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34kurt
3 years ago
10

An object of mass 3.4 kg is moving in a straight line with kinetic energy 59.177 J. A force is applied in the direction of its m

otion for 2.7 seconds, and as a result, its kinetic energy is multiplied by a factor of 1.57. By what factor is its momentum multiplied?
Physics
1 answer:
rusak2 [61]3 years ago
3 0

Answer:

Its momentum is multiplied by a factor of 1.25

Explanation:

First, we <u>calculate the initial velocity of the object</u>:

  • K = 0.5 * m * v₁²
  • 59.177 J = 0.5 * 3.4 kg * v₁²
  • v₁ = 5.9 m/s

With that velocity we can <u>calculate the initial momentum of the object</u>:

  • p₁ = v₁ * m
  • p₁ = 20.06 kg·m/s

Then we <u>calculate the velocity of the object once its kinetic energy has increased</u>:

  • (59.177 J) * 1.57 = 0.5 * 3.4 kg * v₂²
  • v₂ = 7.4 m/s

And <u>calculate the second momentum of the object</u>:

  • p₂ = v₂ * m
  • p₂ = 25.16 kg·m/s

Finally we <u>calculate the factor</u>:

  • p₂/p₁ = 1.25
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The electric field inside the sphere is given by

E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}

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The potential at the surface, V(R), is that of a point charge, so

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore we can find the potential inside the sphere, V(r):

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The graph of V versus r can be found in attachment.

We observe the following:

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- Between r = R and r = 3R, the potential decreases as \frac{1}{R}, therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 (\frac{1}{3}V(R))

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