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Dmitry [639]
2 years ago
11

Reception of signals from a radio facility, located off the airway being flown, may be inadequate at the designated mea to ident

ify the fix. In this case, which altitude is designated for the fix?
Engineering
1 answer:
lakkis [162]2 years ago
3 0

The altitude ensures acceptable navigational signal coverage only within 22 NM of a VOR.

<h3>What is altitude?</h3>

Altitude or height exists as distance measurement, usually in the vertical or "up" approach, between a reference datum and a point or object. The exact meaning and reference datum change according to the context.

The MOCA exists in the lower published altitude in effect between fixes on VOR airways, off-airway routes, or route segments that satisfy obstacle support conditions for the whole route segment. This altitude also ensures acceptable navigational signal coverage only within 22 NM of a VOR.

The altitude ensures acceptable navigational signal coverage only within 22 NM of a VOR.

Therefore, the correct answer is 22 NM of a VOR.

To learn more about altitudes refer to:

brainly.com/question/1159693

#SPJ4

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You were hoping to receive a scholarship for volleyball and are entering your personal information and statistics on a secure we
kozerog [31]
I think it is fair to say 5 and 3/4 inches.
6 0
3 years ago
Read 2 more answers
For each of the following combinations of parameters, determine if the material is a low-loss dielectric, a quasi-conductor, or
Alborosie

Answer:

Glass: Low-Loss dielectric

  α = 8.42*10^-11 Np/m

  β = 468.3 rad/m

  λ = 1.34 cm

  up = 1.34*10^8 m/s

  ηc = 168.5 Ω

Tissue: Quasi-Conductor

  α = 9.75 Np/m

  β = 12.16 rad/m

  λ = 51.69 cm

  up = 0.52*10^8 m/s

  ηc = 39.54 + j 31.72 Ω        

Wood: Good conductor

  α = 6.3*10^-4 Np/m

  β = 6.3*10^-4 Np/m

  λ = 10 km

  up = 0.1*10^8 m/s

  ηc = 6.28*( 1 + j )

Explanation:

Given:

Glass with µr = 1, εr = 5, and σ = 10−12 S/m at 10 GHz

Animal tissue with µr = 1, εr = 12, and σ = 0.3 S/m at 100 MHz.

Wood with µr = 1, εr = 3, and σ = 10−4 S/m at 1 kHz

Find:

Determine if  the material is a low-loss dielectric, a quasi-conductor, or a good conductor, and then  calculate α, β, λ, up, and ηc:

Solution:

- We need to determine the loss tangent to determine category of the medium as follows:

                                σ / w*εr*εo

Where, w is the angular speed of wave

            εo is the permittivity of free space = 10^-9 / 36*pi

- Now we classify as follows:

    Glass = \frac{10^-^1^2 }{2*\pi * 10*10^9 * \frac{5*10^-^9}{36\pi } } = 3.6*10^-^1^3\\\\Tissue = \frac{0.3 }{2*\pi * 100*10^6 * \frac{12*10^-^9}{36\pi } } = 4.5\\\\Wood = \frac{10^-^4 }{2*\pi * 1*10^3 * \frac{3*10^-^9}{36\pi } } = 600\\  

- For σ / w*εr*εo < 0.01 --- Low-Loss dielectric and σ / w*εr*εo > 100 --- Good conducting material.

    Glass: Low-Loss dielectric

    Tissue: Quasi-Conductor

    Wood: Good conductor

- Now we will use categorized material base equations from Table 17-1 as follows:

     Glass: Low-Loss dielectric

          α = (σ / 2)*sqrt(u / εr*εo) = (10^-12 / 2)*sqrt( 4*pi*10^-7/5*8.85*10^-12)

          α = 8.42*10^-11 Np/m

          β = w*sqrt (u*εr*εo) = 2pi*10^10*sqrt (4*pi*10^-7*5*8.85*10^-12)

