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jarptica [38.1K]
3 years ago
10

To test the performance of its tires, a car

Physics
1 answer:
Rom4ik [11]3 years ago
5 0

The coefficient of static friction is 0.222

Explanation:

In order for the car to remain in circular motion, the frictional force must be able to provide the necessary centripetal force. Therefore, the car will start skidding when the two forces are equal:

\mu mg=m\frac{v^2}{r}

where the term on the left is the frictional force, while the term on the right is the centripetal force, and where

\mu is the coefficient of static friction

m is the mass of the car

g is the acceleration of gravity

v is the speed of the car

r is the radius of the track

In this problem, we have:

r = 564 m

v = 35 m/s

g=9.8 m/s^2

And re-arranging the equation for \mu, we can find the coefficient of static friction:

\mu = \frac{v^2}{gr}=\frac{35^2}{(9.8)(564)}=0.222

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

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starting point.

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energy is conserved

        Em₀ = Em_f

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        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

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          r2 = \sqrt{1 + 2.25 10^{-9}}

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