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DedPeter [7]
3 years ago
13

Explain when single-stranded or multistranded wire should be used.

Engineering
1 answer:
lisabon 2012 [21]3 years ago
8 0
Single stranded wires are best suited for products that won't encounter much movement;this type of wiring is often only used in smaller gauge wiring applications as it can be difficult to maneuver and utilize a heavy gauge, single conductor wire.
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You must create a Low Pass Filter for an audio amplifier. You must pass 10khz and block
Katena32 [7]

Answer:

) You must create a Low Pass Filter for an audio amplifier. You must pass 10khz and block

60khz. In designing this, you must choose a cutoff frequency exactly half way between

these two frequencies. In addition, you must use a 0.2uF capacitor in you LPF circuit.

Given these criteria, analyze this circuit, and determine the necessary resistor value in

Ohms. Find the output Voltage at both ends of the spectrum. Based on your results, is

this a good design? NOTE: You MUST show your work (when using MS Word, choose

“Insert”--- “Equation”). (6 marks)

a. Solve for fc

b. Solve for R

c. Solve for Vout at 10khz

d. Solve for Vout at 60khz

e. Good design> (yes or no with an explanation)

Explanation:

6 0
3 years ago
For some metal alloy, the following engineering stresses produce the corresponding engineering plastic strains prior to necking.
kirza4 [7]

Answer:

203.0160

Explanation:

Because you add then subtract then multiply buy 7 the subtract then divide then you add that to the other numbers you got than boom

7 0
2 years ago
A well-insulated rigid tank contains 5 kg of a saturated liquid–vapor mixture of water at 200 kPa. Initially, three-quarters of
vlabodo [156]

Answer:

S_{gen} = 18.519\,\frac{kJ}{K}

Explanation:

Given that rigid tank is a closed system, the following model is constructed after the First Law of Thermodynamics:

W_{heater} + U_{sys,1} - U_{sys,2} = 0

W_{heater} = U_{sys,2} - U_{sys,1}

W_{heater} = m\cdot (u_{2}-u_{1})

The entropy generation inside the rigid tank is determined by appropriate application of the Second Law of Thermodynamics:

S_{sys,1} - S_{sys,2} + S_{gen} = 0

S_{gen} = S_{sys,2} - S_{sys,1}

S_{gen} = m\cdot (s_{2}-s_{1})

The properties of the steam are obtained from steam tables:

Intial State

P = 200\,kPa

T = 120.21\,^{\textdegree}C

\nu = 0.2222\,\frac{m^{3}}{kg}

u = 1010.7\,\frac{kJ}{kg}

s = 2.9294\,\frac{kJ}{kg\cdot K}

x = 0.25

Final State

P = 869.567\,kPa

T = 173.88\,^{\textdegree} C

\nu = 0.2222\,\frac{m^{3}}{kg}

u = 2578.6\,\frac{kJ}{kg}

s = 6.6332\,\frac{kJ}{kg\cdot K}

x = 1.00

The entropy change of the steam during the process is:

S_{gen} = (5\,kg)\cdot \left(6.6332\,\frac{kJ}{kg\cdot K} - 2.9294\,\frac{kJ}{kg\cdot K} \right)

S_{gen} = 18.519\,\frac{kJ}{K}

8 0
4 years ago
The reason for the nature of the signal's value is: Not Linear Humans are Continuous The stock market closes once per day, and s
Serga [27]

Answer:

Explanation:

* Daily close of the stock market

  • The nature of the signal in value is <u>Discrete</u>, as the stock market closes once per day,and so is discrete time.
  • And the nature of the signal in time is continuous as the company are continuous.
8 0
3 years ago
Purely resistive loads of 24kW, 18kW and 12kW are connected between the neutral and the red, yellow and the blue respectively of
seraphim [82]

Answer:

The phase current in each line conductor are;

I_{R} = 100.17 < 0A

I_{Y} = 75.13< - 120A

I_{B} = 50.08

Explanation:

Given the following data;

Red phase = 24kW,

Yellow phase = 18kW

Blue phase = 12kW

Line voltage = 415V

For a star connected system, we have;

Phase voltage (V_{p} ) = \frac{Line voltage}{\sqrt{3}}

Phase voltage (V_{p} ) = \frac{415}{\sqrt{3}}

Phase voltage (V_{p} ) = 239.6V

The phase sequence for RYB is given by;

V_{R} = 239.6

Phase current (I) = \frac{Phase power}{Phase voltage}

Hence, I = \frac{P}{V}

<em>For the Red phase;</em>

I_{R} = \frac{24000}{239.6

I_{R} = 100.17 < 0A

<em>For the Yellow phase;</em>

I_{Y} = \frac{18000}{239.6

I_{Y} = 75.13< - 120A

<em>For the Blue phase;</em>

I_{B} = \frac{12000}{239.6

I_{B} = 50.08

For the line neutral;

I_{N} =\sqrt{ (I_{R}^{2} +I_{Y}^{2}+I_{B}^{2}-I_{R}I_{Y}-I_{Y}I_{B}-I_{R}I_{B}

Substituting we have, I_{N} = 43.29A

6 0
3 years ago
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