Answer: gravitational potential energy is converted into kinetic energy
Explanation:
When the diver stands on the platform, at 20 m above the surface of the water, he has some gravitational potential energy, which is given by
![E=mgh](https://tex.z-dn.net/?f=E%3Dmgh)
where m is the man's mass, g is the gravitational acceleration and h is the height above the water. As he jumps, the gravitational potential energy starts decreasing, because its height h above the water decreases, and he acquires kinetic energy, which is given by
![K=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
where v is the speed of the diver, which is increasing. When he touches the water, all the initial gravitational potential energy has been converted into kinetic energy.
Answer:
specific purpose statement
Explanation:
It is a specific purpose statement made for a persuasive speech on the question of fact.
A specific purpose statement helps to build on the general purpose (that is to inform) and to make it more specific to the audience. So if the first speech is an informative speech, our general purpose is to inform our audience about a very specific realm of knowledge.
A specific purpose statement is given to audience to persuade on specific information.
Answer:
Work done by external force is given as
![Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2](https://tex.z-dn.net/?f=Work_%7Bexternal%7D%20%3D%20mgLsin%5Ctheta%20%2B%20%5Cmu%20mgLcos%28%5Ctheta%29%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv_2%5E2%20-%20%5Cfrac%7B1%7D%7B2%7Dmv_1%5E2)
Explanation:
As per work energy Theorem we can say that work done by all force on the car is equal to change in kinetic energy of the car
so we will have
![Work_{external} + Work_{gravity} + Work_{friction} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2](https://tex.z-dn.net/?f=Work_%7Bexternal%7D%20%2B%20Work_%7Bgravity%7D%20%2B%20Work_%7Bfriction%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv_2%5E2%20-%20%5Cfrac%7B1%7D%7B2%7Dmv_1%5E2)
now we have
![W_{gravity} = -mg(Lsin\theta)](https://tex.z-dn.net/?f=W_%7Bgravity%7D%20%3D%20-mg%28Lsin%5Ctheta%29)
![W_{friction} = -\mu mgcos(\theta) L](https://tex.z-dn.net/?f=W_%7Bfriction%7D%20%3D%20-%5Cmu%20mgcos%28%5Ctheta%29%20L)
so from above equation
![Work_{external} - mgLsin\theta - \mu mgLcos(\theta) = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2](https://tex.z-dn.net/?f=Work_%7Bexternal%7D%20-%20mgLsin%5Ctheta%20-%20%5Cmu%20mgLcos%28%5Ctheta%29%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv_2%5E2%20-%20%5Cfrac%7B1%7D%7B2%7Dmv_1%5E2)
so from above equation work done by external force is given as
![Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2](https://tex.z-dn.net/?f=Work_%7Bexternal%7D%20%3D%20mgLsin%5Ctheta%20%2B%20%5Cmu%20mgLcos%28%5Ctheta%29%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv_2%5E2%20-%20%5Cfrac%7B1%7D%7B2%7Dmv_1%5E2)
Boiii you better be more pasfic when you ask a question I don’t have any information about it I wanna was wyour one of the day