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Nutka1998 [239]
4 years ago
5

Which of the following impulse responses correspond(s) to stable LTI systems? (a) h1(t) = e-(1-2j)ru(t) (b) h2(t) = e-r cos(2t)u

(t)
Engineering
1 answer:
Illusion [34]4 years ago
6 0

Answer:

LTI system is stable if Impulse response is finite.

so the correct answer is "b"

(b) h2(t) = e-r cos(2t)u(t)

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Oxygen enters an insulated 14.2-cm-diameter pipe with a velocity of 60 m/s. At the pipe entrance, the oxygen is at 240 kPa and 2
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Answer:

Entropy generation==0.12 KW/K

Explanation:

s_2-s_1=C_p\ln \frac{T_2}{T_1}-R\ln \frac{P_2}{P_1}

s_2-s_1=0.891\ln \frac{291}{293}-0.2598\ln \frac{200}{240}

s_2-s_1=0.0412\frac{KJ}{kg-K}

Mass flow rate= \rho\times\dfrac{\pi}{4}d^2V

\rho_1=\dfrac {P_1}{RT_1}

\rho_1=\dfrac {240}{0.2598\times 293}

\rho_1=3.51\frac{kg}{m^3}

mass flow rate=\rho_1A_1V_1

So by putting the values

Mass flow rate=2.97 kg/s

So entropy generation=(2.97)(0.0412)

                                    =0.12 KW/K

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4 years ago
Which two statements about professional technical jobs in the energy industry are correct?
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The answer is both B and D
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1. What is resilience engineering and how does it pertain to the current<br>corona virus crisis? ​
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the ability to absorb or avoid damage without suffering complete failure and is an objective of design, maintenance and restoration for buildings and infrastructure

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When entering a freeway you should always:
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The system is initially moving with the cable taut, the 15-kg block moving down the rough incline with a speed of 0.080 m/s, and
garik1379 [7]

Solution :

The spring is expanded by 2 times of the block when it moves down an inclined by x times.

Here, $x_1$ = 39 mm

        x_2 = 225 mm

a). From the work energy principal,

   Work forces = kinetic energy

$(mg \sin 50^\circ)\times \frac{99}{1000}-(\mu_k mg \cos 50^\circ) \times \frac{99}{1000} -\frac{1}{2}k(0.225^2 - 0.039^2)=\frac{1}{2}m(V^2_2-0.08^2)$

$(112.6 \times 0.099)-(14.17 \times 0.099)-4.91= 7.5(V^2_2-0.08^2)$

$9.75= 7.5(V^2_2-0.08^2)$

$1.3= V^2_2-0.08^2$

$V_2=1.14\ m/s$

b). calculating the distance travelled by the block before it comes to rest.

Substitute the value of V_2 in (1),

$-(\mu_kmg \cos 50^\circ)x + (mg \sin 50^\circ)x-\frac{1}{2}k\left( ( 2x+0.039)^2 - 0.039^2\right)= -\frac{1}{2}m(0.08)^2$

$-14.17x+112.6x - 100(4x^2+0.156x)=-0.048$

$98.43x - 100(4x^2+0.156x)+0.048=0$

$98.43x - 400x^2-15.6x+0.048=0$

$82.83x - 400x^2+0.048=0$

$  400x^2- 82.83x-0.048=0$

x = 0.20 m

4 0
3 years ago
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