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natulia [17]
3 years ago
15

Complex poles and zeros. Sketch the asymptotes of the Bode plot magnitude and phase for each of the listed open-loop transfer fu

nctions, and approximate the transition at the second-order break point, based on the value of the damping ratio. After completing the hand sketches, verify your result using Matlab.
L(s)= 1/s(s^2+ 3s+ 10)
Engineering
1 answer:
Sergio [31]3 years ago
7 0
The answer in image

Don't forget put heart ♥️

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Saturated liquid water at 150 F is put under pressure to decrease the volume by 1% while keeping the temperature constant. To wh
Romashka-Z-Leto [24]

Answer:

Between 5 & 10 MPa in Table B.1.4

Explanation:

150F = 65°C

State 1:

T = 65°C , x = 0.0; Table B.1.1: v = 0,001020 m^3 /kg

Process: T = constant = 65°C

State 2:

T, v = 0.99 x v f (65°C)  = 0.99 x 0,001020 = 0.0010098 m^3 /kg

Between 5 & 10 MPa in Table B.1.4

5 0
3 years ago
Which of the following statements about cylinder placement are true?
Alenkinab [10]
What is the following?
3 0
3 years ago
A pin fin of uniform cross-sectional area is fabricated of an aluminum alloy (k = 160W m-1 K-1 ). The fin diameter is D = 4 mm,
disa [49]

Answer: (a) 36.18mm

(b) 23.52

Explanation: see attachment

4 0
4 years ago
Here, we want to become proficient at changing units so that we can perform calculations as needed. The basic heat transfer equa
netineya [11]

Answer:

9500 kJ; 9000 Btu

Explanation:

Data:

m = 100 lb

T₁ = 25 °C

T₂ = 75 °C

Calculations:

1. Energy in kilojoules

ΔT = 75 °C - 25 °C = 50 °C  = 50 K

m = \text{100 lb} \times \dfrac{\text{1 kg}}{\text{2.205 lb}} \times \dfrac{\text{1000 g}}{\text{1 kg}}= 4.54 \times 10^{4}\text{ g}\\\\\begin{array}{rcl}q & = & mC_{\text{p}}\Delta T\\& = & 4.54 \times 10^{4}\text{ g} \times 4.18 \text{ J$\cdot$K$^{-1}$g$^{-1}$} \times 50 \text{ K}\\ & = & 9.5 \times 10^{6}\text{ J}\\ & = & \textbf{9500 kJ}\\\end{array}

2. Energy in British thermal units

\text{Energy} = \text{9500 kJ} \times \dfrac{\text{1 Btu}}{\text{1.055 kJ}} = \text{9000 Btu}

7 0
4 years ago
A piston cylinder containing 0.2 kg of air undergoes a process where pressure and volume are related by the expression P=CV wher
Black_prince [1.1K]

Answer:

Work = 51,33 kJ

Explanation:

According to the ideal gas equation, the initial volume is calculated as follows:

V_{i} =RT_{i}m/P_{i}=(0,2870)kJ/kg K*(25+273,15)K*(0,2)kg/(100)kN/m^{2}=0,1711m^{3}

Then the work for an isobaric process can be calculated using the following expression:

W=P(V_{f}-V_{i} )

And considering that,

V_{f}=4V_{i}

The work can be calculated as follows:

W=P(4V_{i}-V_{i})=3PV_{i} =(3)*(100)kn/m^{2}*(0,1711)m^{3}=51,33kJ

3 0
3 years ago
Read 2 more answers
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