Answer:
Closely fits into the connector.
Explanation:
It's one of the steps used for the splicing of aluminium conductors in the underground connections. Where we do the strip insulation to splice the conductors by using compression type connectors.
To solve this problem we will apply the concepts related to electric potential and electric potential energy. By definition we know that the electric potential is determined under the function:

= Coulomb's constant
q = Charge
r = Radius
At the same time

The values of variables are the same, then if we replace in a single equation we have this expression,

If we replace the values, we have finally that the charge is,




Therefore the potential energy of the system is 
Answer:
Maharashtra - mashru or himroo / dhoti and lugda
Gujarat - patola / ghagra choli
Punjab - pat / kurta and pajama
Odisha - ikat / Sadi
West Bengal - tossa / kurta
Karnataka - Mysore silk / mundu
Acute health effects such as skin burns or acute radiation syndrome can occur when doses of radiation exceed certain levels.
Answer:
d) 0 V
Explanation:
It can be showed that the potential due to a point charge q, to a distance d from the charge, can be expressed as follows:

where k = 
As the potential is an scalar, and is linear with the charge, we can apply the superposition principle, which means that we can find the potential due to one of the charges, as if the other were not present.
By symmetry, all four charges are at the same distance from the center, so we can write the total potential, as follows:

where d, is the semi-diagonal of the square, that we can find applying Pythagorean theorem, as follows:

Replacing by the values in (1) we have:

which is equal to the option d).