Answer:
The reactances vary with frequency, with large XL at high frequencies and large Xc at low frequencies, as we have seen in three previous examples. At some intermediate frequency fo, the reactances will be the same and will cancel, giving Z = R; this is a minimum value for impedance and a maximum value for Irms results. We can get an expression for fo by taking
XL=Xc
Substituting the definitions of XL and XC,
2
foL=1/2foC
Solving this expression for fo yields
fo=1/2
where fo is the resonant frequency of an RLC series circuit. This is also the natural frequency at which the circuit would oscillate if it were not driven by the voltage source. In fo, the effects of the inductor and capacitor are canceled, so that Z = R and Irms is a maximum.
Explanation:
Resonance in AC circuits is analogous to mechanical resonance, where resonance is defined as a forced oscillation, in this case, forced by the voltage source, at the natural frequency of the system. The receiver on a radio is an RLC circuit that oscillates best at its {f} 0. A variable capacitor is often used to adjust fo to receive a desired frequency and reject others is a graph of current versus frequency, illustrating a resonant peak at Irms at fo. The two arcs are for two dissimilar circuits, which vary only in the amount of resistance in them. The peak is lower and wider for the highest resistance circuit. Thus, the circuit of higher resistance does not resonate as strongly and would not be as selective in a radio receiver, for example.
A current versus frequency graph for two RLC series circuits that differ only in the amount of resistance. Both have resonance at fo, but for the highest resistance it is lower and wider. The conductive AC voltage source has a fixed amplitude Vo.
This question is incomplete, the complete question is;
(Laminar flow) A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length 2
. The flow is laminar and fully developed. The pressure drop for the first pipe is 1.657 times greater than it is for the second pipe. If the diameter of the first pipe is D, determine the diameter of the second pipe.
D₃ = _____D.
{ the tolerance is +/-3% }
Answer:
the diameter of the second pipe D₃ is 1.13D
Explanation:
Given the data in the question;
Length = 2
pressure drop in the first pipe is 1.657 times greater than it is for the second pipe.
Now, we know that for Laminar Flow;
V' = πD⁴ΔP / 128μL
where V'₁ = V'₂ and ΔP₁₋₂ = 1.657 ΔP₂₋₃
Hence,
V'₁ = πD⁴ΔP₁₋₂ / 128μL = V'₃ = πD₃⁴ΔP₂₋₃ / 128μL
so
D₃ = D
ΔP₁₋₂ / ΔP₂₋₃ 
we substitute
D₃ = D 1.657
D₃ = D( 1.134568 )
D₃ = 1.13D
Therefore, the diameter of the second pipe D₃ is 1.13D
Answer:
Tech B
Explanation:
Bleeding the master cylinder on the bench does nothing for the air in the brake lines on the vehicle. The bench bleeding is a preferred first step, but bleeding the rest of the brake system is also required. A final check of proper operation on the vehicle should also be accomplished.
Tech B is correct.
Answer:
79 kW.
Explanation:
The equation for enthalpy is:
H2 = H1 + Q - L
Enthalpy is defined as:
H = G*(Cv*T + p*v)
This is specific volume.
The gas state equation is:
p*v = R*T (with specific volume)
The specific gas constant for air is:
287 K/(kg*K)
Then:
T1 = 60 + 273 = 333 K
T2 = 200 + 273 = 473 K
p1*v1 = 287 * 333 = 95.6 kJ/kg
p2*v2 = 287 * 473 = 135.7 kJ/kg
The Cv for air is:
Cv = 720 J/(kg*K)
So the enthalpies are:
H1 = 0.8*(0.72 * 333 + 95.6) = 268 kW
H2 = 0.8*(0.72 * 473 + 135.7) = 381 kW
Ang the heat is:
Q = 34 kW
Then:
H2 = H1 + Q - L
381 = 268 + 34 - L
L = 268 + 34 - 381 = -79 kW
This is the work from the point of view of the air, that's why it is negative.
From the point of view of the machine it is positive.
Answer:
Fatigue factor of safety is 2.0267
Explanation:
Solution is attached below.
