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
Answer:
The final velocity of the object is, 
= 27 m/s
Explanation:
Given,
The acceleration of the object, a = 1000 m/s²
The initial displacement of the object, 
= 0 m
The final displacement of the object, 
= 0.75 m
The initial velocity of the object will be, 
= o m/s
The final velocity of the object, = ?
The average velocity of the object,
v = ( - )/ t
= 0.75 / t
The acceleration is given by the relation
a = v / t
1000 m/s² = 0.75 / t²
t² = 7.5 x 10⁻⁴
t = 0.027 s
Using the I equation of motion,
= u + at
Substituting the values
= 0 + 1000 x 0.027
= 27 m/s
Hence, the final velocity of the object is, = 27 m/s
Answer:
two 5kg objects 2 meters apart✨
Answer:
The work done on the box is 80 J.
Explanation:
Given that,
Weight of box = 40 N
Distance = 2 meter
We need to calculate the work done
Using formula of work done
Where, x = distance
mg = weight
Put the value into the formula
Hence, The work done on the box is 80 J.
Answer:
this is a difficult question but I will try to answer it answer for this is 3220 a + b b u s y d l new
