Answer:
9.81 × 10 = 98.1 meters
vertical displacement is s=1/2 at^2 + vt
initial vertical velocity is 0 so s=1/2 at^2
a in this instance is gravitational acceleration so 60m= 1/2 (9.81)t^2
solve for t, t = 3.497s. //I corrected this answer as just now I misread horizontal as vertical.
It would be B, like charges repel and unlike charges attract.
The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.
Answer:
Explanation:
So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.
So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then
v = 25.8 + (-1.66×8.3)
v =12.022 m/s.
So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.
