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schepotkina [342]
2 years 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]2 years 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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Part F - Example: Finding Two Forces (Part I)
Temka [501]

Answer: F=28.936 kg/m s^{2}

Explanation:

According to the given information (and figure attached), the block with mass m=10 kg has the following forces acting on it:

In the X component:

F cos(30\°) - F_{s}=0 (1)

Where:

F is the applied force directed 30\° above the horizontal

F_{s}=\mu_{s} N (2) is the force of static friction (which is equal to the coefficient of static friction \mu_{s}=0.3 and the Normal force N

In the Y component:

F sin(30\°) + N - W=0 (3)

Where W=m.g is the weight (the force of gravity) which is proportional to the multiplication of the mass m and gravity g=9.8 m/s^{2}  

Let’s begin by combining (1) and (2):

F cos(30\°) - \mu_{s} N=0 (4)

Isolating N from (3):

N=mg – F sin(30\°) (5)

Substituting (5) in (4):

F cos(30\°) - \mu_{s} (mg – F sin(30\°))=0 (6)

F cos(30\°) - \mu_{s} mg + \mu_{s} F sin(30\°))=0  

((cos(30\°) +\mu_{s} sin(30\°)) F - \mu_{s}mg =0  

Isolating F:

F=\frac{\mu_{s}mg}{(cos(30\°) +\mu_{s} sin(30\°)} (7)

F=\frac{(0.3)(10 kg)(9.8 m/s^{2})}{(cos(30\°) + 0.3 sin(30\°)}  

Finally:

F=28.936 N=8.936 kgm/s^{2} (8) This is the necessary force to overcome static friction and move the block

We can prove it by finding F_{s} and verifying it is less than F:

Substituting (8) in (1):

8.936 kgm/s^{2}cos(30\°) - F_{s}=0 (9)

F_{s}=25.059 kgm/s^{2} (10) This is the static friction force

As we can see F_{s} < F

8 0
2 years ago
A 30-m-long rocket train car is traveling from Los Angeles to New York at 0.5c when a light at the center of the car flashes. Wh
Novosadov [1.4K]

Given that,

Distance =30 m

speed = 0.5c

(A). We need to find the bell and siren simultaneous events for a passenger seated in the car

According to given data

The distance travelled by the light to reach either side of the rocket  train car is same.

So, The two events are simultaneous and the bell and siren are the simultaneous events for a passenger seated in the car.

(B). We need to calculate time interval between the events

Using formula of time dilation

\Delta t=\dfrac{\Delta t'}{\sqrt{1-\dfrac{v^2}{c^2}}}.....(I)

Where, \delta t' = proper time

\delta t = time interval between the events

The time interval between the events measured in a reference frame

The proper time in this case is

\Delta t'=\Delta t_{1}-\dfrac{v\Delta x}{c^2}

For the second interval,

Put the value of \Delta t' in the equation (I)

\Delta t_{2}=\dfrac{\Delta t_{1}-\dfrac{v\Delta x}{c^2}}{\sqrt{1-\dfrac{v^2}{c^2}}}

Put the value in the equation

\Delta t_{2} = \dfrac{0-\dfrac{0.5c\times30}{c^2}}{\sqrt{1-\dfrac{0.5^2c^2}{c^2}}}

\Delta t_{2}=\dfrac{-15}{3\times10^{8}\sqrt{1-0.25}}

\Delta t_{2}=-5.77\times10^{8}\ s

Negative sign shows the siren rings before the bell ring.

Hence, (A). Yes, the bell and siren are simultaneous events.

(B). The siren sounds before the bell rings.

8 0
2 years ago
Look at the two question marks between zinc (Zn) and arsenic (As). At the time, no elements were known
marissa [1.9K]

Answer:

Mendeleev predicted the atomic mass of each element along with compounds they each should form.

Explanation:

Based on other elements in the same group he predicted the existence of eka-aluminum and eka-silicon, later to be named gallium (Ga) and germanium (Ge).

6 0
2 years ago
The speed of sound in air is 320 ms-1 and in water it is 1600 ms-1. It takes 2.5 s for sound to reach a certain distance from th
Nonamiya [84]

Answer:

Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.

Explanation:

Given:

Speed of sound in air = 320 m/s

Speed of sound in water = 1600 m/s

Time taken to reach certain distance in air = 2.5 sec

a.

We have to find the distance traveled by sound in air.

Distance = Product of speed and time.

⇒ Distance = Speed\times time\ taken

⇒ Distance = 320\times 2.5

⇒ Distance = 800 meters.

b.

Now we have to find how much time the sound will take to travel in water.

⇒ Time = Ratio of distance and speed

⇒ Time =\frac{distance}{speed}

⇒ Time =\frac{800}{1600}   <em>   ...distance = 800 m and speed = 1600 m/s</em>

⇒ Time =\frac{1}{2}

⇒ Time =0.5 seconds.

Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.

7 0
2 years ago
The weight of spaceman Speff at the surface of planet X, solely due to its gravitational pull, is 389 N. If he moves to a distan
miv72 [106K]

Answer:

mass of the planet X = 5.6 × 10²³ kg.

Explanation:

According to Newtons law of universal gravitation,

F = GM₁M₂/r²

Where F = gravitational force, M₁ = mass of the speff, M₂ = mass of the planet X, G = gravitational constant r = distance between the speff and the planet X

making M₂ The subject of the equation above,

M₂ = Fr²/GM₁ .......................... equation 2

Where F = 24.31 N, r = 1.08×18⁴km ⇒( convert to m ) =1.08 × 10⁴  × 1000 m

r = 1.08  × 10⁷ m, G = 6.67  × 10 ⁻¹¹ Nm²/kg², M₁ = 75 kg

Substituting this values in equation 2,

M₂ = 24.13(1.08  × 10⁷ )²/75( 6.67  × 10 ⁻¹¹)

M₂ = 24.13 × 1.17 × 10¹⁴/500.25 × 10⁻¹¹

M₂ = (28.23 × 10¹⁴)/(500.25 × 10⁻¹¹)

M₂ = 0.056 × 10²⁵

M₂ = 5.6 × 10²³ kg.

Therefore mass of the planet X = 5.6 × 10²³ kg.

8 0
2 years ago
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