Answer:
h = 3.1 cm
Explanation:
Given that,
The volume of a oil drop, V = 10 m
Radius, r = 10 m
We need to find the thickness of the film. The film is in the form of a cylinder whose volume is as follows :

So, the thickness of the film is equal to 3.1 cm.
The central angle of a circle is 360° or 2π radians.
Therefore
1 radian = (360 degrees)/(2π radians) = 180/π degrees/radian.
4 radians = (4 radians)*(180/π degrees/radian) = 229.18 degrees.
Answer: C. 229.2°
Answer:
The bell has a potential energy of 8550 [J]
Explanation:
Since the belt is 45 [m] above ground level, only potential energy is available. And this energy can be calculated by means of the following equation.
![E_{p}= W*h\\E_{p} = 190*45\\E_{p}=8550[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%3D%20W%2Ah%5C%5CE_%7Bp%7D%20%3D%20190%2A45%5C%5CE_%7Bp%7D%3D8550%5BJ%5D)
Answer:

Explanation:
As given point p is equidistant from both the charges
It must be in the middle of both the charges
Assuming all 3 points lie on the same line
Electric Field due a charge q at a point ,distance r away

Where
- q is the charge
- r is the distance
-
is the permittivity of medium
Let electric field due to charge q be F1 and -q be F2
I is the distance of P from q and also from charge -q
⇒
F1
F2
⇒
F1+F2=