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stiv31 [10]
3 years ago
10

Complex poles cmd zeros. Sketch the asymptotes of the Bode plot magnitude and phase for each of the listed open-loop mmsfer fuoc

tions. and approximate the transitillll at the stX'ond-order break point. based on the value of the ... ,. ,, 410. The Frequency-Response Design Method damping ratio. After completing the hand sketches, verify your result using Mallab. Tum in your hand sketches and the Matlab results on the same scales.
(a) L(s) = s2 + 3~r + 10
(b) L(s) = s(s2 + ~s + 10)
(c) L(s) = (s2 + 2s + 8) s(s2 + 2s + 10)
(d) L(s) = (s2 +I) s(s• +4)
(e) L(s) = (s:? +4) s(.r~ + I) 6
Engineering
1 answer:
Roman55 [17]3 years ago
8 0

Answer: e is the best answer

Explanation:

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An insulated tank having a total volume of 0.6 m3 is divided into two compartments. Initially one compartment contains 0.4 m3 of
IRISSAK [1]

Answer:

a. The final temperature is 69.8°C

b. The final pressure is 2.67bar

c. The amount of entropy produced is 0.38568kJ/K

Explanation:

Pls, check the attached files for detail step by step explanation as typing the solution here can not be explicit.

5 0
3 years ago
Following lockout/tagout (LOTO) procedures is important but not required when working with dangerous
Stella [2.4K]

Answer:

False.

Explanation:

Employees working with dangerous machines and hazardous energy are required to know and follow LOTO procedures in order to ensure the safety of themselves and fellow employees. Employers are also required to train employees in LOTO procedures to ensure they know, understand, and are able to follow the applicable provisions of LOTO procedures.

7 0
3 years ago
A strain gage is mounted at an angle of 30° with respect to the longitudinal axis of the cylindrical pressure. The pressure vess
GuDViN [60]

Answer:

1790 μrad.

Explanation:

Young's modulus, E is given as 10000 ksi,

μ is given as 0.33,

Inside diameter, d = 54 in,

Thickness, t = 1 in,

Pressure, p = 794 psi = 0.794 ksi

To determine shear strain, longitudinal strain and circumferential strain will be evaluated,

Longitudinal strain, eL = (pd/4tE)(1 - 2μ)

eL = (0.794 x 54)(1 - 0.66)/(4 x 1 x 10000)

eL = 3.64 x 10-⁴ radians

Circumferential strain , eH = (pd/4tE)(2-μ)

eH = (0.794 x 54)(2 - 0.33)/(4 x 1 x 10000)

eH = 1.79 x 10-³ radians

The maximum shear strain is 1790 μrad.

4 0
3 years ago
A sleeve made of SAE 4150 annealed steel has a nominal inside diameter of 3.0 inches and an outside diameter of 4.0 inches. It i
irga5000 [103]

Answer:

Class of fit:

Interference (Medium Drive Force Fits constitute a special type of Interference Fits and these are the tightest fits where accuracy is important).

Here minimum shaft diameter will be greater than the maximum hole diameter.

Medium Drive Force Fits are FN 2 Fits.

As per standard ANSI B4.1 :

Desired Tolerance: FN 2

Tolerance TZone: H7S6

Max Shaft Diameter: 3.0029

Min Shaft Diameter: 3.0022

Max Hole Diameter:3.0012

Min Hole Diameter: 3.0000

Max Interference: 0.0029

Min Interference: 0.0010

Stress in the shaft and sleeve can be considered as the compressive stress which can be determined using load/interference area.

Design is acceptable If compressive stress induced due to selected dimensions and load is less than compressive strength of the material.

Explanation:

4 0
3 years ago
Read 2 more answers
Give two causes that can result in surface cracking on extruded products.
Andreas93 [3]

Answer:

1. High friction

2. High extrusion temperature

Explanation:

Surface cracking on extruded products are defects or breakage on the surface of the extruded parts. Such cracks are inter granular.

           Surface cracking defects arises from very high work piece temperature that develops cracks on the surface of the work piece. Surface cracking appears when the extrusion speed is very high, that results in high strain rates and generates heat.

          Other factors include very high friction that contributes to surface cracking an d chilling of the surface of high temperature billets.

6 0
3 years ago
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