The coefficient of friction between the Tyre and the ground is 0.11
<u>Explanation:</u>
Given:
Radius of the track (r)=125 m.
Speed with which the car travels (v) =42 km/hr
=11.67 m/s
To Find:
Coefficient of friction between the Tyre and the ground.
Formula to be used:
We know that,Frictional force is equal to centripetal force
Frictional force=μmg
therefore 1.08 m=μmg
Cancelling "m" on both sides we get,
μ=1.08/g=1.08/9.8
=0.11
Thus the coefficient of friction between the Tyre and the ground is 0.11
Potential energy = (mass) · (gravity) · (height)
PE = (60 kg) · (9.8 m/s²) · (10 m)
PE = (60 · 9.8 · 10) · (kg · m²/s² )
PE = 5,880 joules
When the two sides of a composition correspond to one another in size, shape and placement of form, the work has symmetrical balance. Symmetrical balance refers to the balance that is achieved when elements are arranged on either sides of a center of a composition in an equally weighted manner.
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