Question

Physics ❤

A car takes a circular turn of radius 20 m it a speed of 54 km/hr . what is the least coefficient of friction between tyres and road that can prevent slide slipping?

Need Explaination !​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

ANSWER:-

Logic:- Here frictional force will provide us necessary centripetal acceleration.

So for Car not to slip or go away from track, $f \geq Centripetal \: Acceleration$

$54 km {h}^{-1} = 54 \times \frac{5}{18} m {s}^{-1} = 15 m {s}^{-1}$

$\leadsto f \geq \frac{m{v}^{2}}{R} \\ \\ \implies \mu \times N \geq \frac{m{v}^{2}}{R} \\ \\ \implies \mu \times \cancel{m} \times g \geq \frac{\cancel{m}{v}^{2}}{R} \\ \\ \implies \mu \times 10 \geq \frac{15 \times 15}{20} \\ \\ \implies \mu_{min} = \frac{15 \times 15}{20 \times 10} \\ \\ \implies \mu_{min} = \frac{225}{200} \\ \\ \huge\boxed{\bold{\implies \mu_{min} = 1.125}}$