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algol [13]
2 years ago
10

Technician A says that a radio may be able to receive AM signals, but not FM signals if the antenna is defective. Technician B s

ays that a good antenna should give a reading of about 500 ohms when
tested with an ohmmeter between the center antenna wire and ground. Which technician is correct?
A. Technician A only
B. Technician B only
C. Both Technician A and Technician B
D. Neither Technician A nor Technician B
Engineering
1 answer:
DIA [1.3K]2 years ago
3 0

The response to whether the statements made by both technicians are correct is that;

D: Neither Technician A nor Technician B are correct.

<h3>Radio Antennas</h3>

In radios, antennas are the means by which signals to the sought frequency be it AM or FM are received.

Now, if the antenna is bad, it means it cannot pick any radio frequency at all and so Technician A is wrong.

Now, most commercial antennas usually come around a resistance of 60 ohms and so it is not required for a good antenna to have as much as 500 ohms resistance and so Technician B is wrong.

Read more about Antennas at; brainly.com/question/25789224

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Calculate the wire pressure for a round copper bar with an original cross-sectional area of 12.56 mm2 to a 30% reduction of area
dybincka [34]

Answer:153.76 MPa

Explanation:

Initial Area\left ( A_0\right )=12.56 mm^2

Final Area\left ( A_f\right )=0.7\times 12.56 mm^2=8.792 mm^2

Die angle=30^{\circ}

\alpha =\frac{30}{2}=15^{\circ}

\mu =0.08

Yield stress\left ( \sigma _y \right )=350 MPa

B=\mu cot\left ( \aplha\right )=0.2985

\sigma _{pressure}=\sigma _y\left [\frac{1+B}{B}\right ]\left [ 1-\frac{A_f}{A_0}\right ]^B

\sigma _{pressure}=350\left [\frac{1+0.2985}{0.2985}\right ]\left [ 1-\frac{8.792}{12.56}\right ]^{0.2985}

\sigma _{pressure}=153.76 MPa

8 0
3 years ago
Tech A says that horsepower is a measurement simply of the amount of work being performed. Tech B says that horsepower can be ca
snow_tiger [21]

Answer:

Tech B

Explanation:

Horsepower (hp) refers to a unit of measurement of power in respect of the output of engines or motors.

Horsepower is the common unit of power. It indicates the rate at which work is done.

The formula \frac{rpm*T}{5252}, where rpm is the engine speed, T is the torque, and 5,252 is radians per second.

So,

Tech B is correct

6 0
3 years ago
The tolerance for a geometric can be easily identified by reading the _________.
baherus [9]

Answer:

C

Explanation:

4 0
3 years ago
Read 2 more answers
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.6 mm; the spe
ankoles [38]

Answer:

F =  8849 N

Explanation:

Given:

Load at a given point = F =  4250 N

Support span = L = 44 mm

Radius = R = 5.6 mm

length thickness of tested material = 12 mm

First compute the flexural strength for circular cross section using the formula below:

σ_{fs} = F_{f} L / \pi  R^{3}

σ = FL / π R³

Putting the given values in the above formula:

σ = 4250 ( 44 x 10⁻³ ) / π  ( 5.6 x 10⁻³ ) ³

  = 4250 ( 44 x 10⁻³ )  / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 4250 ( 11 / 250  ) / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³

  = 187 / 3.141593 (0.0056)³

  = 338943767.745358

  = 338.943768 x 10⁶

σ = 338 x 10⁶ N/m²

Now we compute the load i.e. F from the following formula:

F_{f} = 2 σ_{fs} d³/3 L

F = 2σd³/3L

  = 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)

  = 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 676000000 ( 12  x  1/1000  )³ / 3 ( 44 x 10⁻³)

  = 676000000 (  3  / 250  )³ / 3 ( 44 x 10⁻³)

  = 676000000 (  27  / 15625000 )  / 3 ( 44 x 10⁻³)

  = 146016  / 125 / 3 ( 44 x 1 / 1000  )

  = ( 146016  / 125 ) /  (3 ( 11 /  250 ))

  =  97344  / 11

F =  8849 N

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Rosita is planning an investigation to determine how a lifeboat's shape affects its
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