          β = 468.3 rad/m

          λ = 2*pi / β = 2*pi / 468.3

          λ = 1.34 cm

          up = λ*f = 0.0134*10^10

          up = 1.34*10^8 m/s

          ηc = sqrt ( u / εr*εo ) = sqrt( 4*pi*10^-7/12*8.85*10^-12)

          ηc = 168.5 Ω

     Tissue: Quasi-Conductor

          α = (σ / 2)*sqrt(u / εr*εo) = (0.3 / 2)*sqrt( 4*pi*10^-7/12*8.85*10^-12)

          α = 9.75 Np/m

          β = w*sqrt (u*εr*εo) = 2pi*100*10^6*sqrt (4*pi*10^-7*12*8.85*10^-12)

          β = 12.16 rad/m

          λ = 2*pi / β = 2*pi / 12.16

          λ = 51.69 cm

          up = λ*f = 0.5169*100*10^6

          up = 0.52*10^8 m/s

          ηc = sqrt ( u / εr*εo )*( 1 - j (σ / w*εr*εo))^-0.5

          ηc = sqrt (4*pi*10^-7*12*8.85*10^-12)*( 1 - j 4.5)^-0.5

          ηc = 39.54 + j 31.72 Ω

     Wood: Good conductor

          α = sqrt (pi*f*σ u) = sqrt( pi* 10^3 *4*pi* 10^-7 * 10^-4 )

          β = α = 6.3*10^-4 Np/m

          λ = 2*pi / β = 2*pi / 6.3*10^-4

          λ = 10 km

          up = λ*f = 10,000*1*10^3

          up = 0.1*10^8 m/s

          ηc = α*( 1 + j ) / б = 6.3*10^-4*( 1 + j ) / 10^-4

          ηc = 6.28*( 1 + j )

         

           

         

8 0
3 years ago
This test should be performed on all cord sets, receptacles that aren't part of a building or structure's permanent wiring, and
vova2212 [387]

Answer:

A continuity test

Explanation:

A continuity test is used to verified that current will flow in an electrical circuit, it performed by placing a small voltage across the chosen path. continuity test ensure that the equipment grounding conductor is electrically continuous and this test is perform on all the cord sets, receptacles that aren't part of a building or structure's permanent wiring, and cord-and-plug connected equipment required to be grounded. example of equipment used in testing current flow in continuity test are Analog multi-meter, voltage/continuity tester etc.

Continuity test and terminal connection test are the two test required by OSHA on all electrical equipment

8 0
3 years ago
Describe the refrigeration cycle &amp; how it transfers heat, including all the components &amp; lines. be sure to discuss why t
Amanda [17]

Heat can be diverted away from the area you want to cool using the refrigeration cycle, also known as a heat pump cycle. This is done by repeatedly compressing and expanding the working refrigerant (air, water, synthetic refrigerants, etc.) to change its pressure.

<h3>What is the cycle of a heat pump?</h3>
  • In order to heat the air within a building, a heat pump warms the air outside by drawing heat from it and adding it to it. The following procedure is used to achieve this: A gas is created when liquid refrigerant absorbs heat from the ambient air in the "evaporator."
  • The pressure buildup from the compressor raises the temperature of the refrigerant, just like it does with everything that is put under pressure. The refrigerant is a heated vapor, ranging in temperature from 120° to 140°F, when it exits the compressor.
  • The chemical refrigerant, once transformed from a gas to a liquid state, has a remarkable capacity to absorb heat. The heat inside your house might be absorbed and released outside when the refrigerant undergoes a state shift as a result of compression and pressurization.  

To learn more about Compressor refer to:

brainly.com/question/29345949

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6 0
1 year ago
____accelerates the rusting process A lubricant B improper storage CMoisture DOil​
Dmitriy789 [7]

Answer:

i would go with-B

Explanation:

because  Exposure to the elements has resulted in the formation of the blue-green patina

6 0
3 years ago
Read 2 more answers
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