Since they are both positively charged, then the force would in fact be repulsion, because they are both the same. If you were to have one positively charged, and one negatively charged, then they would go towards eachother.
I hope this helps, if you need help on this question or any other questions, just ask. I am here to help.
It would be kinetic and potential energy, potential from the force of the kick, and kinetic from energy exerted.
Who was the proponent of the Neo-classicism?
a) Claude Debussy
b) Joseph Maurice Ravel
c) Igor Stravinsky
d) Arnold Schoenberg
Answer:
The value of heat transfer during the process Q = - 29.49 KJ
Explanation:
Given data
= 50
= 344.7 k pa
= 0.113 
F = 366.4 K
= 477.6 K
Poly tropic index n = 1.2
gas constant for oxygen = 0.26 
From ideal gas equation
= m R 
Put all the values in above equation we get
⇒ 344.7 × 0.113 = m × 0.26 × 366.4
⇒ m = 0.408 kg
Heat transfer in poly tropic process is given by
Q = ![\frac{\gamma - n}{( \gamma - 1)( n - 1)} [ {m R (T_{1} - T_{2} ) ]](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cgamma%20-%20n%7D%7B%28%20%5Cgamma%20-%201%29%28%20n%20-%201%29%7D%20%5B%20%7Bm%20R%20%28T_%7B1%7D%20-%20T_%7B2%7D%20%20%29%20%5D)
Put all the values in above formula we get
⇒ Q = ![\frac{1.4 - 1.2}{( 1.4 - 1)( 1.2 - 1)} [ {m R (T_{1} - T_{2} ) ]](https://tex.z-dn.net/?f=%5Cfrac%7B1.4%20-%201.2%7D%7B%28%201.4%20-%201%29%28%201.2%20-%201%29%7D%20%5B%20%7Bm%20R%20%28T_%7B1%7D%20-%20T_%7B2%7D%20%20%29%20%5D)
⇒ Q = 2.5 × 0.408 × 0.26 × ( 366.4 - 477.6 )
⇒ Q = - 29.49 KJ
This is the value of heat transfer during the process & negative sign shows that heat is lost during the process.