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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;

20log₁₀ $e^{-\alpha _1l_{min$ = -100dB

we substitute

20log₁₀ $e^{-(310.7np/m)l_{min$ = -100dB

$l_{min$ = 3.71 cm

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

6 0

3 years ago

How to tell if something is an electrolyte

schepotkina [342]

Answer: If it has ions, it is an electrolyte

Explanation:

Let's start by explaining that electrolytes are compounds that contain charged particles or<u> ions</u>, which can be cations (positive ions) or anions (negative ions).

So, it is this composition that makes an electrolytic material conduct electricity.

In this sense, the way to identify if a material is an electrolyte or not, is knowing whether it is composed of ions or not.

8 0

3 years ago

A heat engine running backward is called a refrigerator if its purpose is to extract heat from a cold reservoir. The same engine

Goryan [66]

Answer:

a) 2.85 kW

b) $ 432

c) $ 76.95

Explanation:

Average price of electricity = 1 $/40 MJ

Q = 20 kW

Heat energy production = 20.0 KJ/s

Coefficient of performance, K = 7

also

K=(QH)/Win

Now,

Coefficient of Performance, K = (QH)/Win = (QH)/P(in) = 20/P(in) = 7

where

P(in) is the input power

Thus,

P(in) = 20/7 = 2.85 kW

b) Cost = Energy consumed × charges

Cost = ($1/40000kWh) × (16kW × 300 × 3600s)

cost = $ 432

c) cost = (1$/40000kWh) × (2.85 kW × 200 × 3600s) = $76.95

4 0

3 years ago

This formula equation is unbalanced.

Grace [21]

The answer is: [C]: "4" .
___________________________________________________
Note: To balance this equation, the coefficient, "4", should be placed in front of the PCl₃ ; and the coefficient, "6", should be placed in front of the Cl₂ .
________________________________________________________
The balanced equation is:
__________________________________________________
P₄ (s) + 6 Cl₂ (g) <span>→ 4 </span>PCl₃ (l) .
______________________________________________________

6 0

3 years ago

Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4

Mademuasel [1]

Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

3 0

3 years ago