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schepotkina [342]
1 year ago
12

The vocal tract is a resonator, and the transmission of a sound through an acoustic resonator is highly dependent upon frequency

. Sounds at the resonance frequency are less attenuated than other sounds, and are therefore radiated with a higher relative amplitude, or with a greater relative loudness than other sounds. The vocal tract has four or five important resonances called formants. Each formant is associated with a standing wave, that is, with a static pattern of pressure oscillations whose amplitude is at a maximum at the glottal end and near a minimum at the lip opening. The lowest formant corresponds to a quarter of a wavelength, which is to say that a quarter of its wavelength fits within the vocal tract. Similarly, the second, third and fourth formants correspond respectively to a three-quarters wavelength, one and a quarter wavelengths and one and three-quarters wavelengths. Which of the following is not a formant wavelength?
A. 3A/4.
B. A/4.
Physics
1 answer:
andriy [413]1 year ago
3 0

Answer:

Your question is not complete. this option completes your question

<em> Which of the following is not a formant wavelength?  </em>

<em>A. 3λ/4. </em>

<em>B.λ /4.</em>

<em>c.  λ/2 </em>

<em>D. 7λ/4</em>

Explanation:

A formant wavelength will be such that the total length of pipe = nλ/2 + λ/4.

When the above condition is satisfied then only a node is formed at the closed end and antinode at the open end.

Among the given options:

(a) 7λ/4 = nλ/2 + λ/4, satisfied for n = 3

(b) 3λ/4 = nλ/2 + λ/4, satisfied for n = 1

(c) λ/2  λ/2 = nλ/2 + λ/4, not satisfied for any value of n, HENCE NOT A FORMANT WAVELENGTH (ANSWER)

(d) λ/4 = nλ/2 + λ/4, satisfied for n = 0

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Two students walk in the same direction along a straight path, at a constant speed one at 0.90 m/s and the other at 1.90 m/s. a.
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Therefore the time difference is 866.66-410.52= 456.14 s

In order to calculate the distance that faster student should  walk

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Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. the heat capacity of object a is
Harman [31]
Given:
Ca = 3Cb                      (1)
where
Ca =  heat capacity of object A
Cb =  heat capacity f object B

Also,
Ta = 2Tb                     (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.

Let
Tf =  final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.

Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb)           (3)

Substitute (1) and (2) into (3).
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Define k = Ma/Mb, the ratio f the masses.
Then
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Tf = [(1+6k)/(1+3k)]*Tb

Answer:
T_{f} =( \frac{1+6k}{1+3k} )T_{b}= \frac{1}{2}( \frac{1+6k}{1+3k})T_{a}
where
k= \frac{M_{a}}{M_{b}} 
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