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Svet_ta [14]
2 years ago
9

Provide three examples of situations in which mass is the main factor determining an object's momentum

Physics
1 answer:
Ratling [72]2 years ago
3 0

The three examples of situations in which mass is the main factor determining an object's momentum are, a moving truck, a moving trailer and a rolling block.

<h3>What is momentum?</h3>
  • Momentum is the product of mass and velocity of an object

P = mv

where;

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

The three examples of situations in which mass is the main factor determining an object's momentum is below;

  • A moving truck.
  • A moving trailer.
  • A rolling block.

The three examples given above shows three objects with heavy mass.

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

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You and your friends are having a discussion about weight. He/she claims that he/she weighs less on the 100th floor of a buildin
Viktor [21]

Answer:

if the weight theoretically decreases at this height, but in a fraction of 10⁻⁵, which is not appreciable in any scale, therefore, the reading of the scale in the two places is the same.

Explanation:

The weight of a person in the force with which the Earth attracts the person, therefore can be calculated using the law of universal attraction

          F = G m M / r²

Where m is the mass of the person, M the masses of the earth

Let's call the person's weight at ground level as Wo and suppose the distance to the center of the Earth is Re

            W₀ = G m M / Re²

In the calculation of the weight of the person on the 100th floor the only thing that changes is the distance

          r = Re + 100 r₀

Where r₀ is the distance between the floors, which is approximately 2.5 m, so the distance is

         r = Re + 250

We substitute

     W = G m M / r²

      W = G m M / (Re + 250)²

The value of Re is 6.37 10⁶ m, so we can take it out as a factor and perform a serial expansion of the remaining fraction

      W = G m M / Re² (1+ 250 / Re)²

      (1 + 250 / Re)⁻² = 1 + (-2) 250 / Re + (-2 (-2-1)) / 2 (250 / Re)² +….

The value of the expression is

      (1 + 250 / Re)⁻² = 1 -2 250 / 6.37 10⁶ -30 (250 / 6.37)² 10⁻¹² + ...

We can see that the quadratic term is very small, which is why we despise it, we substitute in the weight equation

      W = G m M / Re² (1 - 78.5 10⁻⁶)

Remains

     W = Wo (1 - 7.85  10⁻⁵)

We can see that if the weight theoretically decreases at this height, but in a fraction of 10⁻⁵, which is not appreciable in any scale, therefore, the reading of the scale in the two places is the same.

4 0
3 years ago
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Ymorist [56]

Answer:

(a)2.7 m/s

(b) 5.52 m/s

Explanation:

The total of the system would be conserved as no external force is acting on it.

Initial momentum = final momentum

⇒(4.30 g × 943 m/s) + (730 g × 0) = (4.30 g × 484 m/s) + (730 g × v)

⇒ 730 ×v = (4054.9 - 2081.2) =1973.7

⇒v=2.7 m/s

Thus, the resulting speed of the block is 2.7 m/s.

(b) since, the momentum is conserved, the speed of the bullet-block center of mass would be constant.

V_{COM} = \frac{m_b}{m_b+m_{bl}}v_{bi}=\frac{4.30}{4.30+730}\times 943 m/s = 5.52 m/s

Thus, the speed of the bullet-block center of mass is 5.52 m/s.

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Explanation:

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Answer:

I think is 2.

Explanation:

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Explanation:

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