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frozen [14]
3 years ago
13

A student pushes on a crate with a force of 100 Newtons directed to the right.

Physics
1 answer:
romanna [79]3 years ago
6 0

Answer:

100N

Explanation:

Newton's third law of motion

For every action, there is an equal and opposite reaction.

Therefore 100N of force is exerted by the crate on student as a reaction to his action

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A. 4.64\cdot 10^{11}m

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v=\frac{2\pi r}{T}

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v=30,000 km/s = 3\cdot 10^7 m/s is the orbital speed

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T=27 h \cdot 3600 =97,200 s is the orbital period

Solving for r, we find the distance of the clumps of matter from the centre of the black hole:

r=\frac{vT}{2\pi}=\frac{(3\cdot 10^7 m/s)(97200 s)}{2\pi}=4.64\cdot 10^{11}m

B. 6.26\cdot 10^{36}kg, 3.13\cdot 10^6 M_s

The gravitational force between the black hole and the clumps of matter provides the centripetal force that keeps the matter in circular motion:

m\frac{v^2}{r}=\frac{GMm}{r^2}

where

m is the mass of the clumps of matter

G is the gravitational constant

M is the mass of the black hole

Solving the formula for M, we find the mass of the black hole:

M=\frac{v^2 r}{G}=\frac{(3\cdot 10^7 m/s)^2(4.64\cdot 10^{11} m)}{6.67\cdot 10^{-11}}=6.26\cdot 10^{36}kg

and considering the value of the solar mass

M_s = 2\cdot 10^{30}kg

the mass of the black hole as a multiple of our sun's mass is

M=\frac{6.26\cdot 10^{36} kg}{2\cdot 10^{30} kg}=3.13\cdot 10^6 M_s

C. 9.28\cdot 10^9 m

The radius of the event horizon is equal to the Schwarzschild radius of the black hole, which is given by

R=\frac{2MG}{c^2}

where M is the mass of the black hole and c is the speed of light.

Substituting numbers into the formula, we find

R=\frac{6.26\cdot 10^{36} kg)(6.67\cdot 10^{-11})}{(3\cdot 10^8 m/s)^2}=9.28\cdot 10^9 m

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