Find the equivalent impedance Zeq seen by the source when Vs = 2 cos (5t) v, C = 0.2 F, R = 1 Ω and L = 0.1 H. (Give angles in d
egrees and round final answers to two decimal places.) Calculate the voltage across the resistor. (Give angles in degrees and round final answers to two decimal places.) complex impedance The equivalent impedance seen by the source is + j Ω = at an angle of Ω. The voltage across the resistor is at an angle of v.
1 answer:
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Answer:
Relative density = 0.7 or 70%
Explanation:
The following information was provided by this question
Pd = 1.72mg/mg³
Pd max = 1.81 mg/mg³
Pd min = 1.54 mg/mg³
We substitute into the formula. This formula is contained in the attachment.
[(1/1.54)-(1/1.72)]/[1/1.54 - 1/1.81]
= 0.649350 - 0.581395 / 0.649350 - 0.552486
= 0.067955/0.096864
= 0.7015
= 0.7
The relative density is Therefore 0.7 or 70% when converted to percentage
Answer
given,
a = 2 t - 10
velocity function
we know,
integrating both side
v = t² - 10 t + C
at t = 0 v = 3
so, 3 = 0 - 0 + C
C = 3
Velocity function is equal to v = t² - 10 t + 3
Again we know,
integrating both side
now, at t= 0 s = -4
C = -4
So,
Position function is equal to