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

1.6 × 10^2 teragrams

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

answer: a tank of height h is filled with water and sits on the ground. water squirts from a hole at a height y above the ground and has a range r. for two y values, 0 and h, r is zero. determine the value of y for which the range will be a maximum.

explanation: