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:
a) W = - 1.752 10⁻¹⁸ J, b) U = + 1.752 10⁻¹⁸ J
Explanation:
a) work is defined by
W = F . x
the bold letters indicate vectors, in this case the force is electric
F = q E
we substitute
F = q E x
the charge of the electron is
q = - e
F = - e E x
let's calculate
W = - 1.6 10⁻¹⁹ 365 3 10⁻²
W = - 1.752 10⁻¹⁸ J
b) the change in potential energy is
U = q ΔV
the potential difference is
ΔV = - E. Δs
we substitute
U = - q E Δs
the charge of the electron is
q = - e
U = e E Δs
we calculate
U = 1.6 10⁻¹⁹ 365 3 10⁻²
U = + 1.752 10⁻¹⁸ J
The atomic number of beryllium (Be) is 4, and the atomic number of barium (Ba) is 56. the <span>comparison is best supported by this information is that beryllium has a lower atomic radius than Barium</span>
