The final speed of the orange is 7.35 m/s
Explanation:
The motion of the orange is a free fall motion, since there is only the force of gravity acting on it. Therefore, it is a uniformly accelerated motion with constant acceleration 
towards the ground. So we can use the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time elapsed
For the orange in this problem, we have
u = 0 (it is dropped from rest)
is the acceleration
Substituting t = 0.75 s, we find the final velocity (and speed) of the orange:
Learn more about free fall:
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The answer is B - Current Y has a greater potential difference, and the charges flow at a slower rate.
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To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.
If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,
So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
