F = 130 revs/min = 130/60 revs/s = 13/6 revs/s
t = 31s
wi = 2πf = 2π × 13/6 = 13π/3 rads/s
wf = 0 rads/s = wi + at
a = -wi/t = -13π/3 × 1/31 = -13π/93 rads/s²
wf² - wi² = 2a∅
-169π²/9 rads²/s² = 2 × -13π/93 rads/s² × ∅
∅ = 1209π/18 rads
n = ∅/2π = (1209π/18)/(2π) = 1209/36 ≈ 33.5833 revolutions.
Answer:
that one i know only pe not that sorry again
Answer:
1)
2)
3)
4)
Explanation:
1)
We can use the following equation:

Here, the initial velocity in the y-direction is zero, the final y position is zero and the initial y position is 25 m.


2)
The equation of the motion in the x-direction is:



3)
The velocity in the y-direction of the stone will be:



Now, the velocity in the x-direction is 15 m/s then the velocity will be:

4)
The angle of this velocity is:
Then α=55.92° negative from the x-direction.
I hope it helps you!