Sound waves travel faster
Answer:



Explanation:
From the question we are told that
Electric field of intensity 
Rectangle parameter Width
Length 
Angle to the normal 
Generally the equation for Electric flux at parallel to the yz plane
is mathematically given by



Generally the equation for Electric flux at parallel to xy plane
is mathematically given by



Generally the equation for Electric flux at angle 30 to x plane
is mathematically given by




The resultant displacement between the two vectors will increase.
The resultant of the two vectors is given by parallelogram law of vectors.
The parallelogram law of vector addition states that if two vectors are represented in magnitude and direction by the adjacent sides of a parallelogram, the diagonal of the parallelogram drawn from the point of intersection of the vectors represents the resultant vector in magnitude and direction.
The resultant of these vectors, say vector A, and B, is given as;

When;
θ = 90°

When;
θ = 120°

Thus, the resultant displacement between the two vectors will increase.
Learn more here: brainly.com/question/20885836
C- Stratus clouds are the rainy clouds.
Hope this helps!
Work formula:

F = 50N, d = 1.0 m
When you lift something straight up, the angle of the force is 90º
cos(90º) is 0, so there's no work done when you lift the microwave off the ground

F = 50N, d = 1.0 m
When you push the microwave, the angle is 0º and cos(0º) is 1. So there is work done here:


total work = 50 joules