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DochEvi [55]
3 years ago
15

I really don’t know the answer for this

Physics
1 answer:
EleoNora [17]3 years ago
6 0
The correct answer is the reverse wave I took the test
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In an experiment, a ringing bell is placed in a vacuum jar that does not have any air in it. What best describes why the bell is
madam [21]

Answer:

Light does not need a medium to travel travel through, but since waves must have a medium to vibrate, sound is not created where no air is present.

Explanation:

3 0
3 years ago
Which is a characteristic of all ions? They are made of one type of atom. They have one overall charge. They are made of two or
Ira Lisetskai [31]

Answer:

They are made of one type of atom.

Explanation:

8 0
3 years ago
Read 2 more answers
Consider the following list of numbers. 129, 685, 125, 511, 601, 52, 46 The height of a binary search tree is the maximum number
NemiM [27]

Answer:

The height of the tree is three (3) deep

Explanation:

It's 3 deep

Under 129, comes 125 and 685;

Under 125, comes 52 : Under 685, comes 511;

Under 52, comes 46 : Under 511, is 601.

This is illustrated below.

129

∧

125,685

|,|

52,511

|,|

46,601

5 0
3 years ago
When you observe a distant galaxy whose photons have traveled for 10 billion years before reaching earth we are seeing that gal?
expeople1 [14]
In reality we don't see the galaxy we see it's reflection .. the light hits or got emitted by the star travel all the way long to hit our eyes .. we see their reflection . everything around you that you see is it's reflection
8 0
3 years ago
The aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10−5T, collide with molecules of t
ollegr [7]

Answer:

8.79*10^6 rad/s

Explanation:

To find the frequency of the circular orbit for an electron you use the following expression, for the radius of the trajectory of an electron, that travels trough a constant magnetic field:

r=\frac{mv}{qB}         (1)

r: radius of the trajectory

m: mass of the electron = 9.1*10^-31 kg

v: speed of the electron = 1.0*10^6 m/s

q: charge of the electron = 1.6*10^-19 C

B: magnitude of the magnetic field = 5.0*10^-5 T

You use the fact that the angular frequency in a circular motion is given by:

\omega=\frac{v}{r}

Then, you solve the equation (1) in order to obtain v/r:

\frac{v}{r}=\omega=\frac{qB}{m}

Finally, you replace the values of the parameters:

\omega=\frac{(1.6*10^{-19}C)(5.0*10^{-5}T)}{9.1*10^{-31}kg}\\\\\omega=8.79*10^6\frac{rad}{s}

hence, the angular frequency is 8.79*10^6 rad/s

The frequency is:

f=2\pi \omega=5.5*10^7Hz

5 0
3 years ago
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