Complete Question
Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approximately 57.0 m . If the track is completely flat and the race car is traveling at a constant 30.5 m/s (about 68 mph ) around the turn.
Required:
a. What is the race car's centripetal (radial) acceleration?
b. What is the force responsible for the centripetal acceleration in this case?
O normal
O gravity
O friction
O weight
Answer:
question a

question b
correct option is option 3
Explanation:
From the question we are told that
The radius is 
The constant speed at which the race car is travelling is 
Generally from the question we are told that the track is completely flat so the only force pulling the car to the middle is the frictional force hence the centripetal force is due to the frictional force
Generally the centripetal acceleration is mathematically represented as

=> 
=> 
Answer:11 km/s
Explanation:
Given
Escape velocity at the surface of earth is 11 km/s
Escape velocity is given by

Escape velocity at the surface of earth
--------------------1
If Escape velocity is three times and the radius is also the three times


i.e. 
Answer:
T = 4 sec / 2 = 2 sec period of revolution
S = 2 pi R = 2 * pi * 1.75 m = 11 m
V = S / T = 11 m / 2 sec = 5.5 m/s speed of object
Answer:
option A
Explanation:
given,
Force = F
angle = θ
weight on suitcase = mg
distance = d
constant velocity so, acceleration a = 0
coefficient of friction = µ
Work done = ?
Work done is equal to force into displacement.
Friction act opposite to the force acting so, work done by frictional force will be negative.
frictional force will act into horizontal direction opposite to force.
here displacement is equal to d
now,
W = -F d cos θ
Hence,the correct answer is option A