(7%) Problem 8: A car is driving along a circular track of diameter d = 0.95 km
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Question
(7%) Problem 8: A car is driving along a circular track of diameter d = 0.95 km at a constant speed of v = 27.5 m/s. - - - - - - Otheexpertta.com A 25% Part (a) Write an expression for the magnitude of the acceleration a of the car in terms of the given parameters. a= | Grade Summary Deductions 0% Potential 100% | b | (D 7 8 9 HOME | AN 4 5 6 | 1 2 || 3 || P + - 0. END VO BACKSPACE CLEAR Submissions Alternmpts remaining: 7 (0% per alteript) detailed view j n m O END E Hint | Hint Submit Hints: 0% deduction per hint. Hints remaining: 1 Feedback I give up! Feedback: 0% deduction per feedback. A 25% Part (b) What is the magnitude, in meters per second squared, of the acceleration a of the car? DA 25% Part (c) Write an expression for the minimum coefficient of friction a between the car's tires and the road that is required in order to keep the car going in a circle in terms of the given parameters. 25% Part (d) What is the value of the minimum coefficient of friction u between the car's tires and the road that is required in order to keep the car going in a circle?Explanation / Answer
here,
diameter , d = 0.95 km = 950 m
radius , r = d/2 = 475 m
speed , v = 27.5 m/s
a)
the magnitude of accelration of the car , ac = v^2 /r
b)
the magnitude of accelration of the car , ac = v^2 /r
ac = 27.5^2 /475 = 1.59 m/s^2
c)
let the minimum coefficient of friction be uk
equating the forces
uk * m * g - m * v^2 /r = 0
uk = v^2 /( r * g)
d)
the minimum coefficient of friction , uk = 27.5^2 /( 475 * 9.81)
uk = 0.16
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