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2 years ago

8. A 740 kg car traveling 19 m/s comes to a complete stop in 2.0 s. What is the force exerted on the car during this stop?

2 answers:

6 0

Force = mass * acceleration

acceleration = change_in_velocity / time

so:

force = 740 kg * (19 m/s - 0 m/s) / 2.0 s
= 740 * 19 / 2 kg m per second^{2}
= 7030 kg m per second^{2}
= 7030 newtons of force

svet-max [94.6K]2 years ago

3 0

Answer:

Force exerted on the car is 7030 N.

Explanation:

It is given that,

Mass of the car, m = 740 kg

Initial speed of the car, u = 19 m/s

Final speed of the car, v = 0

Time taken, t = 2 s

Let F is the force exerted on the car during this stop. We know that it is equal to the product of force and acceleration. Mathematically, it is given as :

$F=m\times \dfrac{v-u}{t}$

$F=740\times \dfrac{0-19}{2}$

F = -7030 N

So, the force exerted on the car during this stop is 7030 N. Hence, this is the required solution.

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Answer:

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$P_{2} V_{2} = P_{3} V_{3}$

$P_{3} = \frac{P_{2} V_{2}}{V_{3}}$

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