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vfiekz [6]
3 years ago
8

A microwave transmitter has an output of 0.1 W at 2 GHz. Assume that this transmitter is used in a microwave communication syste

m where the transmitting and receiving antennas are parabolas, each 1.2 m in diameter. a. What is the gain of each antenna in decibels
Engineering
1 answer:
ehidna [41]3 years ago
5 0

Answer:

The gain of each antenna in decibels  is 353 .33 dB.

Explanation:

Given:

Frequency = 2 GHz

Diameter  =  1.2 m

To Find:

The gain of each antenna in decibels = ?

Solution:

Relationship between antenna gain and effective area

= \frac{4 \pi f^2 A_e}{c^2}-----------------------------------(1)

Where

f is frequency

A_e  is  effective area

c is the speed of light

A_e = effective area of a parabolic antenna with a face area of A is 0.56A

A_e =  0.56(\pi r^2)

r  is the radius

r = \frac{1.2}{2}

r =  0.6

A_e = 0.56(\pi (0.6)^2)

A_e = 0.56( 0.36 \pi)

A_e = 0.2016 \pi

Substituting the values in eq(1)

= \frac{4 \pi (2 \times 10^9)^2 \times 0.2016\pi}{(3 \times 10^8)^2}

= \frac{4 \pi (4 \times 10^{18} ) \times 0.2016\pi}{(9 \times 10^{16})}

= \frac{4 \pi (4 \times 10^{2} ) \times 0.2016\pi}{9}

= \frac{31.80 \times 10^{2}}{9}

= 353 .33 dB

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The inlet and exhaust flow processes are not included in the analysis of the Otto cycle. How do these processes affect the Otto
lara31 [8.8K]

Answer:

Suction and exhaust processes do not affect the performance of Otto cycle.

Explanation:

Step1

Inlet and exhaust flow processes are not including in the Otto cycle because the effect and nature of both the process are same in opposite direction.

Step2

Inlet process or the suction process is the process of suction of working fluid inside the cylinder. The suction process is the constant pressure process. The exhaust process is the process of exhaust out at constant pressure.

Step3

The suction and exhaust process have same work and heat in opposite direction. So, net effect of suction and exhaust processes cancels out. The suction and exhaust processes are shown below in P-V diagram of Otto cycle:

Process 0-1 is suction process and process 1-0 is exhaust process.

7 0
3 years ago
The advantage of using rose bud tips is that they:
Jlenok [28]

Answer:

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Explanation:

Hope this helped Mark BRAINLIEST!!

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2 years ago
A gas stream flowing at 1000 cfm with a particulate loading of 400 gr/ft3 discharges from a certain industrial plant through an
Makovka662 [10]

<u>Solution and Explanation:</u>

Volume of gas stream = 1000 cfm (Cubic Feet per Minute)

Particulate loading = 400 gr/ft3 (Grain/cubic feet)

1 gr/ft3 = 0.00220462 lb/ft3

Total weight of particulate matter = 1000 \mathrm{cfm} \times 400 \mathrm{gr} / \mathrm{tt} 3 \times .000142857 \mathrm{lb} / \mathrm{ft} 3 \times 60=3428.568 \mathrm{lb} / \mathrm{hr}

Cyclone is to 80 % efficient

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emissions from this stack be limited to = 10.0 lb/hr

Particles to be remaining after wet scrubber = 10.0 lb/hr

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4 0
3 years ago
Which system of linear inequalities is represented by the graph? y &gt; x – 2 and y x + 1 y x + 1 y &gt; x – 2 and y &lt; x + 1
KatRina [158]

Answer:

The graph representing the linear inequalities is attached below.

Explanation:

The inequalities given are :

y>x-2   and y<x+1

For tables for values of x and y and get coordinates to plot for both equation.

In the first equation;

y>x-2

y=x-2

y-x = -2

The table will be :

x    y

-2  -4

-1    -3

0     -2

1      -1

2      0

The coordinates to plot are : (-2,-4) , (-1,-3), (0,-2), (1,-1) ,(2,0)

Use a dotted line and shade the part right hand side of the line.

Do the same for the second inequality equation and plot then shade the part satisfying the inequality.

The graph attached shows results.

5 0
3 years ago
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neonofarm [45]

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3 years ago
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