Answer:
il(t) = e^(-100t)
Explanation:
The current from the source when the switch is closed is the current through an equivalent load of 15 + 50║50 = 15+25 = 40 ohms. That is, it is 80/40 = 2 amperes. That current is split evenly between the two parallel 50-ohm resistors, so the initial inductor current is 2/2 = 1 ampere.
The time constant is L/R = 0.20/20 = 0.01 seconds. Then the decaying current is described by ...
il(t) = e^(-t/.01)
il(t) = e^(-100t) . . . amperes
Attached is the solution to the above question.
Answer:
A good design for a portable device to mix paint minimizing the shaking forces and vibrations while still effectively mixing the paint. Is:
The best design is one with centripetal movement. Instead of vertical or horizontal movement. With a container and system of holding structures made of materials that could absorb the vibration effectively.
Explanation:
First of all centripetal movement would be friendlier to our objective as it would not shake the can or the machine itself with disruptive vibrations. Also, we would have to use materials with a good grade of force absorption to eradicate the transmission of the movement to the rest of the structure. Allowing the reduction of the shaking forces while maintaining it effective in the process of mixing.
The complete Question is:
Airflow through a long, 0.15-m-square air conditioning duct maintains the outer duct surface temperature at 10°C. If the horizontal duct is uninsulated and exposed to air at 35°C in the crawlspace beneath a home, what is the heat gain per unit length of the duct? Evaluate the properties of air at 300 K. For the sides of the duct, use the more accurate Churchill and Chu correlations for laminar flow on vertical plates.
What is the Rayleigh number for free convection on the outer sides of the duct?
What is the free convection heat transfer coefficient on the outer sides of the duct, in W/m2·K?
What is the Rayleigh number for free convection on the top of the duct?
What is the free convection heat transfer coefficient on the top of the duct, in W/m2·K?
What is the free convection heat transfer coefficient on the bottom of the duct, in W/m2·K?
What is the total heat gain to the duct per unit length, in W/m?
Answers:
- 7709251 or 7.709 ×10⁶
- 4.87
- 965073
- 5.931 W/m² K
- 2.868 W/m² K
- 69.498 W/m
Explanation:
Find the given attachments for complete explanation
Answer:
a) it is periodic
N = (20/3)k = 20 { for K =3}
b) it is Non-Periodic.
N = ∞
c) x(n) is periodic
N = LCM ( 5, 20 )
Explanation:
We know that In Discrete time system, complex exponentials and sinusoidal signals are periodic only when ( 2π/w₀) ratio is a rational number.
then the period of the signal is given as
N = ( 2π/w₀)K
k is least integer for which N is also integer
Now, if x(n) = x1(n) + x2(n) and if x1(n) and x2(n) are periodic then x(n) will also be periodic; given N = LCM of N1 and N2
now
a) cos(2π(0.15)n)
w₀ = 2π(0.15)
Now, 2π/w₀ = 2π/2π(0.15) = 1/(0.15) = 1×20 / ( 0.15×20) = 20/3
so, it is periodic
N = (20/3)k = 20 { for K =3}
b) cos(2n);
w₀ = 2
Now, 2π/w₀ = 2π/2) = π
so, it is Non-Periodic.
N = ∞
c) cos(π0.3n) + cos(π0.4n)
x(n) = x1(n) + x2(n)
x1(n) = cos(π0.3n)
x2(n) = cos(π0.4n)
so
w₀ = π0.3
2π/w₀ = 2π/π0.3 = 2/0.3 = ( 2×10)/(0.3×10) = 20/3
∴ N1 = 20
AND
w₀ = π0.4
2π/w₀ = 2π/π0. = 2/0.4 = ( 2×10)/(0.4×10) = 20/4 = 5
∴ N² = 5
so, x(n) is periodic
N = LCM ( 5, 20 )
