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5.) The density of a black hole is estimated to be as high as 6 x 1015 kg/m?. Using this density, what would be the

mass in teragrams of a piece of black hole the size of a golf ball, 2.50 in in volume? 1 inch = 2.54 cm
6x1018

User Aloso
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1 Answer

4 votes

Answer:

1.6 × 10^2 teragrams

Step-by-step explanation:

  • Density is the ratio of the mass of object to its volume.
  • It is calculated by;

Density = Mass ÷ Volume

we are given;

Density of a black hole = 6×10^15 kg/m³

Volume of the black hole = 2.5 in or 6.35 cm in volume

But; 1 g/cm³ = 1000 kg/m³

Therefore; 6×10^15 kg/m³ =6×10^12 g/cm³

We are required to get mass;

Volume = 4/3(πR³)

Assuming the radius is 2.5 inches or 6.35 cm

= 4/3 × 3.14 × 6.35 ³

= 26.585 cm³

But; Mass = Density × volume

= 6×10^12 g/cm³ × 26.585 cm³

= 1.6 × 10^14 g

but; 1 telegram = 1 teragram = 10^12 g

Therefore; 1.6 × 10^14 g ÷10^12 g

= 1.6 × 10^2 teragrams

Therefore, the mass of the black hole is 1.6 × 10^2 teragrams

User Mohammed Sabbah
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