Fastest
(Known as the fast lane)
Answer:
A) 282.34 - j 12.08 Ω
B) 0.0266 + j 0.621 / unit
C)
A = 0.812 < 1.09° per unit
B = 164.6 < 85.42°Ω
C = 2.061 * 10^-3 < 90.32° s
D = 0.812 < 1.09° per unit
Explanation:
Given data :
Z ( impedance ) = 0.03 i + j 0.35 Ω/km
positive sequence shunt admittance ( Y ) = j4.4*10^-6 S/km
A) calculate Zc
Zc =
=
=
= 282.6 < -2.45°
hence Zc = 282.34 - j 12.08 Ω
B) Calculate gl
gl =
d = 500
z = 0.03 i + j 0.35
y = j4.4*10^-6 S/km
gl = 
= 
= 0.622 < 87.55 °
gl = 0.0266 + j 0.621 / unit
C) exact ABCD parameters for this line
A = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )
B = Zc sin h (gl) Ω = 164.6 < 85.42°Ω ( as calculated )
C = 1/Zc sin h (gl) s = 2.061 * 10^-3 < 90.32° s ( as calculated )
D = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )
where : cos h (gl) = 
sin h (gl) = 
Answer:
the elongation of the metal alloy is 21.998 mm
Explanation:
Given the data in the question;
K = σT/ (εT)ⁿ
given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,
strain-hardening exponent n = 0.22
we substitute
K = 345 / 
K = 815.8165 Mpa
next, we determine the true strain
(εT) = (σT/ K)^1/n
given that σT = 412 MPa
we substitute
(εT) = (412 / 815.8165 )^(1/0.22)
(εT) = 0.04481 mm
Now, we calculate the instantaneous length
= 
given that
= 480 mm
we substitute
=
× 
= 501.998 mm
Now we find the elongation;
Elongation = 
we substitute
Elongation = 501.998 mm - 480 mm
Elongation = 21.998 mm
Therefore, the elongation of the metal alloy is 21.998 mm
Answer:
q1q1 ⇒ 01
Explanation:
The outputs of a positive edge triggered register will match the inputs after a rising clock edge.
q1q1 ⇒ 01 . . . . matching d1d0 = 01