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rosijanka [135]
3 years ago
14

What is the momentum of a 950 kg car moving at 10.0 m/s?

Physics
2 answers:
pentagon [3]3 years ago
8 0

Answer:

<h3>The answer is 9500 kgm/s</h3>

Explanation:

The momentum of an object can be found by using the formula

<h3>momentum = mass × velocity</h3>

From the question

mass = 950 kg

velocity = 10.0 m/s

We have

momentum = 950 × 10

We have the final answer as

<h3>9500 kgm/s</h3>

Hope this helps you

Nesterboy [21]3 years ago
8 0
<h2><u>Given</u> :</h2>
  • A 950 kg car moving at 10.0 ms
<h2><u>To Find</u> :</h2>
  • Momentum
<h2><u>Solution</u> :</h2>

<u>We have,</u>

  • Mass = 950 kg
  • Velocity = 10.0 m/s = 10 m/s

<u>We know that,</u>

<h3>→ Momentum = Mass × Velocity</h3>

<u>Substituting the values we get,</u>

\impliesMomentum = 950 × 10

\impliesMomentum = 9500 kgm/s

∴ The momentum is 9500 kgm/s.

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Answer:

A. Sedimentary

Explanation:

I took the test and got this correct. Hopefully this helps you!

3 0
3 years ago
Water raises a boat once every 3.0 seconds. What is the frequency (f) of the waves passing the boat?
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I think its A

Explanation:

if its every 3 seconds wouldnt it be 3.0 Hz

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If a football player collides with a goal post, what forces are at work?
AlekseyPX
When a footballer collides with the goal post, the forces at work are the action and reaction forces. The player will exert an action force on the goal post, and then a reaction force from the goal post will stop the player. The reaction force call will cause pain and even injury to the player.
7 0
3 years ago
A series of pulses, each of amplitude 0.1 m, is sent down a string that is attached to a post at one end. The pulses are reflect
AleksAgata [21]

Answer:

 A_resulting = 0.2 m

Explanation:

Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.

With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is

           A_res = 2A

           A_resultant = 2 .01

           A_resulting = 0.2 m

8 0
2 years ago
An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b
lys-0071 [83]

Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

             S = ut + 0.5at²

     Here u = 0, and a = g

              S = 0.5gt²

    Distance traveled during the first second ( t =1 )

              S = 0.5 x 9.81 x 1² = 4.905 m

   Distance traveled during the first second = 4.905 m.

b)  We have equation of motion

            v² = u² + 2as

      Here u = 0, s= 75 m and a = g

           v² = 0² + 2 x g x 75 = 150 x 9.81

           v = 38.36 m/s

      Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

                   75 = 0.5 x 9.81 x t²

                    t = 3.91 s

   We need to find distance traveled last second

   That is

          S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

   Distance traveled during the last second of motion before hitting the ground = 33.45 m

       

3 0
3 years ago
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