Answer:
Explanation:
Given
average speed of train
Maximum acceleration=0.05g
Now centripetal acceleration is


r=7346.93 m
(b)Radius of curvature=900 m
therefore 



Special relativity led the path for general relativity; special relativity is in a sense a special application of the rules of general relativity. While general relativity is in position to tackle all of these problems, special relativity can tackle only problems in inertial frames. Inertial frame means that the frame of reference is inot accelerating. So, we disqualify answers A and D. However, remember that moving in a circle means that there is an acceleration, the centrifugal one, even if the speed does not change. Hence C is also incorrect.
The correct answer is B, since if there is no change in velocity, the frame does not accelerate and it is inertial.
Answer:
The sum of the initial and final velocity is divided by 2 to find the average. The average velocity calculator uses the formula that shows the average velocity (v) equals the sum of the final velocity (v) and the initial velocity (u), divided by 2.
The height of the ball above the ground is 38.45 m
First we will calculate the velocity of the ball when it touch the ground by using first equation of motion
v=u+gt
v=0+9.81×2.8
v=27.468 m/s
now the height of the ground can be calculated by the formula
v=√2gh
27.468=√2×9.81×h
h=38.45 m