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mamaluj [8]
3 years ago
10

A 2.5 x 10^3 kilogram truck with rubber tires moves

Physics
1 answer:
ArbitrLikvidat [17]3 years ago
3 0

Answer:

The question implies that force of friction is 2.1E4 N

F = M v^2 / R       frictional force needed to supply this centripetal force

2.1E4 = M v^2 / R

v^2 = R * 2.1E4 / M

v^2 = 120 * 2.1E4 / 2.5E3 = 1200 * 2.1 / 2.5 = 1008 m^2/s^2

v = 31.7 m/s    max speed of truck     (about 70 MPH)

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