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2 years ago

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Wave traveling at 330 m/sec has a wavelength of 4.3 meters. What is the frequency of this wave?

1 answer:

Rasek [7]2 years ago

6 0

Answer:

76.74 Hz

Explanation:

Given:

Wave velocity ( v ) = 330 m / sec

wavelength ( λ ) = 4.3 m

We have to calculate Frequency ( f ):

We know:

v = λ / t [ f = 1 / t ]

v = λ f

= > f = v / λ

Putting values here we get:

= > f = 330 / 4.3 Hz

= > f = 3300 / 43 Hz

= > f = 76.74 Hz

Hence, frequency of sound is 76.74 Hz.

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The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m

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Answer:

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Explanation:

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Working a little the equation, we can take:

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And this equation will give us the distance between the charges. Taking the values of the problem

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$d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}$

$d = \sqrt{ 28,462,500 \ m^2}}$

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3 years ago

Read 2 more answers

A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl

kondaur [170]

Answer:

Part a)

$y_m = 0.157 mm$

part b)

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

Explanation:

As we know that the speed of wave in string is given by

$v = \sqrt{\frac{T}{m/L}}$

so we have

$T = 17.5 N$

$m/L = 5.4 g/cm = 0.54 kg/m$

now we have

$v = \sqrt{\frac{17.5}{0.54}}$

$v = 5.69 m/s$

now we have

Part a)

$y_m$ = amplitude of wave

$y_m = 0.157 mm$

part b)

$k = \frac{\omega}{v}$

here we know that

$\omega = 2\pi f$

$\omega = 2\pi(92.2) = 579.3 rad/s$

so we have

$k = \frac{579.3}{5.69}$

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

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3 years ago