The flow of Direct current (DC) is constant and flows in one direction. Most digital electronics make use of DC. Alternating current (AC) periodically flows in reverse and is mostly used to deliver power to houses, buildings and the like. With that alone, you can already rule out A, C and D.
The answer would then be B. constant, periodically reversing.
If the distance between two charges is halved, the electrical force between them increases by a factor 4.
In fact, the magnitude of the electric force between two charges is given by:

where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges
We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:

the magnitude of the force changes as follows:

so, the force increases by a factor 4.
Answer:

Explanation:
As we know that resistance of one copper wire is given as

here we know that

now we have


now we know that such 17 resistors are connected in parallel so we have


Now if a single copper wire has same resistance then its diameter is D and it is given as

now from above two equations we have


now we have

Answer: 
Explanation:
Given
Wavelength of light 
Screen is
away
Distance between two adjacent bright fringe is 
When same experiment done in water, wavelength reduce to 
So, the distance between the two adjacent bright fringe is 
Keeping other factor same, distance becomes

Answer:
Explanation:
This does not violate Newton's 1st law because the net force would still be 0 in order to produce uniform motion (aka constant velocity). The other forces acting on the vehicles is air resistance which is non-zero. So we need car internal force to counter balance this force, which require extra gas for the car.