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EleoNora [17]
2 years ago
11

A 3.0-kg ball with an initial velocity of (4.0i + 3.0j) m/s collides with a wall and rebounds with a

Physics
1 answer:
irga5000 [103]2 years ago
3 0

The impulse exerted on the ball by the wall is 24i kgm/s.

<h3>Impulse exerted by the ball on the wall</h3>

The impulse exerted by the ball on the wall is the change in the linear momentum of the ball.

J = ΔP

ΔP = Pf - Pi

P = mv

where;

  • m is mass of the ball
  • v is the velocity

ΔP = 3(4.0i + 3.0j) - 3(-4.0i + 3.0j)

ΔP = (12i + 9j) - (-12i - 9j)

ΔP = 24i kgm/s

Thus, the impulse exerted on the ball by the wall is 24i kgm/s.

Learn more about impulse here: brainly.com/question/904448

#SPJ1

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 At point B (located at the bottom of the wedge), the energy is:
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SP 15The magnetic field in a region of space is measured to be:This field is known to be caused by a cluster of long-straight wi
tino4ka555 [31]

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 i = 0.477 10⁴ B

the current flows in the  counterclockwise

Explanation:

For this exercise let's use the Ampere law

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Where the path is closed

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