I., II., and IV. are examples of acceleration. III. isn't.
Answer:
C
Explanation:
the formula is a + b = ab
Answer:
The distance from the charge is 3.35 m.
Explanation:
Given that,
Electric potential, V = 635 V
Magnitude of electric field, E = 189 N/C
We need to find the distance from the charge. We know that the relation between electric field and electric potential is given by :
![E=\dfrac{V}{d}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7BV%7D%7Bd%7D)
d is the distance from charge
![d=\dfrac{V}{E}\\\\d=\dfrac{635}{189}\\\\d=3.35\ m](https://tex.z-dn.net/?f=d%3D%5Cdfrac%7BV%7D%7BE%7D%5C%5C%5C%5Cd%3D%5Cdfrac%7B635%7D%7B189%7D%5C%5C%5C%5Cd%3D3.35%5C%20m)
So, the distance from the charge is 3.35 m. Hence, this is the required solution.
Acted upon by an unbalanced force