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TiliK225 [7]
3 years ago
7

A 2020 kg car traveling at 14.2 m/s collides with a 2940 kg car that is initally at rest at a stoplight. The cars stick together

and move 2.12 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.
Physics
1 answer:
velikii [3]3 years ago
4 0

Answer:

The answer is 3,064xe^{-4}

Explanation:

When the collision happens, the momentum of the first car is applied to the both of them.

So we can calculate the force that acts on both cars as:

  • The momentum of the first car is P = 2020 kg x 14.2 m/s = 28,684 kg.m/s
  • The acceleration of both cars after the crash is going to be a = P / mtotal which will give us a = 28,684 / (2020+2940) = 5.78 m/s
  • Since the second car was initially not moving, the final acceleration was calculated with the momentum of the first car.

Now we can find the force that acts on both of them by using the formula F = m.a which will give us the result as:

  • F = (2020+2940) x 5.78 = 28,684

The friction force acts in the opposite direction and if they stop after moving 2.12 meters;

  • Friction force is Ff = μ x N where μ is the friction coefficient and the N is the normal force which is (2020+2940) x 10 if we take gravitational force as 10, equals to 49,600.
  • F - Ffriction = m x V
  • 28,684 - μ x 49,600 = 4960 x 5.78
  • μ = 3,064xe^{-4}

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An electron moves at a speed of 1.0 x 104 m/s in a circular path of radius 2 cm inside a solenoid. The magnetic field of the sol
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Answer:

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Explanation:

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b. Given that; N/L = 25 turns per centimetre, then the current, I, can be determined by;

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where B is the magnetic field,  μ is the permeability of free space = 4.0 ×10^{-7}Tm/A, N/L is the number of turns per length.

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