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:
brainly.com/question/4084552
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Answer:
h1 = 290.16kj/kg
P = 1.2311
Prandil expression at 8
P=p1/p7×pr
=8(1.2311)
=9.85
Enthalpy state at 8 corresponding to 9.85
h1 = 526.13kj/kg
Now prandtl state at 9 that correspond to 1400k.
h9 = 1515.42kj/kg
Pr = 450.5
Prandtl expression at state 10
P= p10/p9×pr
=1/8(450.5)
=56.31
Enthalpy at state 10 corresponding to prandtl 56.31
h10 = 860.39kj/kg
At 520k
h11 = 523.63kj/kg
Answer: Static Balancing is when a stationary object has the the ability to balance because its core centre of gravity is on the axis of rotation.
Answer:
A. Loose concrete
Explanation:
Concrete is a solid composite structural material.It is the most used component in setting up a structure. Concrete is known for its very good ability to bear load, but can sometimes fail, or can be damaged. To rehabilitate or repair a concrete, the loose concrete must first be removed, so that the surface can be cleaned and well set before the material used for repair is applied on it.
