A cyclist of mass 30 kg exerts a force of 250 N to move his cycle. The acceleration is 4 ms−2. The force of friction between the
road and tires will be
130 N
120 N
150N
115 N
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Answered by
12
Answer:
F = ma so a= 250/30 a = 8.34 now the result is 4mt/secsqure so reduction is 4.34 so friction force will b 4.34 ×30 so F is 130.2 n.
Answered by
0
fricitional force is 130N
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