Answer:
The kinetic energy is: 50[J]
Explanation:
The ball is having a potential energy of 100 [J], therefore
PE = [J]
The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.
![E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cg%3D%20gravity%5Bm%2Fs%5E%7B2%7D%20%5D%5C%5Cm%20%3D%20mass%20%5Bkg%5D%5C%5Cm%3D%20%5Cfrac%7BE_%7Bp%7D%20%7D%7Bg%2Ah%7D%5C%5C%20m%3D%20%5Cfrac%7B100%7D%7B9.81%2A10%7D%5C%5C%5C%5Cm%3D%201.01%5Bkg%5D%5C%5C%5C%5C)
In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.
When the elevation is 5 [m], we have a potential energy of
![P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\](https://tex.z-dn.net/?f=P_%7Be%7D%20%3Dm%2Ag%2Ah%5C%5CP_%7Be%7D%20%3D1.01%2A9.81%2A5%5C%5C%5C%5CP_%7Be%7D%20%3D%2050%20%5BJ%5D%5C%5C)
This energy is equal to the kinetic energy, therefore
Ke= 50 [J]
<u>Answer</u>
5.the stage before a star becomes a main sequence star
<u>Explanation</u>
A protostar is a small star that is still gathering its masses. When it forms enough masses it make a parent molecular cloud.
This been the case, from the choices given, the correct statement about a protostar is;
5.the stage before a star becomes a main sequence star
Answer:
Explanation:
The change is as follows
P₁ V₁ to 3P₁, V₁ ( constt volume ) --- first process
3P₁,V₁ to 3P₁ , 5V₁ ( constt pressure ) ---- second process
In the first process Temperature must have been increased 3 times . So if initial temperature is T₁ then final temperature will be 3 T₁
P₁V₁ = n R T₁ , n is no of moles of gas enclosed.
nRT₁ = P₁V₁
Heat added at constant volume = n Cv ( 3T₁ - T₁)
= n x 5/3 R X 2T₁ ( for diatomic gas Cv = 5/3 R)
= 10/3 x nRT₁
= 10/3x P₁V₁
In the second process, Temperature must have been increased 5 times . So if initial temperature is 3T₁ then final temperature will be 15 T₁
Heat added at constant pressure in second case
= n Cp ( 15T₁ - 3T₁)
= n x 7/3 R X 12T₁ ( For diatomic gas Cp = 7/3 R)
= 28 x nRT₁
= 28 P₁V₁
