Answer:
The volume flow rate of air is 
Explanation:
A random duct is shown in the below attached figure
The volume flow rate is defined as the volume of fluid that passes a section in unit amount of time
Now by definition of velocity we can see that 'v' m/s means that in 1 second the flow occupies a length of 'v' meters
From the attached figure we can see that
The volume of the prism that the flow occupies in 1 second equals

Hence the volume flow rate is 
At entrance to the nozzle, the pressure is 0.180 MPa and the temperature is 1200 K. The kinetic energy of the gas entering the nozzle is very much smaller than ... The specific heat of the exhaust gas varies with temperature approximately as follows: ... Problem 4P: In an aircraft jet engine at takeoff, the combustion product.
Answer:
The elastic modulus of the steel is 139062.5 N/in^2
Explanation:
Elastic modulus = stress ÷ strain
Load = 89,000 N
Area of square cross section of the steel bar = (0.8 in)^2 = 0.64 in^2
Stress = load/area = 89,000/0.64 = 139.0625 N/in^2
Length of steel bar = 4 in
Extension = 4×10^-3 in
Strain = extension/length = 4×10^-3/4 = 1×10^-3
Elastic modulus = 139.0625 N/in^2 ÷ 1×10^-3 = 139062.5 N/in^2
Answer:
116.3 electrons
Explanation:
Data provided in the question:
Time, t = 2.55 ps = 2.55 × 10⁻¹² s
Current, i = 7.3 μA = 7.3 × 10⁻⁶ A
Now,
we know,
Charge, Q = it
thus,
Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)
or
Q = 18.615 × 10⁻¹⁸ C
Also,
We know
Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C
Therefore,
Number of electrons past a fixed point = Q ÷ q
= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]
= 116.3 electrons