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tangare [24]
3 years ago
7

A 1400kg car moving westward with a velocity of 15m/s collides with a utility pole and is brought to rest in 0.30sec. find the f

orce exerted on the car during the collision.?
Physics
2 answers:
sergejj [24]3 years ago
8 0
Force=(mass*velocity)-(mass*velocity)/time

force=0-(-15*1400)/0.30

force=70000N



HACTEHA [7]3 years ago
8 0

Answer: - 70kN

Explanation: Since the car is moving westward that is to the left it's initial velocity Vi is -15 m/s. (Take note of the negative sign.)

It's final velocity Vf is 0 m/s since the car eventually came to rest. The time taken for this change in velocity is 0.3secs.

Recall that,

Acceleration

= change in velocity/time

= {Vf - Vi}/t

={0 - 15}/0.3

= - 50m/s²

Now force exerted on the car

F = mass* acceleration

= 1400kg*-50m/s²

= -70000N

= -70kiloNewton

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Answer:

We conclude that the total distance traveled by the object in the 4 seconds is 36 m.

Explanation:

Given

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  • Acceleration a = 2.0 m/s²
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To determine

The total distance traveled by the object in the  4.0 seconds is

<u>Important Tip:</u>

We can determine the total distance traveled by the object in the  4.0 seconds by using the equation of motion such as

s=ut+\frac{1}{2}at^2

where

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substituting u = 5.0, a = 2, and t = 4 in the formula

s=ut+\frac{1}{2}at^2

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A force of 680 N is required to get the clock moving, so the maximum static friction is also <em>f</em> ˢ = 680 N. The clock is at rest, so the net vertical force acting on it is 0, and by Newton's second law,

<em>n</em> - <em>mg</em> = 0

where

<em>n</em> = magnitude of the normal force

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<em>g</em> = 9.8 m/s² = magnitude of the acceleration due to gravity

So we have

<em>n</em> = <em>mg</em> = (85 kg) (9.8 m/s²) = 833 N

which means the static friction <em>f</em> ˢ is such that

<em>f</em> ˢ = <em>µ </em>ˢ <em>n</em>

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