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3 years ago

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Sound waves are a longitudinal wave that have a speed of about 340 m/s when traveling through room temperature air. What is the

wavelength of a sound wave that has a frequency of 6,191 Hz?

1 answer:

AlexFokin [52]3 years ago

8 0

The wavelength of the wave is 0.055 m

Explanation:

The relationship between speed, frequency and wavelength of a wave is given by the wave equation:

$v=f\lambda$

where

v is the speed

f is the frequency

$\lambda$ is the wavelength

For the sound wave in this problem we have

v = 340 m/s is the speed

f = 6,191 Hz is the frequency

Solving for $\lambda$ , we find the wavelength:

$\lambda=\frac{v}{f}=\frac{340}{6191}=0.055 m$

Learn more about waves and wavelength:

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Answer:

D) resistance, resistivity

Explanation:

Resistance is a physical quantity that indicates the opposition of an object to conduct electricity, this quantity depends on different factors such as temperature, material, object length, among other things. The resistance of two objects of the same material may be different, because it depends on the specifications of the object.

On the other hand, resistivity is a more general quantity, since it is assigned to materials and depends only on the nature of the material and its temperature.

So the resistivity is related to the meterial rather than the object.

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Explanation:

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$\dfrac{1}{R}=\dfrac{1}{150}+\dfrac{1}{150}+\dfrac{1}{150}\\\\R=50\ \Omega$

5. Three resistors of 20 ohms, 40 ohms and 100 ohms are connected in parallel, So,

$\dfrac{1}{R}=\dfrac{1}{20}+\dfrac{1}{40}+\dfrac{1}{100}\\\\=11.76\ \Omega$

Hence, this is the required solution.

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A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit

Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

$P = \frac{1}{2} \mu \omega^2 A^2 v$

Here,

$\mu$ = Linear mass density of the string

$\omega =$ Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

$v = \sqrt{\frac{T}{\mu}} \rightarrow$ Here T is the Period

For the linear mass density we have that

$\mu = \frac{m}{l}$

And the angular frequency can be written as

$\omega = 2\pi f$

Replacing this terms and the first equation we have that

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})$

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})$

$P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})$

PART A ) Replacing our values here we have that

$P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.2320W$

PART B) The new amplitude A' that is half ot the wavelength of the wave is

$A' = \frac{1.8*10^{-3}}{2}$

$A' = 0.9*10^{-3}$

Replacing at the equation of power we have that

$P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.058W$

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2 years ago