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34kurt
2 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]2 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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Answer:

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Explanation:

a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)

a= ω^2*r

ω= 300rev/min

convert into rev/s

300/60= 5rev/s

a= 18.75m/s^2

b) use Krot= 1/2 Iω^2

plug in gives

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8 0
2 years ago
large amounts of energy are converted to small amounts of mass. small amounts of energy are converted to large amounts of mass.
Norma-Jean [14]

Answer:

small amounts of mass are converted to large amounts of energy

Explanation:

According to the mass-energy equivalence, which Albert Einstein initially proposed as a general principle. It was revealed that mass and energy are connected and that a "small amount of mass can be converted into enormous amounts of energy."

Using the formula E=mc^2. This means Energy equals mass times the speed of light squared.

Hence, it is true that "small amounts of mass are converted to large amounts of energy."

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3 years ago
A source charge of 3 µC generates an electric field of 2.86 × 105 N/C at the location of a test charge. Using k = 8.99 × 109N.m^
Nataliya [291]
Variables:

Source charge, Q = 3 micro C = 3 * 10^ - 6 C

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d = distance = ?

Formula:

E = K * Q / (d^2) => d^2 = K * Q / E

=> d^2 = 8.99 * 10^9 N * m^2 / C * 3 * 10^ -6 C / (2.86 * 10^ 5 N/C)

d^2 = 9.43 * 10 ^ -2  m^2

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7 0
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Answer:

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Explanation:

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workdone = force × distance

From the question we have

workdone = 100 × 0.5

We have the final answer as

<h3>50 J</h3>

Hope this helps you

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Answer:

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