Class Management Help Chapter 4 Begin Date: 9/14/2018 1201:00 AM Due Date: 9/24/

Question

Class Management Help Chapter 4 Begin Date: 9/14/2018 1201:00 AM Due Date: 9/24/2018 11:59 00 PM End Date: 10/1/2018 11:59:00 PM (8%) Problem 10: A car is traveling at constant speed ofys 22.2 ms Its tires have a diameter of d Randomized Variables v= 22.2 m/s 0.99 m. d 0.99 m 50% Part (a) Consider a port on the outer edge of the tre what is the centripetal acceleration, ac, at this port in mi? ac = 1497 81 Grade Summary Deductions 10% Potential 90% sincos0 cotanOasinOacos0 atanO acotansinh0 coshOtanh0cotanh(0 Subnsissions Attempts remaining 4 (10% per attempt) detailed vie END * Degrees Radians NO BACKSPACE DEL CLEAR Submit Hint Feedback Hints: deduction per hint Hints remaining Feedback: deduction per feedback. 50% Part (b) If the car was traveling twice as fast, what would be the numerical value of the ratio of the new centripetal acceleration, a , to the old centripetal acceleration?

Explanation / Answer

Centripetal acceleration is given by, Ac = v^2/r

A)

Acceleration, ac = 22.2^2/0.495 = 995.64 m/s^2

B)

If v' = 2 v

Then (ac) new/ (ac)older = v'^2 / v^2 = (2 v)^2/ v = 4

So, ratio will be 4

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