The Coulomb force between two charges q1 and q2 at separation r in air is F. If half of the separation is filled with medium of
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Answer:
a) The centripetal acceleration of the car is 0.68 m/s²
b) The force that maintains circular motion is 940.03 N.
c) The minimum coefficient of static friction between the tires and the road is 0.069.
Explanation:
a) The centripetal acceleration of the car can be found using the following equation:
Where:
v: is the velocity of the car = 51.1 km/h
r: is the radius = 2.95x10² m
Hence, the centripetal acceleration of the car is 0.68 m/s².
b) The force that maintains circular motion is the centripetal force:
Where:
m: is the mass of the car
The mass is given by:
Where P is the weight of the car = 13561 N
Now, the centripetal force is:
Then, the force that maintains circular motion is 940.03 N.
c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:
Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.
I hope it helps you!
