To make an even better electrical junction, what should you do?

A room has a width of 14.1 feet, a length of 15.5 feet, and a ceiling height of 12.0 ft. The average flow rate for this room's a

prohojiy [21]

Answer:

Your question lacks the time required hence i will calculate the Average flow rate using a general concept and an assumed time value of 25 seconds

ANSWER : 104.904 ft^3/sec

Explanation:

General concept : Average flow rate is the volume of fluid per unit time through an area

Hence the average flow rate of the air conditioning unit of this room

Volume of the room / time taken for the air to cycle the room = v / t

assuming the time taken = 25 seconds

volume of room = width * length * height

= 14.1 * 15.5 * 12 = 2622.6 ft^3

Average flow rate = V/ t

= 2622.6 / 25 = 104.904 ft^3/sec

8 0

3 years ago

Design an Armstrong indirect FM modulator to generate an FM signal with a carrier frequency 98.1 MHz and a frequency deviation △

Bess [88]

Answer:

See explaination

Explanation:

In the Armstrong method of FM generation, the phase of the carrier is directly modulated in the combing network through summation, generating indirect frequency modulation.

Very high frequency stability is achieved through Armstrong method since the crystal oscillator is used as carrier frequency generator.

Please kindly check attachment for the step by step solution of the given problem.

3 0

4 years ago

A motor with an operating resistance of 32 22 is connected to a

Akimi4 [234]

quizlet can help you its very helpful on quizes it some time mostly have the same excact quizes

4 0

3 years ago

A sign structure on the NJ Turnpike is to be designed to resist a wind force that produces a moment of 25 k-ft in one direction.

Jlenok [28]

Solution :

Finding the cohesion of the soil(c) using the relation:

$c = \frac{q_u}{2}$

Here, $q_u$ is the unconfined compression strength of the soil;

$c = \frac{800}{2}$

= 400 psf

∴ The cohesion value is greater than 0

So the use of the angle of internal friction is 0

Referring to the table relation between bearing capacity factors and angle of internal friction.

For the angle of inter friction $0^\circ$

$N_c = 5.14$

$N_q = 1.0$

$N_r = 0$

Therefore,

$q_{ult} = (400 \times 5.14 )+(110 \times 3 \times 1.0)+(0.5 \times 100 \times 13 \times 0)$

= 2386 psf

∴ Allowable bearing capacity $q_{a} = \frac{Q_{allow}}{A}$

$=\frac{30}{B^2}$

∴ $q_a = \frac{q_{ult}}{F.O.S}$

$\frac{30}{B^2} = \frac{2386}{3}$

∴ B = 0.2 ft

Therefore, the dimension of the square footing is 0.2 ft x 0.2 ft

$=0.04 \ ft^2$

7 0

3 years ago

Using the data in the photo write the complex waveform expression

UNO [17]

Answer:

1st Harmonic:

$v(t) = 50\cos(2000\pi t)$

3rd Harmonic:

$v(t) = 9\cos(6000\pi t)$

5th Harmonic:

$v(t) = 6\cos(10000\pi t)$

7th Harmonic:

$v(t) = 2\cos(14000\pi t)$

Explanation:

The general form to represent a complex sinusoidal waveform is given by

$v(t) = A\cos(2\pi f t + \phi)$

Where A is the amplitude in volts of the sinusoidal waveform

Where f is the frequency in cycles per second (Hz) of the sinusoidal waveform

Where $\phi$ is the phase angle in radians of the sinusoidal waveform.

1st Harmonic:

We have A = 50, f = 1000 and φ = 0

$v(t) = 50\cos(2\pi 1000 t + 0) \\\\v(t) = 50\cos(2000\pi t)$

3rd Harmonic:

We have A = 9, f = 3000 and φ = 0

$v(t) = 9\cos(2\pi 3000 t + 0) \\\\v(t) = 9\cos(6000\pi t)$

5th Harmonic:

We have A = 6, f = 5000 and φ = 0

$v(t) = 6\cos(2\pi 5000 t + 0) \\\\v(t) = 6\cos(10000\pi t)$

7th Harmonic:

We have A = 2, f = 7000 and φ = 0

$v(t) = 2\cos(2\pi 7000 t + 0) \\\\v(t) = 2\cos(14000\pi t)$

Note: The even-numbered harmonics have 0 amplitude that is why they are not shown here.

8 0

4 years ago

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