Answer:
Explanation:
The fish is initially at rest and it is also at rest when the spring is fully stretched at the maximum distance.
Change in gravity potential energy = change in spring potential energy
mgh = 1/2kh^2
Assume gravity constant g is 10m/s^2
2.6*10*h = 1/2*200*h^2
100h^2 - 26h = 0
2h(50h - 13) = 0
h = 0 or h = 13/50 = 0.65m
h = 0 is before the spring is stretched
So the maximum distance is 0.65m.
Answer: 3 m.
Explanation:
Neglecting the mass of the seesaw, in order the seesaw to be balanced, the sum of the torques created by gravity acting on both children must be 0.
As we are asked to locate Jack at some distance from the fulcrum, we can take torques regarding the fulcrum, which is located at just in the middle of the length of the seesaw.
If we choose the counterclockwise direction as positive, we can write the torque equation as follows (assuming that Jill sits at the left end of the seesaw):
mJill* 5m -mJack* d = 0
60 kg*5 m -100 kg* d =0
Solving for d:
d = 3 m.
I believe the answer to your question is A. 340 meters/second hope i helped
844J.
Assuming that there were no encumbrances during it's foreswing and it reached it's full potential at apogee.
