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WITCHER [35]
3 years ago
13

Please select ALL THAT APPLY

Physics
1 answer:
miss Akunina [59]3 years ago
6 0

Answer:

B, C, D

Explanation:

Considering the United States is the only country that does not used the metric system; France, England, and Canada all use the metric system.

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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
Which of the following is an example of a Cnidarian?
siniylev [52]
The answer to this is C jellyfish i am positive 

hope this helps
5 0
3 years ago
Read 2 more answers
PLEASE HELP it takes tsooo long! lol ima fail! please help! thanks!
Natasha_Volkova [10]

Answer:

A and c, hope i helped xx

Explanation:

4 0
3 years ago
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Why does a green leaf look green in our eyes
mamaluj [8]
They look green because of the “special pair” of chlorophyll molecules.
3 0
3 years ago
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1.Suppose someone pulls a cart up a ramp a distance of 85cm along the ramp with a force of 15N.
Drupady [299]

1. 12.75 J

Assuming that the force applied is parallel to the ramp, so it is parallel to the displacement of the cart, the work done by the force is

W=Fd

where

F = 15 N is the magnitude of the force

d = 85 cm = 0.85 m is the displacement of the cart

Substituting in the formula, we get

W=(15 N)(0.85 m)=12.75 J


2. 10.6 N

In this part, the cart reaches the same vertical height as in part A. This means that the same work has been done (because the work done is equal to the gain in gravitational potential energy of the object: but if the vertical height reached is the same, then the gain in gravitational potential energy is the same, so the work done must be the same).

Therefore, the work done is

W=Fd=12.75 J

However, in this case the displacement is

d = 120 cm = 1.20 m

Therefore, the magnitude of the force in this case is

F=\frac{W}{d}=\frac{12.75 J}{1.20 m}=10.6 N

3 0
3 years ago
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