An object undergoing <span>uniform circular motion </span>is moving with a constant speed. Nonetheless, it is accelerating due to its change in direction. So I'm thinking velocity
Under water turbans that are placed at the above to middle of the ocean they are used to capture kinetic motion
Answer:
Fx1 (6 m) sin 60 = 300 (3 m) cos 60 balancing torques about floor
Fx1 = 900 * 1/2 / 5.20 = 86.6 N this is the horizontal force that must be supplied by the wall to balance torques about the floor
This is also equal to the static force of friction that must be applied at the point of contact with the floor to balance forces in the x-direction.
Fx1 = Fx2 = 86.6 N
(Hint: the time<span> to rise to the </span>peak<span>is one-half the </span>total hang-time<span>.).</span>
Answer:
v = 5.15 m/s
Explanation:
At constant velocity, the cable tension will equal the car weight of 984(9.81) = 9,653 N
As the cable tension is less than this value, the car must be accelerating downward.
7730 = 984(9.81 - a)
a = 1.95 m/s²
kinematic equations s = ut + ½at² and v = u + at
-5.00 = u(4.00) + ½(-1.95)4.00²
u = 2.65 m/s the car's initial velocity was upward at 2.65 m/s
v = 2.65 + (-1.95)(4.00)
v = -5.15 m/s
