A scooter weights 150kg f. Breaks are applied so that wheels stop rolling and start skidding. Find the force of friction if the coefficient of friction is 0.4 (A) 60kg f (B) 48kg f (C) 25kg f (D) 32kg f
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Answer:
A scooter weight 120kg f. Brakes are applied so that wheels stops rolling and start skidding find the force of friction is 0.4
Explanation:
fs=umg
fs=0.4into120=48
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Scooter weight 150 kg brakes as applied so the digits will stop and start kidding.
Force of friction-:
48kg f
OK dear helps u..
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