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gayaneshka [121]

3 years ago

7

A local am radio station broadcasts at a frequency of 764 khz. calculate the energy of the frequency at which it is broadcasting

. energy = kj/photon

1 answer:

pentagon [3]3 years ago

7 0

The frequency of the emitted radio waves is:
$f=764 kHz = 7.64 \cdot 10^5 Hz$

The energy of a photon is given by:
$E=hf$
where h is the Planck constant and f is the photon frequency. Using the frequency given by the problem, we can find the energy of each photon of this radiation:
$E=hf=(6.6 \cdot 10^{-34} Js)(7.64 \cdot 10^5 Hz)=5.04 \cdot 10^{-28} J=5.04 \cdot 10^{-31} kJ$

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Sergeu [11.5K]

Answer:

The work is in the explanation.

Step-by-step explanation:

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The sine difference identity is:

$\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(a)$ .

The cosine addition identity is:

$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ .

The cosine difference identity is:

$\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$ .

We need to find a way to put some or all of these together to get:

$\sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2}$ .

So I do notice on the right hand side the $\sin(a+b)$ and the $\sin(a-b)$ .

Let's start there then.

There is a plus sign in between them so let's add those together:

$\sin(a+b)+\sin(a-b)$

$=[\sin(a+b)]+[\sin(a-b)]$

$=[\sin(a)\cos(b)+\cos(a)\sin(b)]+[\sin(a)\cos(b)-\cos(a)\sin(b)]$

There are two pairs of like terms. I will gather them together so you can see it more clearly:

$=[\sin(a)\cos(b)+\sin(a)\cos(b)]+[\cos(a)\sin(b)-\cos(a)\sin(b)]$

$=2\sin(a)\cos(b)+0$

$=2\sin(a)\cos(b)$

So this implies:

$\sin(a+b)+\sin(a-b)=2\sin(a)\cos(b)$

Divide both sides by 2:

$\frac{\sin(a+b)+\sin(a-b)}{2}=\sin(a)\cos(b)$

By the symmetric property we can write:

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snow_tiger [21]

Answer:

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arsen [322]

Answer:

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Step-by-step explanation:

Given function:

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the two equation, the x and are breaked apart..
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