A diver having mass m climbs up the diving board.
We know that Gravitational potential energy is given as <span>P<span>EG</span>=mgΔh</span>
What changes is his Gravitational potential energy due to change of height <span>Δh</span> with reference to the ground/water level.
While standing on the diving board his velocity is zero. As such kinetic energy is also zero.
Once he jumps off the springboard we see he gets additional energy from the springboard and falls down under action of gravity g. Due to decrease of height above the ground level Gravitational potential energy decreases and gets converted in to his kinetic energy. <span>1/2m<span>v2</span></span>.
While in air he encounters air resistance. Some of his energy is spent in overcoming this resistance. Gets converted in to kinetic and thermal energy of surrounding air and his body.
Once diver reaches the water, we see water splashing and hear noise of splash. Thereafter the diver comes to rest. Now his potential energy becomes zero. And converted kinetic energy has been converted in to kinetic energy, heat energy and sound energy of water.
As such energy transformation equation looks like
<span><span>Gravitational PE+Elastic PE of springboard</span><span>→Kinetic energy of air and water+Sound energy of splash+thermal energy</span></span>
The object is not accelerating
Explanation:
The given data is as follows.
Mass, m = 75 g
Velocity, v = 600 m/s
As no external force is acting on the system in the horizontal line of motion. So, the equation will be as follows.
where, 
= mass of the projectile
= mass of block
v = velocity after the impact
Now, putting the given values into the above formula as follows.
![75(10^{-3}) \times 600 = [(75 \times 10^{-3}) + 50] \times v](https://tex.z-dn.net/?f=75%2810%5E%7B-3%7D%29%20%5Ctimes%20600%20%3D%20%5B%2875%20%5Ctimes%2010%5E%7B-3%7D%29%20%2B%2050%5D%20%5Ctimes%20v)
= 
v = 0.898 m/s
Now, equation for energy is as follows.
E = 
= 
= 13500 J
Now, energy after the impact will be as follows.
E' = ^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B75%20%5Ctimes%2010%5E%7B-3%7D%20%2B%2050%5D%280.9%29%5E%7B2%7D)
= 20.19 J
Therefore, energy lost will be calculated as follows.
= E E'
= (13500 - 20) J
= 13480 J
And, n = 
= 
= 99.85
= 99.9%
Thus, we can conclude that percentage n of the original system energy E is 99.9%.
Answer:
it appears to be farther away than it actually is, and therefore smaller then the object itself.
