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:
420000N
Explanation:
Given parameters:
Mass of the train = 5.6 x 10⁵kg
Acceleration = 0.75m/s²
Unknown:
Resultant force = ?
Solution:
According to newton's second law, force is the product of mass and acceleration;
Force = mass x acceleration
Resultant force that acts on the train is given below;
Force = 5.6 x 10⁵kg x 0.75m/s² = 420000N
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36 hours after the event.
Coefficient of static friction = tan(a) = 0.4
r = 740 m
g = 9.8 m/s²

v = √(9.8 × 740 × 0.4) m/s
v ≈ 53.85908 m/s