Therefore, if the block moves from its position of maximum spring stretch to maximum spring compression in 0.25 s, the time required for a full cycle is twice as much; T = 0.5 s.
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.
Steel is more elastic than rubber<span> because </span>steel<span> comes back to its original shape faster </span>than rubber<span> when the deforming forces are removed, hope it helps :)</span>
Explanation:
Calculate position vectors in a multidimensional displacement problem. Solve for the displacement in two or three dimensions. Calculate the velocity vector
Answer:
why would you waste points
Explanation:
