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Maslowich
3 years ago
13

Voices of swimmers at a pool travel 400 m/s through the air and 1,600 m/s underwater. The wavelength changes from 2 m in the air

to 8 m underwater. How has the change in media affected the frequency of the wave? It increased. It decreased. It stayed the same. It cannot be calculated.
Physics
2 answers:
brilliants [131]3 years ago
8 0

Answer : Frequency stayed the same.

Explanation :

Voices of swimmers at a pool travel through the air at a speed of 400 m/s and 1600 m/s.

The wavelength changes from 2 m in the air to 8 m underwater.

We know that the relation between speed and the wavelength.

v=\nu\times \lambda

\nu=\dfrac{v}{\lambda}

\nu is the frequency.

\lambda is the wavelength.

Frequency through air, \nu_a=\dfrac{400\ m/s}{2\ m}=200\ Hz

Frequency through water, \nu_w=\dfrac{1600\ m/s}{8\ m}=200\ Hz

So, the frequency of voice of swimmers stayed the same.

Hence, the correct option is (C) " It stayed the same".

Nata [24]3 years ago
5 0

The frequency of a sound is whatever frequency leaves the source. It doesn't change.

Voiced of swimmers at the pool don't change frequency in or out of the water. Only their speed and wavelength change.

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Explanation:

Remark

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The plan is to first of all find C1*C2 and use this information and C1 + C2 to find the individual resistances.

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36*C2 - C2^2 =288                    Multiply by - 1

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<em> </em>

C2 = 24 or

C2 = 12

C1 = 36 - C2

C1 = 36 - 24 = 12

C1 = 36 - 12  = 24

Answer

<u><em>C1 = 12 ohms or 24 ohms</em></u>

<u><em>C2 = 24 ohms or 12 ohms</em></u>

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