume of lead is .

11

step-by-step explanation:

32.50-7.50=25

225/25=11

answer:   28.56 feet (approx)

step-by-step explanation:

since, triangle abc has measure of angle c equal to 55 degrees, measure of angle abc equal to 90 degrees, and length of bc equal to 20 feet.

we have to find out the measurement of segment ab.

therefore, tan 55° =

but bc = 20 feet.

⇒ tan 55 ° =  

⇒ ab = 20 × tan 55 °

⇒ ab = 20 × 1.42814800674

⇒ ab = 28.5629601348   ≈ 28.56 feet