ENGINEERING COLLEGE

. Were you able to observe ???? = 0 in the circuit you constructed during lab? Why or why not? Hint: What value of resistance would be needed for ???? = 0? 2. What feature in the time response of an RLC circuit distinguishes a critically damped response from an underdamped response? 3. Why must an op-amp be powered to be used in a circuit? 4. If you were handed a parts kit with an unknown op-amp, what information would you need to find prior to using it in a circuit?

Answers

Answer 1

Answer:

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.

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Young students show a preference for which modality?

Answers

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:

(d) Arches NP is known for its spectacular arches that develop in the jointed areas of the park. Placemark Problem 2d flies you to Landscape Arch, the arch with the largest span in Arches NP. If the stresses that stretched the rock to form the joints were oriented perpendicular to the joint surfaces and the rock fins that contain the arches, what was the direction that the rocks were stretched? ☐ N-S
☐ E-W
☐ NW-SE
☐ NE-SW

Answers

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.

A insulated vessel s has two compartments separated by a membreane. On one side is 1kg of steam at 400 degC and 200 bar. The other side is evacuated . The membrane ruptures, filling the entire volume. The finial pressure is 100bar. Determine the final temperature of the steam and the volume of the vessel.

Answers

Answer:

See explaination

Explanation:

See attachment for the detailed step by step solution of the given problem.

Webster is giving a speech on the benefits of moving toward the use of windmill energy instead of having to rely on fossil fuels. Most likely, he will select which method of arrangement?

Answers

Answer:

the netuon arrangment

Explanation:

Air enters the first compressor stage of a cold-air standard Brayton cycle with regeneration and intercooling at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The overall compressor pressure ratio is 10, and the pressure ratios are the same across each compressor stage. The temperature at the inlet to the second compressor stage is 300 K. The turbine inlet temperature is 1400 K. The compressor stages and turbine each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%. For k = 1.4, calculate: a. the thermal efficiency of the cycle
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.

Answers

Answer:

a. $\eta _(th)$ = 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

$T_(2s)$ = T₁×(√10)^(0.4/1.4) = 300 × (√10)^(0.4/1.4) = 416.85 K

T₂ = T₁ + ( $T_(2s)$ - T₁)/ $\eta _c$ = 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/ $T_(4s)$ = (P₂/P₁)^((k - 1)/k) = (√10)^(0.4/1.4)

$T_(4s)$ = T₃/((√10)^(0.4/1.4) ) = 300/((√10)^(0.4/1.4)) = 215.905 K

T₄ = T₃ + ( $T_(4s)$ - T₃)/ $\eta _c$ = 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

$T_(7s)$ /T₆ = (1/√10)^(0.4/1.4)

$T_(7s)$ = 1400×(1/√10)^(0.4/1.4) = 1007.6 K

T₇ = T₆ - $\eta _t$ (T₆ - $T_(7s)$ ) = 1400 - 0.8*(1400 - 1007.6) = 1086.08 K

T₈ = 1400 K

T₉ = 1086.08 K

T₅ = T₄ + $\epsilon _(regen)$ (T₉ - T₄) = 446.0625 +0.8*(1086.08 - 446.0625) = 958.0765 K

$\eta _(th)$ =(((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

$\eta _(th)$ = 77.65%

b. Back work ratio, bwr = $bwr = (w_(c,in))/(w_(t,out))$

((446.0625 - 300)+(194.881 - 300))/((1400 - 1086.08) + (1400 -1086.08 ))

40.9435/627.84 = 6.5%

c. $w_(net, out) = c_p[(T_6 -T_7) + (T_8 - T_9)] - [(T_2 - T_1) + (T_4 -T_3)]$

Power developed is given by the relation;

$\dot m \cdot w_(net, out)$

$\dot m \cdot w_(net, out)$ = 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

Determine the voltage across a 2-μF capacitor if the current through it is i(t) = 3e−6000t mA. Assume that the initial capacitor voltage is zero g

Answers

Answer:

$v = 250[1 - {e^(-6000t)}]$ mV

Explanation:

The voltage across a capacitor at a time t, is given by:

$v(t) = (1)/(C) \int\limits^(t)_(t_0) {i(t)} \, dt + v(t_0)$ ----------------(i)

Where;

v(t) = voltage at time t

$t_(0)$ = 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 $e^(-6000t)$ mA

t₀ = 0

v(t₀ = 0) = 0

Substitute these values into equation (i) as follows;

$v = (1)/(2*10^(-6)) \int\limits^(t)_(0) {3e^(-6000t)} \, dt + v(0)$

$v = (1)/(2*10^(-6)) \int\limits^(t)_(0) {3e^(-6000t)} \, dt + 0$

$v = (1)/(2*10^(-6)) \int\limits^(t)_(0) {3e^(-6000t)} \, dt$

$v = (3)/(2*10^(-6)) \int\limits^(t)_(0) {e^(-6000t)} \, dt$ [Solve the integral]

$v = (3)/(2*10^(-6)*(-6000)) {e^(-6000t)}|_0^t$

$v = (-3000)/(12) {e^(-6000t)}|_0^t$

$v = -250 {e^(-6000t)}|_0^t$

$v = -250 {e^(-6000t)} - [-250 {e^(-6000(0))]$

$v = -250 {e^(-6000t)} - [-250]$

$v = -250 {e^(-6000t)} + 250$

$v = 250 -250 {e^(-6000t)}$

$v = 250[1 - {e^(-6000t)}]$

Therefore, the voltage across the capacitor is $v = 250[1 - {e^(-6000t)}]$ mV

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