Answer
Ceres, Pluto, and Eris are classified as DWARF PLANET.
A) Leftover planetesimals inside the frost line are known as ASTEROIDS.
B) METEORITES are the pieces of Asteroids which are fallen on the earth's surface.
C) COMETS are the objects which are visible with long tails.
D) COMETS are also the leftover planetesimals that are occupied by the jovian planets and are formed in the solar system.
E) Meteor showers are associated with debris from COMETS
9 × 10²¹ electrons flow through a cross section of the wire in one hour.
<h3>What is the relation between current and charge?</h3>
- Mathematically, current = charge / time
- In S.I. unit, Charge is written in Coulomb and time in second.
<h3>What is the amount of charge flown through a wire for one hour if it carries 0.4 A current?</h3>
- Charge= current × time
- Current= 0.4 A, time = 1 hour= 3600 s
- Charge= 0.4× 3600
= 1440 C
<h3>How many numbers of electrons present in 1440C of charge?</h3>
- One electron= 1.6 × 10^(-19) C
- So, 1440 C = 1440/1.6 × 10^(-19)
= 9 × 10²¹ electrons
Thus, we can conclude that the 9 × 10²¹ electrons flow through a cross section of the wire in one hour.
Learn more about current here:
brainly.com/question/25922783
#SPJ1
Answer:
(a) 135 kV
(b) The charge chould be moved to infinity
Explanation:
(a)
The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by
where 
Difference in potential between the points is
![kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}](https://tex.z-dn.net/?f=kQ%5Cleft%5B-%5Cdfrac%7B1%7D%7B0.2%5Ctext%7B%20m%7D%7D%20-%5Cleft%28%20-%5Cdfrac%7B1%7D%7B0.1%5Ctext%7B%20m%7D%7D%5Cright%29%5Cright%5D%20%3D%20%5Cdfrac%7BkQ%7D%7B0.2%5Ctext%7B%20m%7D%7D%20%3D%20%5Cdfrac%7B9%5Ctimes10%5E9%5Ctext%7B%20F%2Fm%7D%5Ctimes3%5Ctimes10%5E%7B-6%7D%5Ctext%7B%20C%7D%7D%7B0.2%5Ctext%7B%20m%7D%7D)
(b)
If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.
![270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]](https://tex.z-dn.net/?f=270%5Ctimes10%5E3%20%3D%20kQ%5Cleft%5B-%5Cdfrac%7B1%7D%7Bx%7D-%5Cleft%28-%5Cdfrac%7B1%7D%7B0.1%5Ctext%7B%20m%7D%7D%5Cright%29%5Cright%5D)
The charge chould be moved to infinity
