Answer:
Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

=![\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ]](https://tex.z-dn.net/?f=%5Cfrac%7Bq%7D%7B4%5Cpi%20e_%7B0%7D%20%7D%20%5B%5Cfrac%7B1%7D%7BR%7D%20-%5Cfrac%7Br%5E%7B2%7D-R%5E%7B2%7D%20%20%7D%7B2R%5E%7B3%7D%20%7D%20%5D)
∴NOTE: Graph is attached
Attenuation is the correct answer.
Answer:
Explanation:
1 )
We shall apply conservation of momentum law to solve the problem.
mv = ( M +m) V , m and M are masses of small and large object , v is the velocity of small object before collision and V is the velocity of both the objects together after collision .
.5 x .2 = (1.5 + .5)V
V = .05 m /s
2 ) We shall use formula for velocity of object after elastic collision as follows
v₁ = 
m₁ and m₂ are masses of first and second object u₁ and u₂ are their initial velocity and v₁ and v₂ are their final velocity.
Putting the values
= 
= - .66 m /s
Since the sign is negative so it will be in opposite direction .
Because,
In left image pin is not touch to the wire.
In right image pin is touch to the wire.
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