Let's assume that ground level is the height 0 meters. The change in potential energy is going to be gravitational potential energy, which is given by PE=mgh.
ΔPE=mgh-mgy
=mg(h-y)
=50(28-0)
=1400 J
No. That's the description of the wave's 'frequency'.
Refraction of a wave is its behavior when it crosses
the boundary between two different media.
Answer:
31 m/s
Explanation:
As both the monkey and the darts are subjected to constant gravitational acceleration g = 9.8 m/s2 and both start from rest (vertically speaking). Their vertical position will always be the same. For the dart to hit the monkey, its horizontal position must be the same as the monkey's, which is unchanged before reaching the ground. Therefore, the time it takes for the dart to travel across 70 m must be less than the time it takes for the monkey to drop 25m to the ground. We can find it out using the following equation of motion
![s_m = gt_m^2/2](https://tex.z-dn.net/?f=s_m%20%3D%20gt_m%5E2%2F2)
![25 = 9.8t_m^2/2](https://tex.z-dn.net/?f=25%20%3D%209.8t_m%5E2%2F2)
![t_m^2 = 50/9.8 = 5.1](https://tex.z-dn.net/?f=t_m%5E2%20%3D%2050%2F9.8%20%3D%205.1)
![t_m = \sqrt{5.1} = 2.26 s](https://tex.z-dn.net/?f=t_m%20%3D%20%5Csqrt%7B5.1%7D%20%3D%202.26%20s)
For the dart to takes less that 2.26 s to travel 70m, its horizontal speed must at least be 70 / 2.26 = 31 m/s