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NeTakaya
1 year ago
14

a crude approximation of voice production is to consider the breathing passages and mouth to be a resonating tube closed at one

end. what is the fundamental frequency ????1 if the tube is 0.240 m long, by taking air temperature to be 37.0∘c? ????1
Physics
1 answer:
prohojiy [21]1 year ago
7 0

The fundamental frequency of the tube is 0.240 m long, by taking air temperature to be 37^oC is 367.42 Hz.

A standing wave is basically a superposition of two waves propagating opposite to each other having equal amplitude. This is the propagation in a tube.

The fundamental frequency in the tube is given by

f=\frac{v_T}{4L}

where, v_T=v\sqrt{\frac{T}{273} }

Since, T=37+273 K = 310 K

v = 331 m/s

\therefore v_T=331\sqrt{\frac{310}{273} } = 352.72 \ m/s

Using this, we get:

f=\frac{352.72}{4(0.240)} \\f=367.42 \ Hz

Hence, the fundamental frequency is 367.42 Hz.

To learn more about Attention here:

brainly.com/question/14673613

#SPJ4

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in the potential well the potential energy is zero, so completely the electron's energy is in the shape of kinetic energy: 
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Outer the potential well the potential energy is U₀, so 
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3 years ago
A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 1
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Answer:

a)  y= 3.5 10³ m, b)   t = 64 s

Explanation:

a) For this exercise we use the vertical launch kinematics equation

Stage 1

          y₁ = y₀ + v₀ t + ½ a t²

          y₁ = 0 + 0 + ½ a₁ t²

Let's calculate

         y₁ = ½ 16 10²

         y₁ = 800 m

At the end of this stage it has a speed

        v₁ = vo + a₁ t₁

        v₁ = 0 + 16 10

        v₁ = 160 m / s

Stage 2

        y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²

        y₂ = 800 + 150 5 + ½ 11 5²

        y₂ = 1092.5 m

Speed ​​is

        v₂ = v₁ + a₂ t

        v₂ = 160 + 11 5

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The rocket continues to follow until the speed reaches zero (v₃ = 0)

         v₃² = v₂² - 2 g y₃

         0 = v₂² - 2g y₃

         y₃ = v₂² / 2g

         y₃ = 215²/2 9.8

         y₃ = 2358.4 m

The total height is

          y = y₃ + y₂

          y = 2358.4 + 1092.5

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           y= 3.5 10³ m

b) Flight time is the time to go up plus the time to go down

Let's look for the time of stage 3

          v₃ = v₂ - g t₃

          v₃ = 0

          t₃ = v₂ / g

          t₃ = 215 / 9.8

          t₃ = 21.94 s

The time to climb is

          t_{s} = t₁ + t₂ + t₃

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        t_{b} = 26.54 s

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        t = t_{s} + t_{b}

        t = 36.94 + 26.54

        t =63.84 s

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