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Ronch [10]
3 years ago
9

What happens to the motion of an object when the forces acting on it are balanced? A. The motion changes. B. The motion does not

change. C. The motion speeds up. D. The motion slows down.
Physics
2 answers:
dexar [7]3 years ago
8 0

D. The motion slows down. Because If they are two forces putting force on the opposite sides, then the object would start to balance out and slow down

S_A_V [24]3 years ago
4 0

Question: What happens to the motion of an object when the forces acting on it are balanced?

Choices:

A. The motion changes.

B. The motion does not change.

C. The motion speeds up.

D. The motion slows down.

Answer: <u>D) The motion slows down when a motion of an object forces acting on the balance</u>.

<em>Hope this helps!.</em>

<em>~~~~~~~~~~~</em>

<em>~A.W~ZoomZoom44</em>

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When we jump from the truck and accelerate towards the earth surface, the earth also accelerates towards us but it's acceleration is very negligible.

To find the answer, we need to know about the acceleration of earth due to the gravitational attraction.

<h3>What's the gravitational force between the earth and a person?</h3>
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m= mass of the person

r= separation between them.

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2 years ago
Why aren't descriptive investigations repeatable ?
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Because the information cant be out of the investigation
4 0
3 years ago
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A proton that has a mass m and is moving at +164 m/s undergoes a head-on elastic collision with a stationary carbon nucleus of m
Irina18 [472]
The concept of this problem is the Law of Conservation of Momentum. Momentum is the product of mass and velocity. To obey the law, the momentum before and after collision should be equal:

m₁ v₁ + m₂v₂ = m₁v₁' + m₂v₂', where
m₁ and m₂ are the masses of the proton and the carbon nucleus, respectively,
v₁ and v₂ are the velocities of the proton and the carbon nucleus before collision, respectively,
v₁' and v₂' are the velocities of the proton and the carbon nucleus after collision, respectively,

m(164) + 12m(0) = mv₁' + 12mv₂'
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The second equation is the coefficient of restitution, e, which is equal to 1 for perfect collision. The equation is

(v₂' - v₁')/(v₁ - v₂) = 1
(v₂' - v₁')/(164 - 0) = 1
v₂' - v₁'=164 ---> equation 2

Solving equations 1 and 2 simultaneously, v₁' =  -138.77 m/s and v₂' = +25.23 m/s. This means that after the collision, the proton bounced to the left at 138.77 m/s, while the stationary carbon nucleus move to the right at 25.23 m/s.
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