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2 years ago

5

the law of conservation of momentum states that, if left alone what happens to the total momentum of two interacting objects tha

t make up a system

1 answer:

miv72 [106K]2 years ago

4 0

Answer:

It remains the same

Explanation:

At any point in time the total momentum of two objects interacting remains the same

This follows the laws of space(space is uniform)

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What happens in the process of gravitational condensation?

Lerok [7]

Answer:

An object decreases in size due to the collision of materials. An object increases in size due to the addition of materials. Gas particles are formed from solar nebula materials.

3 0

3 years ago

ella [17]

I think the answer maybe C

5 0

3 years ago

Read 2 more answers

A cue stick hits a cue ball with an average force of 22 N for a duration of 0.029 s. If the mass of the ball is 0.15 kg, how fas

Likurg_2 [28]

Answer:

4.25 m/s

Explanation:

Force, F = 22 N

Time, t = 0.029 s

mass, m = 0.15 kg

initial velocity of the cue ball, u = 0

Let v be the final velocity of the cue ball.

Use newton's second law

Force = rate of change on momentum

F = m (v - u) / t

22 = 0.15 ( v - 0) / 0.029

v = 4.25 m/s

Thus, the velocity of cue ball after being struck is 4.25 m/s.

8 0

3 years ago

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In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting a

IRISSAK [1]

A. $4.64\cdot 10^{11}m$

The orbital speed of the clumps of matter around the black hole is equal to the ratio between the circumference of the orbit and the period of revolution:

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

where we have:

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

r is the orbital radius

$T=27 h \cdot 3600 =97,200 s$ 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

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

8 0

2 years ago

Jimmy is cooking pasta in a pot of water on the stove. He uses a stopwatch to see how long the pasta takes to cook depending on

Alex787 [66]

Answer:

stove

Explanation:

The stove is the independent variable in this question as it does not depend on anything like other variables such as pot, water, temperature and of cause pasta.

The pot depends on the stove, the water depends on the pot and stove, the pasta depends on the water and its temperature.

Cheers.

4 0

3 years ago