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Whitepunk [10]
2 years ago
14

65 POINTS! PLEASE ANSWER EVERY QUESTION! NEED HELP ASAP!

Physics
1 answer:
otez555 [7]2 years ago
3 0
Maybe you can split up the questions. I will try to answer your first question.

1. In an elastic collision, momentum is conserved. The momentum before the collision is equal to the momentum after the collision. This is a consequence of Newton's 3rd law. (Action = Reaction)

2. Momentum: p = m₁v₁ + m₂v₂

m₁ mass of ball A
v₁ velocity of ball A
m₂ mass of ball B
v₂ velocity of ball B

Momentum before the collision:
p = 2*9 + 3*(-6) = 18 - 18 = 0

Momentum after the collision:
p = 2*(-9) + 3*6 = -18 + 18 = 0

3: mv + m(-v) = m(-v) + m(v)
the velocities would reverse.

4.This question is not factual since the energy of an elastic collision must also be conserved. The final velocities should be: v₁ = -1 m/s and v₂ = 5 m/s. That said assuming the given velocities were correct:
before collision
p = 10*3 + 5*(-3) = 30 - 15 = 15
after collision:
p = 10*(-2) + 5 * v₂ = 15
v₂ = 7

5.You figure out.



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A stubborn, 100 kgkg mule sits down and refuses to move. To drag the mule to the barn, the exasperated farmer ties a rope around
jolli1 [7]

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Explanation:

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3 years ago
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4 0
2 years ago
PLEASEE HELP!!! LOTS OF POINTS AND BRAINLIEST!!!!!!!!
Dennis_Churaev [7]
A clock and a battery
3 0
2 years ago
Read 2 more answers
What is the magnitude of g at a height above Earth's surface where free-fall acceleration equals 6.5m/s^2?
prohojiy [21]

You've given the answer, right there in your question.

The "magnitude of gravity" is described in terms of the acceleration
due to it, and you just told us what that is.

We can also notice that the figure you gave is about 0.66 of the
acceleration due to gravity on the Earth's surface. That tells us that
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                     (1 / √0.66) = 1.23 times

the Earth's radius, so the height is about  910 miles above the surface.


7 0
3 years ago
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