1. A car is safely negotiating an unbanked circular turn at a speed of 20 m/s. T

Question

1.     A car is safely negotiating an unbanked circular turn at a speed of 20 m/s. The road is dry, and the maximum static frictional force acts on the tires. Suddenly a long wet patch in the road decreases the maximum static frictional force to one-fourth of its dry-road value. If the car is to continue safely around the curve, to what speed must the driver slow the car?

_______________m/s

_______________m/s

A car is safely negotiating an unbanked circular turn at a speed of 20 m/s. The road is dry, and the maximum static frictional force acts on the tires. Suddenly a long wet patch in the road decreases the maximum static frictional force to one-fourth of its dry-road value. If the car is to continue safely around the curve, to what speed must the driver slow the car? What is the average linear speed of Jupiter about the Sun? (The distance from the Sun to Jupiter is 7.78 times 1011 meters. The orbital period around the Sun is 4332.7 days.)

Explanation / Answer

1.

if Fc = m/r *(V^2)and m and r are constant, then

Fc/3 = m/r * (V^2)/3

Since V was initially 20 m/s, V^2 would be 400 and V^2/ 3 would be 133.3333

V = Sqrt (147) = 11.54m/s

2.

I hope by days they mean earth's day (24 hours) and not the time taken by Jupiter to rotate about its axis.
If so, the problem is simple. First find T in seconds by multiplying 4332.7 by 24 and then by 3600.
This will be the denominator.
The nominator will be 2*3.142*7.78E11.
My answer is 13051.69 m/s

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