Question

That minimum coefficient of static friction between the tyres and the road that will allow the car that will round safely.

Answer 1

Answer:

The answer is "$0.13562748$"

Explanation:

Please find the complete question in the attached file.

Using formula:

$\bold{U_s=\frac{v^2}{rg}}$

Given:

$r=1.60 \times 10^2 \ m\\\\g=9.8\\\\v= 52.5\ \frac{km}{hr}= 14.583\ \frac{m}{s}$

put the value into the above-given formula:

$U_s= \frac{14.583^2}{160 \times 9.8}$

    $= \frac{212.663889}{1568}\\\\=0.13562748$