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What is the maximum rate with i beg your pardon a 1200-kg vehicle can ring a turn of radius 94.0 m on a flat road if the coefficient of static friction be


What is the maximum speed with which a 1200-kg automobile can ring a turn of radius 94.0 m ~ above a flat road if the coefficient of static friction in between tires and also road is 0.50?


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Answer:

v= 21.47m/s

Explanation:

For the auto to turn at the about the centripetal force must not be better than the revolution friction between the tires and the road

we will usage the expression relating centripetal force and also static friction below

let U represent the coefficient of static friction

Given that

U= 0.50

mass m= 1200-kg

radius r= 94.0 m

Assuming g= 9.81 m/s^2

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substituting our offered data in to expression we have the right to solve because that the rate V

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making v the topic of formula we have

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v= 21.47m/s

hence the preferably velocity the the vehicle is 21.47m/s


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