A train travels 90 kilometers in 4 hours, and then 82 kilometers in 1 hours. What is its average speed?

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

On a distant planet, the GPE of a 65 kg astronaut is 11,115 j when they are on a 46 m tall cliff. What is the acceleration due t

Ainat [17]

Answer:Learn what gravitational potential energy means and how to calculate it. ... a pulley and rope, so the force due to lifting the box and the force due to gravity, ... would be used by an elevator lifting a 75 kg person through a height of 50 m if the ... When you are close to a planet you are effectively bound to the planet by gravity ..

Explanation:

3 0

3 years ago

What kind of EM waves are most dangerous to humans

zhuklara [117]

The most dangerous frequencies of electromagnetic energy are X-rays, gamma rays, ultraviolet light and microwaves. X-rays, gamma rays and UV light can damage living tissues, and microwaves can cook them. Hope this helps! =^-^=

4 0

3 years ago

A car accelerates from rest at 3.6 m/s 2 . How much time does it need to attain a speed of 5 m/s?

Olenka [21]

car starts from rest

$v_i = 0$

final speed attained by the car is

$v_f = 5 m/s$

acceleration of the car will be

$a = 3.6 m/s^2$

now the time to reach this final speed will be

$t = \frac{v_f - v_i}{a}$

$t = \frac{5 - 0}{3.6}$

$t = 1.39 s$

so it required 1.39 s to reach this final speed

6 0

3 years ago

A train pulls away from a station with a constant acceleration of 0.42 m/s2. A passenger arrives at a point next to the track 6.

Rina8888 [55]

Answer:

2.69 m/s

Explanation:

Hi!

First lets find the position of the train as a function of time as seen by the passenger when he arrives to the train station. For this state, the train is at a position x0 given by:

x0 = (1/2)(0.42m/s^2)*(6.4s)^2 = 8.6016 m

So, the position as a function of time is:

xT(t)=(1/2)(0.42m/s^2)t^2 + x0 = (1/2)(0.42m/s^2)t^2 + 8.6016 m

Now, if the passanger is moving at a constant velocity of V, his position as a fucntion of time is given by:

xP(t)=V*t

In order for the passenger to catch the train

xP(t)=xT(t)

(1/2)(0.42m/s^2)t^2 + 8.6016 m = V*t

To solve this equation for t we make use of the quadratic formula, which has real solutions whenever its determinat is grater than zero:

0≤ b^2-4*a*c = V^2 - 4 * ((1/2)(0.42m/s^2)) * 8.6016 m =V^2 - 7.22534(m/s)^2

This equation give us the minimum velocity the passenger must have in order to catch the train:

V^2 - 7.22534(m/s)^2 = 0

V^2 = 7.22534(m/s)^2

V = 2.6879 m/s

4 0

3 years ago