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marin [14]
3 years ago
14

Any change in the cross section of the vocal tract shifts the individual formant frequencies, the direction of the shift dependi

ng on just where the change in area falls along the standing wave. Constriction of the vocal tract at a place where the standing wave of a formant exhibits minimum-amplitude pressure oscillations generally causes the formant to drop in frequency; expansion of the tract at those same places raises the frequency. Three other major tools for changing the shape of the tract in such a way that the frequency of a particular formant is shifted in a particular direction are the jaw, the body of the tongue and the tip of the tongue. Moving the various articulatory organs in different ways changes the frequencies of the two lowest formants over a considerable range [18].
One way to increase formant frequency is to ________ the vocal tract at a place where the standing wave of a formant frequency exhibits minimum-amplitude pressure oscillations.

a. Stretch
b. Vibrate
c. Contract
d. Expand
Physics
1 answer:
Leona [35]3 years ago
3 0

Answer:

The correct answer is option D.

Explanation:

It is stated in the question that constriction of the vocal tract at a place where the standing wave of a formant exhibits minimum-amplitude pressure oscillations generally causes the formant to drop in frequency so to increase formant frequency, the vocal should expand where the standing wave of a formant exhibits minimum-amplitude pressure oscillations. The answer is D.

I hope this helps.

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The calculator display gives 4.6975 x 10¯9. Rounded off to three significant figures gives 4.70 x 10¯9 as the answer.

3) (8.41 x 106) x (5.02 x 1012)

The calculator display gives 4.2218 x 1019. Rounded off to three significant figures gives 4.22 x 1019 as the answer.

When done as a division, the answer to this problem is 1.68 x 10¯6.

4) (9.21 x 10¯4) ÷ (7.60 x 105)

The calculator display gives 1.2118 x 10¯9. Rounded off to three significant figures gives 1.21 x 10¯9 as the answer.

1) (2.68 x 10¯5) x (4.40 x 10¯8)

The calculator display gives 1.1792 x 10¯12. Rounded off to three significant figures gives 1.18 x 10¯12 as the answer.

2) (2.95 x 107) ÷ (6.28 x 1015)

The calculator display gives 4.6975 x 10¯9. Rounded off to three significant figures gives 4.70 x 10¯9 as the answer.

3) (8.41 x 106) x (5.02 x 1012)

The calculator display gives 4.2218 x 1019. Rounded off to three significant figures gives 4.22 x 1019 as the answer.

When done as a division, the answer to this problem is 1.68 x 10¯6.

4) (9.21 x 10¯4) ÷ (7.60 x 105)

The calculator display gives 1.2118 x 10¯9. Rounded off to three significant figures gives 1.21 x 10¯9 as the answer

7 0
3 years ago
Light is incident on the left face of an isosceles prism; with an apex angle of 49o, such that the light exiting the right face
Sunny_sXe [5.5K]

Answer:

\mu = 1.645

Explanation:

By Snell's law we know at the left surface

\theta_i = 19^o

\theta_r = ?

\mu_1 = 1

\mu_2 = \mu

now we have

1 sin19 = \mu sin\theta_r

0.33 = \mu sin\theta_r

now on the other surface we know that

angle of incidence = \theta_r'

\theta_e = 90

so again we have

\mu sin\theta_r' = 1 sin90

so we have

\theta_r = sin^{-1}\frac{0.33}{\mu}

\theta_r' = sin^{-1}\frac{1}{\mu}

also we know that

\theta_r + \theta_r' = 49

sin^{-1}\frac{0.33}{\mu} + sin^{-1}\frac{1}{\mu} = 49

By solving above equation we have

\mu = 1.645

3 0
3 years ago
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olga_2 [115]

Answer:

250 m/s

Explanation:

4 0
3 years ago
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DedPeter [7]

Answer:

22

Explanation:

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Fudgin [204]
Answer: B) 2.5 m/s

Explanation: Find the average of the time and distance, and see how far they go in only 1 second.


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7.2 divided by 3 = 2.4

Therefore, the answer is technically 2.4m/s
4 0
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