Answer:
an attachment is below
Explanation:
1) the formula for damping coefficient id for RLC series circuit.
For \xi =0 you can make c=0 but inductor will still have some capacitance.
2) the responses of critically damped system and under damped system are shown with comments on their time response.
4) There can be many different answers to this question, but the 4 I have mentioned are the most important parameters we need to know about an unknown op-amp if we are to use it in our circuit.
Hope it answers all your questions.
Answer. In the present study, it was found that 61% students had multimodal learning style preferences and that only 39% students had unimodal preferences. Amongst the multimodal learning styles, the most preferred mode was bimodal, followed by trimodal and quadrimodal respectively
Explanation:
☐ E-W
☐ NW-SE
☐ NE-SW
Answer:
☐ NE-SW
Explanation:
Based on the description, the rock direction is North East - South West (NE-SW). Rocks generally can expand or compress depending on the type and magnitude of stress applied on the rocks. However, if the applied stress is sufficiently high, cracks and fractures will be created on the rock and it can ultimately lead to the formation of particles.
Answer:
See explaination
Explanation:
See attachment for the detailed step by step solution of the given problem.
Answer:
the netuon arrangment
Explanation:
b. the back work ratio
c. the net power developed, in kW
d. the rates of exergy destruction in each compressor stage and the turbine stage as well as the regenerator, in kW, for T 0 = 300 K.
Answer:
a. = 77.65%
b. bwr = 6.5%
c. 3538.986 kW
d. -163.169 kJ
Explanation:
a. The given property are;
P₂/P₁ = 10, P₂ = 10 * 100 kPa = 1000 kPa
p₄/p₁ = 10
P₂/P₁ = p₄/p₃ = √10
p₂ = 100·√10
= T₁×(√10)^(0.4/1.4) = 300 × (√10)^(0.4/1.4) = 416.85 K
T₂ = T₁ + ( - T₁)/
= 300 + (416.85 - 300)/0.8 = 446.0625 K
p₄ = 10×p₁ = 10×100 = 1000 kPa
p₄/p₃ = √10 =
p₃ = 100·√10
T₃ = 300 K
T₃/ = (P₂/P₁)^((k - 1)/k) = (√10)^(0.4/1.4)
= T₃/((√10)^(0.4/1.4) ) = 300/((√10)^(0.4/1.4)) = 215.905 K
T₄ = T₃ + ( - T₃)/
= 300 + (215.905- 300)/0.8 = 194.881 K
The efficiency = 1 - (T₄ - T₁)/(T₃ - T₂) = 1 - (194.881 -300)/(300 -446.0625 ) = 0.28
T₄ = 446.0625 K
T₆ = 1400 K
/T₆ = (1/√10)^(0.4/1.4)
= 1400×(1/√10)^(0.4/1.4) = 1007.6 K
T₇ = T₆ - (T₆ -
) = 1400 - 0.8*(1400 - 1007.6) = 1086.08 K
T₈ = 1400 K
T₉ = 1086.08 K
T₅ = T₄ + (T₉ - T₄) = 446.0625 +0.8*(1086.08 - 446.0625) = 958.0765 K
=(((T₆ - T₇) + (T₈ - T₉)) -((T₂ - T₁) + (T₄ - T₃)))/((T₆ - T₅) + (T₈ - T₇))
(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300)))/((1400 -958.0765 ) + (1400 -1086.08 )) = 0.7765
= 77.65%
b. Back work ratio, bwr =
((446.0625 - 300)+(194.881 - 300))/((1400 - 1086.08) + (1400 -1086.08 ))
40.9435/627.84 = 6.5%
c.
Power developed is given by the relation;
= 6*1.005*(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300))) = 3538.986 kW
d. Exergy destruction = 6*(1.005*(300-446.0625 ) - 300*1.005*(-0.3966766)
-163.169 kJ
Answer:
mV
Explanation:
The voltage across a capacitor at a time t, is given by:
----------------(i)
Where;
v(t) = voltage at time t
= initial time
C = capacitance of the capacitor
i(t) = current through the capacitor at time t
v(t₀) = voltage at initial time.
From the question:
C = 2μF = 2 x 10⁻⁶F
i(t) = 3 mA
t₀ = 0
v(t₀ = 0) = 0
Substitute these values into equation (i) as follows;
[Solve the integral]
Therefore, the voltage across the capacitor is mV