A cyclist on a level road takes a sharp circu | vedclass

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(B) The maximum speed $v_{m}$ at which the cyclist will not skid is given by the formula $v_{m} = \sqrt{\mu r g}$.Substituting the given values: $\mu

Vedclass · Accepted Answer

$7.2 \; km \cdot h^{-1}$

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(B) The maximum speed $v_{m}$ at which the cyclist will not skid is given by the formula $v_{m} = \sqrt{\mu r g}$.Substituting the given values: $\mu = 0.2$,$r = 3 \; m$,and $g = 10 \; m \cdot s^{-2}$.$v_{m} = \sqrt{0.2 	imes 3 	imes 10} = \sqrt{6} \approx 2.45 \; m \cdot s^{-1}$.To convert this speed into $km \cdot h^{-1}$,we multiply by $\frac{18}{5}$:$v_{m} = 2.45 	imes 3.6 = 8.82 \; km \cdot h^{-1}$.The cyclist will not skid if their speed is less than or equal to $8.82 \; km \cdot h^{-1}$.Among the given options,$7.2 \; km \cdot h^{-1}$ is the only speed less than $8.82 \; km \cdot h^{-1}$.

$A$ cyclist on a level road takes a sharp circular turn of radius $3 \; m$ $(g = 10 \; m \cdot s^{-2})$. If the coefficient of static friction between the cycle tyres and the road is $0.2$,at which of the following speeds will the cyclist not skid while taking the turn?

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