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3 years ago

12

Consider a transition at 5000 Å with a width of 1 Å and a cavity 2 cm3 in volume. How many electromagnetic modes exist in this f

requency band for this cavity?

1 answer:

Lena [83]3 years ago

3 0

Answer:

total number of modes is 8

Explanation:

attached here is the calculations

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When is the electric flux on a section of a closed surface positive?.

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When the electric field pierces outward through the section.

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2 years ago

Below are three graphs. The shape of graph 3 shows that the current flowing through this component is __________ proportional to

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The relationship between variables might be either directly porportional to each other or inversely proportional to each other. <em>For instance, the current flowing is _</em><em><u>directly</u></em><em>_ </em><em>proportional to the potential difference across</em> it.

---------------------

Since I do not have the graph, I will propose two potential options and I will describe each of them.

I suggest you to choose the one that matches the shape of graph 3.

<h3>Graphs</h3>

Probably your graph is composed of the X and Y axes, perpendicularly placed, in a way in which four quadrants are formed.

You might see that one of the axes represents the current (I), and the other one represents the potential difference (V).

And a straight line that represents the relationship between the current and the potential difference.

The direction of the straight line reflects the relationship between variables.

The change on each variable is proportional to the change on the other variable. Now we have to analyze how that change occurs.

<h3 /><h3>Option 1 </h3>

<em>The current flowing through this component is __</em><u><em>DIRECTLY</em></u><em>__ proportional to the potential difference across it</em>.

If this is the case of your graph, you should see the straight line crossing from the left inferior quadrant to the right superior one.

The direction of the line suggests that one of the variables increases as the other one increases too. <em>The current increases as the potential increases. </em>

<h3> </h3>

<em>This is the case of the </em><em>Ohm's law.</em><em> </em>

<h3 /><h3 /><h3>Option 2 </h3>

<em>The current flowing through this component is __</em><u><em>INVERSELY</em></u><u><em>_</em></u><em>_ proportional to the potential difference across it.</em>

If this is the case of your graph, you should see the straight line crossing the right quadrant from the superior left corner to the inferior right one.

The direction of the line suggests that the variable represented in the Y ax decreases as the other one increases.

For instance, The current decreases as the potential increases. Or the potential decreases as the current increases.

In the attached files you will find graphs of both options.

-------------------------

You can learn more about the Ohm's law at

brainly.com/question/17286882?referrer=searchResults

You can learn more about the relationship between variables at

brainly.com/question/16937826?referrer=searchResults

brainly.com/question/21627823?referrer=searchResults

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2 years ago

A subway train is traveling at 22.2 m/s when it approaches a slower train 50m ahead traveling in the same direction at 6.94 m/s.

Amiraneli [1.4K]

Answer:

Time that they collide = 4.99s

Relative speed of the trains when they collide: The relative speed of The first train relative to the second, slower train at collision = 4.781 m/s

Explanation:

We will use the equations of motion to obtain the solution required

At time t = 0

speed of first train = 22.2 m/s

Initial space between the two trains = 50 m

Speed of second train = 6.94 m/s

For the first car, distance covered by the first train = y

y = distance covered between the beginning of the deceleration and the point where the the two trains hit one another.

u = initial velocity = 22.2 m/s

t = time taken for all this to happen

a = deceleration = - 2.1 m/s²

y = ut + (1/2)at²

y = 22.2t - 1.05t² (eqn 1)

For the second train,

At t = 0, y = 50 m

Let the new distance moved by the second train before collision = (y - 50)

u = initial velocity = 6.94 m/s

t = time taken = t

a = acceleration of the second train = 0 m/s² (constant velocity)

(y - 50) = ut + (1/2)at²

y - 50 = 6.94t

y = 6.94t + 50 (eqn 2)

substituting for y in eqn 2 using the expression obtained in eqn 1

y = 22.2t - 1.05t²

y = 6.94t + 50

22.2t - 1.05t² = 6.94t + 50

1.05t² - 15.26t + 50 = 0

Solving this quadratic equation

t = 4.99 s or 9.54 s

The position of the two trains are the same at those two times, but the first time is when they hit each other.

t = 4.99 s

At 4.99 s, the the velocity of the first train

v = u + at

v = 22.2 + (-2.1×4.99) = 11.721 m/s in the same direction as the second train.

Relative velocity at this point will be

= 11.721 - 6.94 = 4.781 m/s

Relative speed of the trains when they collide: The relative speed of The first train relative to the second, slower train at collision = 4.781 m/s

Hope this Helps!!!

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2 years ago

Which of the following biomes might be found at a latitude of 33° South?

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The answer is going be desert.

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