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satela [25.4K]
3 years ago
15

How do you suppose the frequency of an electromagnetic wave compares with the frequency of the electrons it sets into oscillatio

n in a receiving antenna?
Physics
1 answer:
FromTheMoon [43]3 years ago
6 0

Answer:

The frequencies are the same.

Explanation:

This is similar to the photoelectric effect. The frequency of the electromagnetic wave has to match a threshold frequency before it can set electrons into oscillation in a receiving antenna. After the frequency of the electromagnetic waves have matched and exceeded this threshold frequency, all of it's energy is converted to the kinetic energy of the electrons it sets into oscillation. And since their energies are similar, the frequencies too, subsequently, will be the same.

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What do the key results indicate?
Svetradugi [14.3K]

Answer:

I can answer it but wats the experiment exactly?

7 0
3 years ago
Read 2 more answers
Two people are trying to push a large box across a floor. Person 1 pushes with a force of 15 N to the right, while person 2 push
ozzi

Answer:

35 N to the right

Explanation:

When calculating net force when both forces are on the same side you add them when they are going against each other you subtract them.

6 0
3 years ago
Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as
MAXImum [283]

Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m

or in kilometers,

h_2 = 47 km

the first altitude in kilometers is

h_1 = 155 km

so the difference between the two altitudes is

\Delta h = 155 km - 47 km = 108 km

8 0
4 years ago
What is the momentum of a 0.145 kg baseball traveling at 40.0 m/s
dimulka [17.4K]
Answer: 5.8 kg m/s

Identify the given information.
m = 0.145 kg
v = 40.0 m/s

Choose which formula you should use. In this case, you are finding momentum.
p = mv, where p is momentum.

Substitute and solve.
p = 0.145 kg x 40.0 m/s
p = 5.8 kg m/s
7 0
3 years ago
Relate the output of energy from a heat engine to the energy put into the heat engine considering the second law of thermodynami
Vinil7 [7]

<u>Answer: </u>

<em>Considering the II law of thermodynamics</em>

<em>From the figure</em>

<em>Out put of energy: </em>

Heat supplied from the source/ reservoir  (Q₁) - Heat rejected to the surroundings from the system (Q) = Q₁ - Q₂. Also known as Net work done on the system.

<em>Input of energy: </em>

Amount of heat energy supplied to the system from the source (Q₁ ).

Efficiency (H.E) = η = Output÷ Input

                             η  = (Q₁ - Q₂) ÷ Q₁

                      OR η = Wnet ÷ Q₁ ;        since Wnet = (Q₁ - Q₂)


3 0
3 years ago
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