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
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Using an appropriate failure theory, find the factor of safety in each case. State the name of the theory that you are using the theory is max stress theory.
<h3>Wat is the max stress theory?</h3>
The most shear strain concept states that the failure or yielding of a ductile fabric will arise whilst the most shear strain of the fabric equals or exceeds the shear strain fee at yield factor withinside the uniaxial tensile test.”
Stress states at various critical locations are f= 2.662.
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Answer:
El principio de arquimedes es un principio de la hidroestática que explica lo que experimenta un cuerpo ... la teoría y después como es costumbre pasaremos a ver problemas resueltos sobre éste principio. ... Si la magnitud del peso del cuerpo es menor a la magnitud de empuje. ... Alguien me ayuda con este Ejercicio :.
Explanation:
Answer:
a. Fraction of Atom = 2.41E-5 when T = 600K
b. Fraction of Atom = 5.03E-10 when T = 298K
Explanation:
a.
Given
T = Temperature = 600K
Qv = Energy for formation = 0.55eV/atom
To calculate the fraction of atom sites, we make use of the following formula
Nv/N = exp(-Qv/kT)
Where k = Boltzmann Constant = 8.62E-5eV/K
Nv/N = exp(-0.55/(8.62E-5 * 600))
Nv/N = 0.000024078672493307
Nv/N = 2.41E-5
b. When T = 298K
Nv/N = exp(-0.55/(8.62E-5 * 298))
Nv/N = 5.026591237904E−10
Nv/N = 5.03E-10 ----- Approximated
Answer:
The required mechanical work is required to reduce each day by 1.05×10^8 Joules.
Explanation:
Coefficient of Performance (COP) = Q/W
Q is thermal energy absorbed by the air conditioner
W is mechanical work done
Q = 3.9×10^8 J
COP of old air conditioner = 2.3
W = Q/COP = 3.9×10^8/2.3 = 1.70×10^8 J
COP of new air conditioner = 6
W = Q/COP = 3.9×10^8/6 = 6.5×10^7 J
Reduction in mechanical work = (1.7×10^8) - (6.5×10^7) = 1.05×10^8 J
