Discrete packets of kinetic energy in light are called
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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
Answer:
τ=0.060 N.m
Explanation:
By kinematics:
Solving for α:
where ωo = 600*2*π/60; ωf = 0; t=10s
The sum of torque is:
The velocity of the ball is 12.5 m/s
Explanation:
The velocity of the ball is given by the ratio between the distance covered by the ball and the time taken:
First, we calculate the distance covered. We know that the radius of the circle is
r = 0.450 m
And the length of the circumference is
The ball makes 25.0 revolutions, so a total distance of
In a time of
t = 9.37 s
So, its velocity is
Learn more about velocity here:
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