60017 is or answer ok you got it
This is just a simple problem finding out the outer surface charge, the inner surface charge and the net charge. Net charge by definition means the difference between two charges. In this case, the formula that is applicable here is outer surface charge = total net charge - inner cavity surface charge. Since we are given already with the net charge equal to 12.0 x10-6 C and the inner charge magnituude f 3.7 x10-6 C, the the total charge must be outer charge is +10x10(-6)) - (-3.0x10(-6)) = +1.3x10(-5) C.
Charges are measured in coloumbs and most likely exist on surfaces of entities like particles, walls etc.
Answer:1.27
Explanation:
Given
incident angle 
refracted angle 
Suppose
is the refractive index of material then using Snell's law we can write

where
=refractive index of air



Answer:
<em>Correct choice: b 4H</em>
Explanation:
<u>Conservation of the mechanical energy</u>
The mechanical energy is the sum of the gravitational potential energy GPE (U) and the kinetic energy KE (K):
E = U + K
The GPE is calculated as:
U = mgh
And the kinetic energy is:

Where:
m = mass of the object
g = gravitational acceleration
h = height of the object
v = speed at which the object moves
When the snowball is dropped from a height H, it has zero speed and therefore zero kinetic energy, thus the mechanical energy is:

When the snowball reaches the ground, the height is zero and the GPE is also zero, thus the mechanical energy is:

Since the energy is conserved, U1=U2
![\displaystyle mgH=\frac{1}{2}mv^2 \qquad\qquad [1]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20mgH%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%20%20%20%5Cqquad%5Cqquad%20%5B1%5D)
For the speed to be double, we need to drop the snowball from a height H', and:

Operating:
![\displaystyle mgH'=4\frac{1}{2}m(v)^2 \qquad\qquad [2]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20mgH%27%3D4%5Cfrac%7B1%7D%7B2%7Dm%28v%29%5E2%20%5Cqquad%5Cqquad%20%5B2%5D)
Dividing [2] by [1]

Simplifying:

Thus:
H' = 4H
Correct choice: b 4H
Answer: Option (a) is the correct answer.
Explanation:
It is known that potential energy is the energy occupied by an object or substance due to its position is known as potential energy.
Therefore, more is the space occupied by an object more will be its position at a particular location. Hence, more will be its potential energy. On the other hand, smaller is the space occupied by an object, smaller will be the position holded by it.
Hence, smaller will be its potential energy.
Thus, we can conclude that for the given situation the statement, potential energy of the larger sphere is greater than that of the smaller sphere, is true.