What is the only "power" unit of measurement in light energy?

An electromagnetic flowmeter is useful when it is desirable not to interrupt the system in which the fluid is flowing (e.g. for

AURORKA [14]

Answer:

2 m/s

Explanation:

The electromagnetic flow-metre work on the principle of electromagnetic induction. The induced voltage is given as

$E = Blv$

where $E$ is the induced voltage = 2.88 mV = 2.88 x 10^-3 V

$l$ is the distance between the electrodes in this field which is equivalent to the diameter of the tube = 1.2 cm = 1.2 x 10^-2 m

$v$ is the velocity of the fluid through the field = ?

$B$ is the magnetic field = 0.120 T

substituting, we have

2.88 x 10^-3 = 0.120 x 1.2 x 10^-2 x $v$

2.88 x 10^-3 = 1.44 x 10^-3 x $v$

$v$ = 2.88/1.44 = 2 m/s

8 0

3 years ago

If we cannot see the whole society what can we see

creativ13 [48]

I believe it's People interacting. I hope i've helped you! (:

8 0

2 years ago

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

2 years ago

Which statement is true for a sound wave entering an area of warmer air

Reika [66]

That waves travel faster than the wave lenght!

8 0

3 years ago

Read 2 more answers

A car is making a 40 mi trip. It travels the first half of the total distance 20.0 mi at 18.00 mph and the last half of the tota

sukhopar [10]

Answer: The average speed is 27,24 mph (exactly 1008/37 mph)

Explanation:

This is solved using a three rule: We know the speeds and the distances, what we can obtain from it is the time used. It is done like this:

1h--->18mi

X ---->20 mi, then X=20mi*1h/18mi= 10/9 h=1,111 h

1h--->56mi

X ---->20 mi, then X=20mi*1h/56mi= 5/14 h=0,35714 h

Then the average speed is calculated by taking into account that it was traveled 40mi and the time used was 185/126 h=1,468 h and since speed is distance over time we get the answer. Average speed= 40mi/(185/126 h)=1008/37 mph=27,24 mph.

5 0

3 years ago