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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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A circular test track for cars has a circumference of 3.5 kmkm . A car travels around the track from the southernmost point to t
elixir [45]

Answer:

(A) Distance will be equal to 1.75 km

(B) Displacement will be equal to 1.114 km

Explanation:

We have given circumference of the circular track = 3.5 km

Circumference is given by 2\pi r=3.5

r = 0.557 km

(a) It is given that car travels from southernmost point to the northernmost point.

For this car have to travel the distance equal to semi perimeter of the circular track

So distance will be equal to =\frac{3.5}{2}=1.75km

(b) If car go along the diameter of the circular track then it will also go from southernmost point to the northernmost point. and it will be equal to diameter of the track

So displacement will be equal to d = 2×0.557 = 1.114 m

8 0
3 years ago
FIRST ANSWER IS BRAINIEST TAKE YOUR TIME I NEED RIGHT ANSWERS FIRST ANSWER IS BRAINIEST ANSWER TAKE YOUR TIME NEED RIGHT ANSWER
V125BC [204]
A = 7
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C = 5
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4 0
3 years ago
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a car company wants to ensure its newest model can stop in 50 m when traveling at 30 m/s (which is about 108 km/h). if we assume
Brut [27]

A decrease in velocity is referred to as deceleration. If car is moving at 30 m/s and stop in 50 m .The value of deceleration is 11.56 ms−2.

<h3>How to calculate deceleration ?</h3>

While acceleration is motion in which an object's speed varies every second, deceleration is motion that causes an object to slow down.

We are aware that acceleration refers to an object's rate of increase in speed, and deceleration refers to an object's rate of decrease in speed. For instance, when we apply the brakes while driving, we benefit from the vehicle's ability to decelerate and slow down.

The Deceleration Formula is the final velocity minus the initial velocity, with a negative sign in the result because the velocity is decreasing, if starting velocity, final velocity, and time taken are given.

velocity of car = 30 m/s

car need to stop in 50m

Deceleration a = v^2 –  u^2 / 2s

                          = 0^2 - 50^2 / 2*30

                          = 11.56

Deceleration of the care = 11.56 ms−2

To learn more about deceleration refer :

brainly.com/question/75351

#SPJ4

8 0
1 year ago
According to newtons law of action reaction what is the most likely to occur if to ice skaters with approximately the same mass
Free_Kalibri [48]

Answer:

They would keep on moving but unless being acted upon or stop slowly because of the friction

Explanation:

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3 years ago
A baseball pitcher throws a ball at 40 ms^-1. if the acceleration is approximately constant over a distance of 2 m, how large is
Reil [10]

We have the equation of motion v^2=u^2+2as, where v i the final velocity, u is the initial velocity, a is the acceleration and s is the displacement

Here final velocity, v = 40m/s

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Substituting 40^2=0^2+2*a*2\\ \\ a=400m/s^2

So the baseball pitcher accelerates at 400m/s^2 to release a ball at 40 m/s.

4 0
3 years ago
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