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Blababa [14]
3 years ago
15

Interactive sites where users write about personal topics an comment to threaded discussion is called

Physics
1 answer:
pishuonlain [190]3 years ago
5 0
Blogs are sites where you can write about whatever you want, ranges from personal topics to general opinions. 
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T cell

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At take off, a plane flies 100 km north before turning to fly 200 km east. How far is its destination from where the plane took
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A less massive moving object has an elastic collision with a more massive object that is not moving. Compare the initial velocit
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Assume that the small-massed particle is m and the heavier mass particle is M.

Now, by momentum conservation and energy conservation:

   mv = mv_{m} + Mv_{M}

   mv^{2} = mv^{2}_{m} + Mv^{2}_{M}

Now, there are 2 solutions but, one of them is useless to this question's main point so I excluded that point. Ask me in the comments if you want the excluded solution too.

   v_{m} = v\frac{m - M}{m + M}\\\\\\v_{M} = v\frac{m + M}{m + M}\\

So now, we see that v_{m} < 0 and v_{M} > 0. So therefore, the smaller mass recoils out.

Hope this helps you!

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3 years ago
What is the safety rules?
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Quick question do you mean what are some safety rules

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A 91.0-kg hockey player is skating on ice at 5.50 m/s. another hockey player of equal mass, moving at 8.1 m/s in the
never [62]

The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of hockey player 1= 91.0-kg

(m₂) is the mass of hockey player 2=  91.0-kg

(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.

(u₂) is the velocity before the collision of hockey player 2=?

a)

Momentum before the collision;

\rm  m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2

Momentum before the collision = 1237.6 kg m/s².

b)

The velocity of the two hockey players after the collision from the law of conservation of the momentum as:

Momentum before collision = Momentum after the collision

1237.6 kg m/s² = (m₁+m₂)V

1237.6 kg m/s² =(2 ×91.0-kg )V

V=6.8 m/sec.

Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

8 0
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