Answer: When the electric field due to one is a maximum, the electric field due to the other is also a maximum, and this relation is maintained as time passes. They alternatively reinforce and cancel each other.
Explanation:
In a wave, the phase, is an arbitrary time reference, used to locate a given point of the wave in time, within a cycle.
Two waves can travel at the same speed, or even have the same wavelength, but this is not enough to be sure that at a given point in time, both waves will be in their maximum, as it only can be determined from the phase of the waves.
So, only when the waves reach at the same point in time at the same amplitude, we can say that they arrive in phase, in a constructive interference.
Answer:B
Explanation:
Given
mass of Sled is m
another package of mass m is thrown on it
Suppose u be the initial velocity of sled
conserving momentum


where v is the final velocity
Initial kinetic energy 
Final Kinetic Energy
Final Kinetic Energy
Final kinetic Energy is half of initial
Yeah i believe it is going to be c
Answer:
Part a)

Part b)

Part c)

Part d)

Explanation:
Part a)
As we know that speed of package is same as that of helicopter in horizontal direction
So after time "t" the velocity in x direction will remain constant while in Y direction it will go free fall
So we have



Part b)
Distance from helicopter is same as the distance of free fall
so we will have

Part c)
If helicopter is rising upwards with uniform speed
then final speed of the package after time t is given as


Part d)
distance from helicopter

ANSWER
My answer is in the photo above