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Nutka1998 [239]

3 years ago

7

A stereo speaker produces a pure \"E\" tone, with a frequency of 329.6 Hz. What is the period of the sound wave produced by the

speaker? What is the wavelength of this sound wave as it travels through air with a speed of about 341 m/s? What is the wavelength of the same sound wave as it enters some water, where it has a speed of about 1480 m/s?

1 answer:

olya-2409 [2.1K]3 years ago

8 0

Answer:

0.003034 s

1.035 m

4.5 m

Explanation:

$f$ = frequency of the tone = 329.6 Hz

$T$ = Time period of the sound wave

we know that, Time period and frequency are related as

$T =\frac{1}{f}\\T =\frac{1}{329.6}\\T = 0.003034 s$

$v$ = speed of the sound in the air = 341 ms⁻¹

wavelength of the sound is given as

$\lambda =\frac{v}{f} \\\lambda =\frac{341}{329.6}\\\lambda = 1.035 m$

$v$ = speed of the sound in the water = 1480 ms⁻¹

wavelength of the sound in water is given as

$\lambda =\frac{v}{f} \\\lambda =\frac{1480}{329.6}\\\lambda = 4.5 m$

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Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

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Given that,

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Using formula of electric field

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For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

4 0

3 years ago

What does the electric field strength tell about the electric firld?

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Answer:

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The strength of electric field is defined as the force experienced by the unit positive test charge.

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Where, F is the force of attraction or repulsion between the two charges and q is the test charge on which the electric field strength is to be calculated.

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Answer:

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we said that:

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therefore, the new electric field will be

$E'=\frac{2V}{2d}=\frac{V}{d}=E$

So, the electric field is unchanged. And since the force on the particle at the center is directly proportional to the electric field:

F = qE

Then the force on the particle will stay the same.

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3 years ago