Answer: A.). Be equal to the time the ball hits the group
Explanation:
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
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Answer:
42.05 m/s
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 0 m/s
Height (h) = 239 m
Acceleration due to gravity (g) = 3.7 m/s²
Final velocity (v) =?
The velocity with which the camera hits the ground can be obtained as follow:
v² = u² + 2gh
v² = 0² + 2 × 3.7 × 239
v² = 0 + 1768.6
v² = 1768.6
Take the square root of both side
v = √(1768.6)
v = 42.05 m/s
Therefore, the velocity with which the camera hits the ground is 42.05 m/s