Answer:
When energy contained in coal is turned into heat, and then into electrical energy. As boiling water heated by the burning coal is cooled, steam forges from these cone-shaped cooling towers.
<span>The weightlifter does no work. Although he has exerted force, work is the product of force over distance. Since he has not moved the wall he has done no work.</span>
Explanation:
When bullet is shot towards the monkey then let say the distance of monkey from the bullet is "d"
so we can find the time to reach the bullet to the monkey
![t = \frac{d}{vcos\theta}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bd%7D%7Bvcos%5Ctheta%7D)
Now similarly we can find the vertical displacement of the bullet in the same time
![\Delta y = vsin\theta t - \frac{1}{2}gt^2](https://tex.z-dn.net/?f=%5CDelta%20y%20%3D%20vsin%5Ctheta%20t%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
![\Delta y = v sin\theta (\frac{d}{vcos\theta}) - \frac{1}{2}gt^2](https://tex.z-dn.net/?f=%5CDelta%20y%20%3D%20v%20sin%5Ctheta%20%28%5Cfrac%7Bd%7D%7Bvcos%5Ctheta%7D%29%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
so it is given as
![\Delta y = d tan\theta - \frac{1}{2}gt^2](https://tex.z-dn.net/?f=%5CDelta%20y%20%3D%20d%20tan%5Ctheta%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
here if the monkey is initially at height H above the ground at given angle then we can say
![H = dtan\theta](https://tex.z-dn.net/?f=H%20%3D%20dtan%5Ctheta)
so we can say that
![\Delta y = H - \frac{1}{2}gt^2](https://tex.z-dn.net/?f=%5CDelta%20y%20%3D%20H%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
So if at the same time monkey will fall down then the height of monkey from ground after time "t" is given as
![\Delta y = H - \frac{1}{2}gt^2](https://tex.z-dn.net/?f=%5CDelta%20y%20%3D%20H%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
so here bullet will hit the monkey as both monkey and bullet are at same position.
Since f=m(v^2/r),or fnet is equal to ma.
force = unknown
velocity=22m/s
radius=75m
f=m(v^2/r)
f=925(22^2/75)
f=5969.333N