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Sveta_85 [38]
3 years ago
15

Design a digital integrator using the impulse invariance method. Find and give a rough sketch of the amplitude response, and com

pare it with that of the ideal integrator. If this integrator is used primarily for integrating audio signals (whose bandwidth is 20 kHz), determine a suitable value for T.

Engineering
1 answer:
mars1129 [50]3 years ago
8 0

Answer:

50 μsec

Explanation:

See the attached pictures for detailed answer.

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You are investigating surface hardening in iron using nitrogen gas. Two 5 mm thick slabs of iron are separately exposed to nitro
Luden [163]

Answer and Explanation:

The explanation is attached below

4 0
3 years ago
What is the force in kN of work done is 1.2 ms moves through 120m​
Semmy [17]

Answer:

\frac{1.2}{120}

0.01

5 0
2 years ago
A cylinder with a 6.0 in. diameter and 12.0 in. length is put under a compres-sive load of 150 kips. The modulus of elasticity f
jeka94

Answer:

Final Length = 11.992 in

Final Diameter = 6.001 in

Explanation:

First we calculate the cross-sectional area:

Area = A = πr² = π(3 in)² = 28.3 in²

Now, we calculate the stress:

Stress = Compressive Load/Area

Stress = - 150 kips/28.3 in²

Stress = -5.3 ksi

Now,

Modulus of Elasticity = Stress/Longitudinal Strain

8000 ksi = -5.3 ksi/Longitudinal Strain

Longitudinal Strain = -6.63 x 10⁻⁴

but,

Longitudinal Strain = (Final Length - Initial Length)/Initial Length

-6.63 x 10⁻⁴ = (Final Length - 12 in)/12 in

Final Length = (-6.63 x 10⁻⁴)(12 in) + 12 in

<u>Final Length = 11.992 in</u>

we know that:

Poisson's Ratio = - Lateral Strain/Longitudinal Strain

0.35 = - Lateral Strain/(- 6.63 x 10⁻⁴)

Lateral Strain = (0.35)(6.63 x 10⁻⁴)

Lateral Strain = 2.32 x 10⁻⁴

but,

Lateral Strain = (Final Diameter - Initial Diameter)/Initial Diameter

2.32 x 10⁻⁴ = (Final Diameter - 6 in)/6 in

Final Diameter = (2.32 x 10⁻⁴)(6 in) + 6 in

<u>Final Diameter = 6.001 in</u>

8 0
3 years ago
The in-situ dry density of a sand is 1.72Mg/m3. The maximum and minimum drydensities, determined by standard laboratory tests, a
Stells [14]

Answer:

Relative density = 0.7 or 70%

Explanation:

The following information was provided by this question

Pd = 1.72mg/mg³

Pd max = 1.81 mg/mg³

Pd min = 1.54 mg/mg³

We substitute into the formula. This formula is contained in the attachment.

[(1/1.54)-(1/1.72)]/[1/1.54 - 1/1.81]

= 0.649350 - 0.581395 / 0.649350 - 0.552486

= 0.067955/0.096864

= 0.7015

= 0.7

The relative density is Therefore 0.7 or 70% when converted to percentage

8 0
3 years ago
SMAW and GMAW use constant current arc welding machines.<br><br> True<br><br> False
Oksi-84 [34.3K]
It is true. Good luck
4 0
3 years ago
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