Which of newton’s laws best explains why motorists should buckle up?

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

Any magnetic properties occur _____________

lyudmila [28]

Answer:

wen you stick to mangnetits togater

Explanation:

7 0

3 years ago

Read 2 more answers

A uniform disk with a mass of 5.0 kg and diameter 30 cm rotates on a frictionless fixed axis through its center and perpendicula

igomit [66]

Answer:

Angular acceleration of the disk will be $\alpha =10.714rad/sec^2$

Explanation:

We have given mass of the disk m = 5 kg

Diameter of the disk d = 30 cm = 0.3 m

So radius $r=\frac{d}{2}=\frac{0.3}{2}=0.15m$

Moment of inertia of disk is given by $I=\frac{1}{2}mr^2=\frac{1}{2}\times 5\times 0.15^2=0.056kgm^2$

Force is given by F=4 N

Torque is given as $\tau =Fr=4\times 0.15=0.6N-m$

We also know that torque is given by $\tau =I\alpha$

$0.6=0.056\times \alpha$

$\alpha =10.714rad/sec^2$

5 0

3 years ago

Plz help

katrin2010 [14]

The particles of the medium (slinky in this case) move up and down (choice #2) in a transverse wave scenario.

This is the defining characteristic of transverse waves, like particles on the surface of water while a wave travels on it, or like particles in a slack rope when someone sends a wave through by giving it a jolt.

The other kind of waves is longitudinal, where the particles of the medium move "left-and-right" along the direction of the wave propagation. In the case of the slinky, this would be achieved by giving a tensioned slinky an "inward" jolt. You would see that such a jolt would give rise to a longitudinal wave traveling along the length of the tensioned slinky. Another example of longitudinal waves are sound waves.

4 0

3 years ago

A ray of light traveling through air strikes a piece of diamond at an angle of incidence equal to 56 degrees. Calculate the angu

Montano1993 [528]

Answer:

The angle of separation is $\Delta \theta = 0.93 ^o$