Hasgill857
Answered

Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Discover solutions to your questions from experienced professionals across multiple fields on our comprehensive Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

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?

Sagot :

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
aachen

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 :

[tex]F=m\times \dfrac{v-u}{t}[/tex]

[tex]F=740\times \dfrac{0-19}{2}[/tex]

F = -7030 N

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

Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.