Add force to unbalance the forces
Answer:
At point A, the cart has high potential energy. At point b, the cart is pulled down by gravity. At point c, the cart gains its highest kinetic energy. At point d, the cart returns back to the same state but with lower potential energy.
Answer:
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Explanation:
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Answer:
change in momentum, 
Average Force, 
Explanation:
Given:
angle of kicking from the horizon, 
velocity of the ball after being kicked, 
mass of the ball, 
time of application of force, 
We know, since body is starting from the rest
.....................(1)


Now the components:


similarly


also, impulse
.........................(2)
where F is the force applied for t time.
Then from eq. (1) & (2)



Now, the components


&


Since my givens are x = .550m [Vsub0] = unknown
[Asubx] = =9.80
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0])
Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in
0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)
0 = [Vsub0x]^2 -10.78
10.78 = [Vsub0x]^2
Sqrt(10.78) = 3.28 m/s