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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Power = Force * Distance/ time
P = 1,250 * 2/3
P = 2,500/3
P = 833.33 Watts
So, your final answer is 833.33 Watts
Answer:
At the closest point
Explanation:
We can simply answer this question by applying Kepler's 2nd law of planetary motion.
It states that:
"A line connecting the center of the Sun to any other object orbiting around it (e.g. a comet) sweeps out equal areas in equal time intervals"
In this problem, we have a comet orbiting around the Sun:
- Its closest distance from the Sun is 0.6 AU
- Its farthest distance from the Sun is 35 AU
In order for Kepler's 2nd law to be valid, the line connecting the center of the Sun to the comet must move slower when the comet is farther away (because the area swept out is proportional to the product of the distance and of the velocity: 
, therefore if r is larger, then v (velocity) must be lower).
On the other hand, when the the comet is closer to the Sun the line must move faster (, if r is smaller, v must be higher). Therefore, the comet's orbital velocity will be the largest at the closest distance to the Sun, 0.6 A.
False, its up to you to decide what you belive is right and what is wrong.
