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------------------------------------------------ A 126 kg motorcycle is driving around a turn with a radius of 79.1 m at a constant speed of 19.7 m/s. What is the minimum force of friction required to prevent the cycle from sliding off of the road?

Answer :

Final answer:

The minimum force of friction required to prevent the motorcycle from sliding off the road is 499.36 N.

Explanation:

The minimum force of friction required to prevent the motorcycle from sliding off the road can be calculated using centripetal force. The centripetal force is given by the equation Fc = mv^2 / r, where Fc is the centripetal force, m is the mass of the motorcycle, v is the velocity, and r is the radius of the turn. Plugging in the given values, we have Fc = (126 kg)(19.7 m/s)^2 / 79.1 m = 499.36 N.

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