Answer: If saturation effects are ignored, the magnetization curve of Fig. 7.27 becomes a straight line with a constant slope of 150 volts per ampere of field current.
Explanation:
The asymptotes of the open loop transfer are:
- Horizontal: y = 0
- Vertical: x = -10 and x = -100
<h3>How to plot the
asymptotes?</h3>
The open loop transfer function is given as:
f(s) = 100(s + 1)/((s + 10)(s + 100))
Set the numerator of the function to 0.
So, we have:
f(s) = 0/((s + 10)(s + 100))
Evaluate
f(s) = 0
This means that, the vertical asymptote is y = 0
Set the denominator of the function to 0.
(s + 10)(s + 100) 0
Split
s + 10 = 0 and s + 100 = 0
Solve for s
s = -10 and s = -100
This means that, the horizontal asymptotes are s = -10 and s = -100
See attachment for the graph of the asymptotes
Read more about asymptotes at:
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Answer:
1. Move as far right as possible, check traffic, and signal a left turn.
2. Turn the steering wheel sharply to the left and move forward slowly.
3. Shift to reverse, turn your wheels sharply to the right, check traffic, and back your vehicle to the right curb, or edge of roadway.
To develop the problem it is necessary to apply the concepts related to the ideal gas law, mass flow rate and total enthalpy.
The gas ideal law is given as,

Where,
P = Pressure
V = Volume
m = mass
R = Gas Constant
T = Temperature
Our data are given by




Note that the pressure to 38°C is 0.06626 bar
PART A) Using the ideal gas equation to calculate the mass flow,




Therfore the mass flow rate at which water condenses, then

Re-arrange to find 



PART B) Enthalpy is given by definition as,

Where,
= Enthalpy of dry air
= Enthalpy of water vapor
Replacing with our values we have that



In the conversion system 1 ton is equal to 210kJ / min


The cooling requeriment in tons of cooling is 437.2.