The retarding acceleration of $7.35\, ms^{-2}$ due to frictional force stops the car of mass $400\, kg$ travelling on a road. The coefficient of friction between the tyre of the car and the road is
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As we know,coefficient of friction mu $= \frac{F}{N}\\$
$ \Rightarrow \mu = \frac{{ma}}{{mg}} = \frac{a}{g}\left( {a = 7.35m{s^{ - 2}}\,given} \right)\\$
$\therefore \mu = \frac{{7.35}}{{9.8}} = 0.75$
$ \Rightarrow \mu = \frac{{ma}}{{mg}} = \frac{a}{g}\left( {a = 7.35m{s^{ - 2}}\,given} \right)\\$
$\therefore \mu = \frac{{7.35}}{{9.8}} = 0.75$
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