SEPARATION OF VARIABLES - SPHERICAL The potential on the surface of a sphere (radius R) is given by V=V0 cos(2). (Assume V(r=[i
nfinity])=0. Also assume there is no charge in or outside, it's ALL on the surface!) a) Find the potential everywhere inside and outside this sphere. (Hint: Can you express cos(2) as a simple linear combination of Legendre polynomials? ) b) Find the charge density () on the surface of the sphere in part a. Sketch or draw it!
1 answer:
You might be interested in
Answer:
Explanation:
<em>The top 5 winners of NASA's design contest winner received amount of </em><em> </em>(Dollar One hundred thousand)<em>. </em>The respected amount was divided into 5 parts as there were 5 winners as well.
So, the amount was divided among the winners are as follows:
<em>Winner 1 ⇒ </em>
<em>Winner 2 ⇒ </em>
<em>Winner 3 ⇒ </em>
<em>Winner 4 ⇒ </em>
<em>Winner 5 ⇒</em>
Hence by adding all of the 5, we will get the answer, which is .
(a)
The velocity of the meteorite just before hitting the ground is:
The loss of energy of the meteorite corresponds to the kinetic energy the meteorite had just before hitting the ground, so:
(b) 1 megaton of tnt is equal to
To find to how many megatons the meteorite energy loss
corresponds, we can set the following proportion
And so we find
So, 0.162 megatons.
(c) 1 Hiroshima bomb is equivalent to 13 kilotons (13 kT). The impact of the meteorite had an energy of
. So, to find to how many hiroshima bombs it corresponds, we can set the following proportion:
And so we find
So, the energy released by the impact of the meteorite corresponds to the energy of 12.46 hiroshima bombs.
The velocity of the meteorite just before hitting the ground is:
The loss of energy of the meteorite corresponds to the kinetic energy the meteorite had just before hitting the ground, so:
(b) 1 megaton of tnt is equal to
To find to how many megatons the meteorite energy loss
corresponds, we can set the following proportion
And so we find
So, 0.162 megatons.
(c) 1 Hiroshima bomb is equivalent to 13 kilotons (13 kT). The impact of the meteorite had an energy of
And so we find
So, the energy released by the impact of the meteorite corresponds to the energy of 12.46 hiroshima bombs.