Answer:
No. of revolutions before the car slides off is 2
Solution:
As per the question:
Mass of the car, m = 1500 kg
Diameter of the track, d = 50 m
Force, F = 525 N
Coefficient of static friction, 
Now,
The acceleration in the tangential direction is given by:
Here, the centripetal force is given by the friction force:
Thus
Time taken by the car is given by:
v = v' + at
v' = initial velocity = 0
Thus
Total Distance covered, d is given by:
d = 341.99 m
Distance covered in 1 revolution is the circumference of the circle, d' = 
Now, the no. of revolutions is given by:
The answer is A. Velocity. This is because velocity is obtained using position and time
Answer:
a) v = 2,152 10⁸ m / s b) t = 2.71 10⁸ s or t = 85.93 year
Explanation:
a) In this special relativity exercise we have that time is measured in the same ship, so it is the proper time,
v = d / t
Let's reduce the distance to the SI system
d = 4.3 l and (9.46 1015 m / 1ly) = 40.678 10¹⁵ m
t = 5.0 y (365 day / 1 y) (24 h / 1 day) (3600s / 1h) = 1.89 10⁸ s
Let's calculate
v = 40.678 10¹⁵ / 1.89 10⁸
v = 2,152 10⁸ m / s
b) The time seen from the ground for which the ship moves is given by
t = t₀ / √ (1- (v/c)²)
Let's calculate
t = 1.89 10⁸ / √ (1 - (2.152 / 2.998)²)
t = 1.89 10⁸ / 0.6962
t = 2.71 10⁸ s
Let's reduce this time to years
t = 2.71 10⁸ s (1h / 3600s) (1day / 24h) (1 and / 365 d)
t = 85.93 year
We know, F = m*a
Here, F = 600-400 = 200 N downwards
m = 60 Kg
Substitute their values,
a = 200/60
a = 20/6
a = 10/3
a = 3.33 m/s² downwards
In short, Your Answer would be 3.33 m/s² downwards
Hope this helps!
