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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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Why do the quadrants in coordinate plane go anti-clockwise?.
tia_tia [17]

Answer:

Quadrants are counter-clockwise because angles are measured counter-clockwise; and angles are measured counter-clockwise so that Cross Product of unit vector in X direction with that in the Y direction has to be the unit vector in the Z direction (coming towards us from the origin).

Explanation:

7 0
2 years ago
A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur
Amanda [17]

Question in order:

A silicon carbide plate fractures in bending when a blunt load was applied to the plate center. The distance between the fracture origin and the mirror/mist boundary on the fracture surface was 0.796 mm. To determine the stress used to break the plate, three samples of the same material were tested and produced the following. What is the estimate of the stress present at the time of fracture for the original plate?

Mirror Radius (mm) Bending Failure Stress (MPa)

0.603                         225

0.203                         368

0.162                         442

Answer:

191 MPa

Explanation:

Failure stress of bending is Inversely proportional to the mirror radius

Bending Stress = \frac{1}{(Mirror Radius)^{n}}

At mirror radius 1 = 0.603 mm   Bending stress 1 = 225 Mpa..............(1)

At mirror radius 2 = 0.203 mm  Bending stress 2 = 368 Mpa...............(2)

At mirror radius 3 = 0.162 mm   Bending stress 3 = 442 Mpa...............(3)

comparing case 1 and 2 using the above equation

\frac{Stress 1}{Stress 2} = ({\frac{Radius 2}{Radius 1}})^{n_1}

\frac{225}{368} = ({\frac{0.203}{0.603}})^{n_1}

0.6114 = (0.3366)^{n_1}

Taking the natural logarithm of both side

ln(0.6114) = n ln(0.3366)

n₁ = ln(0.6114)/ln(0.3366)

n₁ = 0.452

comparing case 2 and 3 using the above equation

\frac{Stress 2}{Stress 3} = ({\frac{Radius 3}{Radius 2}})^{n_2}

\frac{368}{442} = ({\frac{0.162}{0.203}})^{n_2}

0.8326 = (0.7980)^{n_2}

Taking the natural logarithm of both side

ln(0.8326) = n ln(0.7980)

n₂ = ln(0.8326)/ln(0.7980)

n₂ = 0.821

comparing case 1 and 3 using the above equation

\frac{Stress 1}{Stress 3} = ({\frac{Radius 3}{Radius 1}})^{n_3}

\frac{225}{442} = ({\frac{0.162}{0.603}})^{n_3}

0.5090 = (0.2687)^{n_3}

Taking the natural logarithm of both side

ln(0.5090) = n ln(0.2687)

n₃ = ln(0.5090)/ln(0.2687)

n₃ = 0.514

average for n

n = \frac{n_1 + n_2 + n_3}{3}

n = \frac{0.452 +0.821 + 0.514}{3}

n = 0.596

Hence to get bending stress x at mirror radius 0.796

\frac{Stress x}{Stress 3} = ({\frac{Radius 3}{Radius x}})^{0.596}

\frac{Stress x}{225} = ({\frac{0.603}{0.796}})^{0.596}

\frac{Stress x}{225} = 0.8475

stress x = 191 MPa

3 0
3 years ago
A structural component in the shape of a flat plate 25.0 mm thick is to be fabricated from a metal alloy for which the yield str
balandron [24]

Answer:

The critical length of surface flaw = 6.176 mm

Explanation:

Given data-

Plane strain fracture toughness Kc = 29.6 MPa-m1/2

Yield Strength = 545 MPa

Design stress. =0.3 × yield strength

= 0.3 × 545

= 163.5 MPa

Dimensionless parameter. Y = 1.3

The critical length of surface flaw is given by

= 1/pi.(Plane strain fracture toughness /Dimensionless parameter× Design Stress)^2

Now putting values in above equation we get,

= 1/3.14( 29.6 / 1.3 × 163.5)^2

=6.176 × 10^-3 m

=6.176 mm

5 0
3 years ago
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Answer:

the surface heat-transfer coefficient due to natural convection during the initial cooling period.  = 4.93 w/m²k

Explanation:

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A segment of four-lane freeway (two lanes in each direction) has a 3% upgrade that is 1500 ft long followed by a 1000-ft 4% upgr
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Answer:

The level of service of of compound grade freeway is LOSB.

Explanation:

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3 0
3 years ago
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