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alina1380 [7]
2 years ago
5

A stationary skateboarder I with a mass of 50 kg pushes a stationary skateboarder II with a mass of 75 kg. After the push the sk

ateboarder II moves with a velocity of 2 m/s to the right. What is the velocity of the skateboarder I?
Physics
1 answer:
Mice21 [21]2 years ago
8 0

The final velocity of the skateboarder I after the collision is 3 m/s to the left.

<h3>Velocity of skateboarder I</h3>

The velocity of skateboarder I is determined from the principle of conservation of linear momentum.

m1u1 + m2u2 = m1v1 + m2v2

50(0) + 75(0) = 50v1 + 75(2)

0 = 50v1 + 150

v1 = (-150)/50

v1 = - 3

v1 = 3 m/s to the left

Thus, the final velocity of the skateboarder I after the collision is 3 m/s to the left.

Learn more about linear momentum here: brainly.com/question/7538238

#SPJ1

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The electric force between two charged particles can be increased by decreasing the distance between the two particles.

<h3>How to increase electric force between two charged particles.</h3>

The technique of decreasing the separation distance between objects increases the force of attraction or repulsion between the objects. while

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1 year ago
Arrange the balls in order from greatest amount of gravitational potential energy to least.
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All of them have the same potential energy <span />
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3 years ago
Read 2 more answers
Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s
frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀)  we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t²- 2 t to + to²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to.  The rock y has not left and the distance increases

For t> = to.  the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t   v₁> v₂

3 0
3 years ago
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