Traveling against currents usually takes longer. Kinda like walking against the wind, you feel the heaviness against your jacket as you push through it. Where when you walking with the wind, it kind of gives your a push. Same for with currents.
The electrostatic potential energy, U, of one point charge q at position d in the presence of an electric field E is defined as the negative of the work W done by the electrostatic force to bring it from the reference position d to that position

Thus, to double the electric potential energy U we need to reduce the distance of separation by half (1/2) because they are inversely proportion
In some early mornings , dew drops can be found on grass or a car parked outside, but not on other materials such as the sidewalk because the night -time temperature on grass and the car went below the dew point, but the temperature of the concrete did not drop enough to reach the dew point level
Dew can be formed on any object when the temperature of the object drop. When this happen, the object will be cool which will eventually cool the surrounding air around the object.
Dew drops is as a result of condensation in the air. When the cool air causes the air vapor to convert to liquid. The dew will form when the temperature of the object balances with the dew point in the surrounding environment.
In some early mornings , dew drops can be found on grass or a car parked outside, but not on other materials such as the sidewalk because the night -time temperature on grass and the car went below the dew point, but the temperature of the concrete did not drop enough to reach the dew point level
Therefore the correct option is therefore A
Learn more here : brainly.com/question/13834972
Explanation:
It is given that,
Frequency of monochromatic light, 
Separation between slits, 
(a) The condition for maxima is given by :

For third maxima,



(b) For second dark fringe, n = 2





Hence, this is the required solution.