Answer:

Explanation:
The inlet specific volume of air is given by:

The mass flow rates is expressed as:

The energy balance for the system can the be expresses in the rate form as:
![E_{in}-E_{out}=\bigtriangleup \dot E=0\\\\E_{in}=E_{out}\\\\\dot m(h_1+0.5V_1^2)=\dot W_{out}+\dot m(h_2+0.5V_2^2)+Q_{out}\\\\\dot W_{out}=\dot m(h_2-h_1+0.5(V_2^2-V_1^2))=-m({cp(T_2-t_1)+0.5(V_2^2-V_1^2)})\\\\\\\dot W_{out}=-(10.42lbm/s)[(0.25\frac{Btu}{lbm.\textdegree F})(300-900)\textdegree F+0.5((700ft/s)^2-(350ft/s)^2)(\frac{1\frac{Btu}{lbm}}{25037ft^2/s^2})]\\\\\\\\=1486.5\frac{Btu}{s}](https://tex.z-dn.net/?f=E_%7Bin%7D-E_%7Bout%7D%3D%5Cbigtriangleup%20%5Cdot%20E%3D0%5C%5C%5C%5CE_%7Bin%7D%3DE_%7Bout%7D%5C%5C%5C%5C%5Cdot%20m%28h_1%2B0.5V_1%5E2%29%3D%5Cdot%20W_%7Bout%7D%2B%5Cdot%20m%28h_2%2B0.5V_2%5E2%29%2BQ_%7Bout%7D%5C%5C%5C%5C%5Cdot%20W_%7Bout%7D%3D%5Cdot%20m%28h_2-h_1%2B0.5%28V_2%5E2-V_1%5E2%29%29%3D-m%28%7Bcp%28T_2-t_1%29%2B0.5%28V_2%5E2-V_1%5E2%29%7D%29%5C%5C%5C%5C%5C%5C%5Cdot%20W_%7Bout%7D%3D-%2810.42lbm%2Fs%29%5B%280.25%5Cfrac%7BBtu%7D%7Blbm.%5Ctextdegree%20F%7D%29%28300-900%29%5Ctextdegree%20F%2B0.5%28%28700ft%2Fs%29%5E2-%28350ft%2Fs%29%5E2%29%28%5Cfrac%7B1%5Cfrac%7BBtu%7D%7Blbm%7D%7D%7B25037ft%5E2%2Fs%5E2%7D%29%5D%5C%5C%5C%5C%5C%5C%5C%5C%3D1486.5%5Cfrac%7BBtu%7D%7Bs%7D)
Hence, the mass flow rate of the air is 1486.5Btu/s
Technician A and B are correct . Because according to technician A, the cause written on the repair order is a diagnosis. Here, by diagnosis, he means that the problem is identified after examining the device and hence the judgement is made.
And according to B, you have to write the cause of the problems in the device that have been identified and the concern measures, which is also kind of diagnosis.
So, option D is correct.
Answer:
The magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.
Explanation:
Given;
distance half way between the parallel wires, r = ¹/₂ (40 cm) = 20 cm = 0.2 m
current carried in opposite direction, I₁ and I₂ = 10 A and 20 A respectively
The magnitude of the magnetic field halfway between the wires can be calculated as;

where;
B is magnitude of the magnetic field halfway between the wires
I₁ is current in the first wire
I₂ is current the second wire
μ₀ is permeability of free space
r is distance half way between the wires

Therefore, the magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.