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MrRa [10]
2 years ago
8

Is it point A,B,C or D

Physics
1 answer:
e-lub [12.9K]2 years ago
5 0

the answer is (c) i think

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A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc
adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

f_{c_1 = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

f_{c_1 = 3 × 10¹⁰ / 2( 0.0711  )

f_{c_1 = 3 × 10¹⁰ cm/s / 0.1422  cm

f_{c_1 =  21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

so

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( (f_{c_1 / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀e^{-\alpha _1l_{min = -100dB

we substitute

20log₁₀e^{-(310.7np/m)l_{min = -100dB

l_{min = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

6 0
3 years ago
1. Predict whether the energy required to remove an electron from magnesium and potassium would be more or less than that requir
Inga [223]

In both cases less energy is required

But comparetively Mg require more energy than K

Let's see the electron configuration of Both

  • [Mg]=1s²2s²2p⁶3s²=[Ne]3s²
  • [K]=1s²2s²2p⁶3s²3p⁶4s¹=[Ar]4s¹

K has only one valence electron so very less ionization enthalpy so less energy required

Mg has 2 so more IE hence more energy required

8 0
2 years ago
Order the speed of sound through these materials from the slowest to the fastest.
Sholpan [36]

Speed of sound in cold air < Speed of sound in Warm air < Speed of sound in hot molten lead < Speed of sound in water

Explanation:

Step 1:

Speed of sound in water varies from 1450 to 1498 meters per second

Speed of sound in Hot Molten lead is approximately 1210 meters per second

Speed of sound in warm air is approximately 338.89 meters per second

Speed of sound in cold air is approximately 293.33 meters per second

Step 2:

In warm air sound travels faster than that of sound travelling nature in cold air.

∴ Speed of sound in cold air < Speed of sound in Warm air < Speed of sound in hot molten lead < Speed of sound in water

Speed of sound in cold air the slowest while Speed of sound in water is the fastest mean.

8 0
3 years ago
Calculate the frequency if the number of revolutions is 300 and the paired poles are 50.
Andrei [34K]

Answer: A

Explanation: We know that f=p*n

f=50*300=15000 Hz = 15kHz.

Have a great day! <3

3 0
2 years ago
How do I remember the Mohs hardness scale?
yanalaym [24]
It's a hardness scale from 1-10 determining how easy or hard it is to scratch the mineral.

Remember that talc (like chalk or baby powder) is the softest and easiest to scratch then diamond being 10 is the hardest mineral to scratch or break or cut

6 0
3 years ago
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