Answer with Explanation:
We are given that
Mass of rock=m
Maximum height=h
a.At maximum height, velocity,v=0
We know that
Height,h=h/4
Again,
Where
b.When height,h=3h/4
Answer:
So, if the sphere has net amount of charge on it, such that the test charge will experience its intensity in the given region of influence. As, the charge particles size plays a very important role in this regards, as the total distribution of the net charge occurs on the surface of the sphere.
So, we have the following formula in order to make the calculations and have the results, which is as follows:
- E=F/q, (while "F" is the net amount of force applied by the charged body when the test charge,q is placed in its field)
- And if we analyze the Gauss's law we have, Ф=q/∈,
- While, we can obtain the required results by putting all the data into the following equation,as follows:
- E=kq/r².
True, most likely because of the postion of the earth- maybe? I dunno
10^9 giga, 10^6 mega, 10^3 kilo, 10^-3 milli, 10^-6 micro, 10^-9 nano, 10^-12 pico
Potentially they might want centi which is 10^-2