To solve the problem, it is necessary to apply the concepts related to the change of mass flow for both entry and exit.
The general formula is defined by

Where,
mass flow rate
Density
V = Velocity
Our values are divided by inlet(1) and outlet(2) by





PART A) Applying the flow equation we have to



PART B) For the exit area we need to arrange the equation in function of Area, that is



Therefore the Area at the end is 
Answer:
m = 0.0125 kg
Explanation:
Let us apply the formula for the speed of a wave on a string that is under tension:

where F = tension force
μ = mass per unit length
Mass per unit length is given as:
μ = m / l
where m = mass of the string
l = length of the string
This implies that:

Let us make mass, m, the subject of the formula:

From the question:
F = 20 N
l = 4.50 m
v = 85 m/s
Therefore:

Answer:
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