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3 years ago

What is the frequency of a wave?

Physics

2 answers:

kirill [66]3 years ago

7 0

The Correct answer is A, I even provided proof from my own test. Hope this helps!

Paladinen [302]3 years ago

6 0

The answer should be the number, or amount, of cycles that occur in a given time. Frequency is basically Hertz. Frequency is the number of waves that pass through a certain point every second. I hope that I was able to answer your question.
Peace out!

- Hershy103

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Which of the following relationships is most like the predator-prey relationship?

saw5 [17]

Answer:

symbiosis

Explanation:

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3 years ago

PLEASE HELP ME ASAP!!!!

Angelina_Jolie [31]

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4 years ago

What does activation energy has to do with a chemical reaction.

asambeis [7]

Explanation:

Activation energy and reaction rate

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3 years ago

An Alpha particle moving in north direction give reasons

Mazyrski [523]

Your question has been heard loud and clear.

An alpha particle , can move in any direction randomly. But with a magnetic field , we can deflect the alpha particle in any direction we want.

So , the magnetic field must be placed to the west of the alpha particle , so that the particle gets deflected and moves towards the north direction.

Thank you.

7 0

3 years ago

Read 2 more answers

Star a has an absolute magnitude of 8. what is its apparent magnitude at a distance of 100 pc

tresset_1 [31]

From the definition of apparent magnitude, we know that:
$m_{1} - m_{2} = -2.5Log( \frac{F_{1}}{F_{2}} )$
where:
m = apparent magnitude
F = corresponding flux

We also know that the flux is given by the formula:
$F = \frac{L}{4 \pi d^{2} }$
where:
L = luminosity
d = distance

Therefore:
$\frac{F_{1} }{F_{2}} = (\frac{L_{1} }{4 \pi d_{1}^{2} })(\frac{4 \pi d_{2}^{2} }{L_{2} }) \\ = \frac{L_{1}d_{2}^{2}}{L_{2}d_{1}^{2}}$

Now, let's apply these formulae to the same star (therefore, same luminosity), using apparent magnitude and absolute magnitude (which is defined as the apparent magnitude the star would have if it were at a distance of 10pc):
$m - M = 2.5 Log ( \frac{d}{10pc})^{2}$

Now, let's solve for m:
$m = M + 2.5 Log ( \frac{d}{10})^{2}$
= <span> $8 + 2.5 Log ( \frac{100}{10})^{2}$ </span>
= 13

Hence, the apparent magnitude of the star would be m = +13

3 0

4 years ago