At a classic auto show, a 830 kg 1955 Nash Metropolitan motors by at 8.6 m/s , f

Question

At a classic auto show, a 830 kg 1955 Nash Metropolitan motors by at 8.6 m/s , followed by a 1600 kg 1957 Packard Clipper purring past at 4.8 m/s .

Part C

Let FN be the net force required to stop the Nash in time t, and let FP be the net force required to stop the Packard in the same time. What is the ratio FN/FP of these two forces?

SubmitMy AnswersGive Up

Part D

Now let FN be the net force required to stop the Nash in a distance d, and let FP be the net force required to stop the Packard in the same distance. What is the ratio FN/FP?

Explanation / Answer

According to the question we need to find the

Part C

Let FN be the net force required to stop the Nash in time t, and let FP be the net force required to stop the Packard in the same time. What is the ratio FN/FP of these two forces?

Answer) F(N) / F(P) = 8.6 m/s * 830 kg / 4.8 m/s * 1600

F(N) / F(P) = 0.929427083 ======================answer)

Part D

Now let FN be the net force required to stop the Nash in a distance d, and let FP be the net force required to stop the Packard in the same distance. What is the ratio FN/FP?

Answer) As we can see the distance are same so for that we will use

d=KE/F = (8.6 m/s)2 * 830 kg / ( 4.8 m/s )2 * 1600 = 1.665223524 ================ANSWER)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.