The average power output is the ratio between the work done to compress the spring, W, and the time taken, t:

(1)
The work done is equal to the elastic energy stored by the compressed spring:

where

is the spring constant and

is the compression of the spring. If we substitute the numbers, we find:

And now we can use eq.(1) to calculate the average power output:
Answer:
A. The bomb will take <em>17.5 seconds </em>to hit the ground
B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it
Explanation:
Maverick and Goose are flying at an initial height of
, and their speed is v=688 m/s
When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement
The equation for the height y with respect to ground in a horizontal movement (no friction) is
[1]
With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released
The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time
The range (horizontal displacement) of the bomb x is
[2]
Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:
Setting y=0 and isolating t we get

Since we have 


Replacing in [2]


A. The bomb will take 17.5 seconds to hit the ground
B. The bomb will land 12040 meters on the ground ahead from where they released it
answer is :D it would be a great answer
Explanation:
Let us assume that the maximum allowable horizontal distance be represented by "d".
Therefore, torque equation about A will be as follows.

d = ![\frac{[2 \times 75 \times (0.7+0.15+0.15) - 60 \times 0.15 - 252 \times 0.15 \times 2]}{252}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2%20%5Ctimes%2075%20%5Ctimes%20%280.7%2B0.15%2B0.15%29%20-%2060%20%5Ctimes%200.15%20-%20252%20%5Ctimes%200.15%20%5Ctimes%202%5D%7D%7B252%7D)
d = 0.409 m
Thus, we can conclude that the maximum allowable horizontal distance from the axle A of the wheelbarrow to the center of gravity of the second bag if she can hold only 75 N with each arm is 0.409 m.