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cestrela7 [59]
2 years ago
11

Pls help this is due today.

Physics
1 answer:
bixtya [17]2 years ago
6 0

Answer:

11 moments docx has the answer

Explanation:

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How radio wave carry information?​
Allisa [31]

Answer:

At the sending end, the information to be sent, in the form of a time-varying electrical signal, is applied to a radio transmitter. ... The radio waves carry the information to the receiver location.

Explanation:

5 0
3 years ago
Please help me there​
frozen [14]

Answer:

when switch is off no electricity will flow and then the circuit is called an open circuit

Explanation:

Electricity will not flow in open circuit

7 0
1 year ago
Which situations might cause two observers (A and B) to measure different frequencies for the same vibrating object? Select the
Alex787 [66]

We want to explain why two different observes may measure different frequencies for the same vibrating object.

We will see that the two correct options are:

  • <em>Observer A is stationary and Observer B is moving.</em>
  • <em>Observer A and Observer B are moving at different speeds relative to each other.</em>

<em />

Let's assume that the vibrating object is a guitar string. Thus, the string makes a noise, and from that noise, we can estimate the frequency at which the string vibrates.

Now there appears a really cool effect, called the Doppler Effect. It says that the apparent change of frequency is <u>due to the motion of the observer or the source of the frequency (or both).</u>

For example, if you move towards the vibrating string, the perceived frequency will be larger, and you will hear a "higher" sound.

While if you move away from the string, the opposite happens, and you will hear a "lower" sound.

Then the only thing that impacts in how we perceive the frequency is our velocity relative to the source.

So, why do observers A and B measure different frequencies?

The two correct answers are:

  • <em>Observer A is stationary and Observer B is moving.</em>
  • <em>Observer A and Observer B are moving at different speeds relative to each other.</em>

If you want to learn more, you can read:

brainly.com/question/17107808

6 0
2 years ago
please help with questions
kaheart [24]

Answer:

Please selection one of the following

3 0
2 years ago
The amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is________
maksim [4K]

Answer:

d) 2Fr

Explanation:

We know that the work done in moving the charge from the right side to the left side in the k shell is W = ∫Fdr from r = +r to -r. F = force of attraction between nucleus and electron on k shell. F = qq'/4πε₀r² where q =charge on electron in k shell  -e and q' = charge on nucleus = +e. So, F = -e × +e/4πε₀r² = -e²/4πε₀r².

We now evaluate the integral from r = +r to -r

W = ∫Fdr

= ∫(-e²/4πε₀r²)dr

= -∫e²dr/4πε₀r²

= -e²/4πε₀∫dr/r²

= -e²/4πε₀ × -[1/r] from r = +r to -r

W = e²/4πε₀[1/-r - 1/+r] = e²/4πε₀[-2/r} = -2e²/4πε₀r.

Since F = -e²/4πε₀r², Fr = = -e²/4πε₀r² × r = = -e²/4πε₀r and 2Fr = -2e²/4πε₀r.

So W = -2e²/4πε₀r = 2Fr.

So, the amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is W = 2Fr

6 0
3 years ago
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