Answer:
Part a)
f = 1911.5 Hz
Part b)

Explanation:
Here the source and observer both are moving towards each other
so we know that the apparent frequency is given as

here we know that



now we will have


Part b)
Apparent wavelength is given by the formula

here we will have


Answer:
The magnitude of the electric field at a point equidistant from the lines is 
Explanation:
Given that,
Positive charge = 24.00 μC/m
Distance = 4.10 m
We need to calculate the angle
Using formula of angle



We need to calculate the magnitude of the electric field at a point equidistant from the lines
Using formula of electric field

Put the value into the formula



Hence, The magnitude of the electric field at a point equidistant from the lines is 
If you use the next formula with the data given in the exercise you are asking:
Ey[3.4] - F[1.7] = 0
<span>Ey = F/2
</span>and after that what you need to do is sum the moments of the handle about D to zero asumming it is a positive moment and we proceed like this
Ey[1.5sin19] – P[21 – 1.5sin19] = 0
<span>(F/2)[1.5sin19] = P[21 – 1.5sin19] </span>
<span>F = 2P[21 – 1.5sin19] / [1.5sin19] </span>
<span>F = 84P </span>
Answer:
a

b

Explanation:
From the question we are told that
The charge on each water drop is 
The distance of separation is
Generally the electrostatic force between the water drops is mathematically represented as

Here k is the coulombs constant with value 
So


Generally the quantity of charge is mathematically represented as

Here n is the number of electron present
and e is the charge on one electron with value 
So

