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serious [3.7K]
2 years ago
7

The end of a stopped pipe is to be cut off so that the pipe will be open. If the stopped pipe has a total length L, what fractio

n of L should be cut off so that the fundamental mode of the resulting open pipe has the same frequency as the fifth harmonic (and=5) of the original stopped pipe?
Physics
1 answer:
Alexxandr [17]2 years ago
7 0

Answer:

4/10 of L.

Explanation:

A stopped pipe is a pipe with one closed end and one opened end. it is also called a closed pipe.

The fundamental mode of a stopped pipe is also called its fundamental frequency, and is f₁=v/4L.

Where f₁=fundamental frequency, v= velocity of sound, L= Length of pipe.

The fifth harmonic of the stopped pipe f₅ =5v/4L .................(1)

For an open pipe,

Fundamental  mode is also called fundamental frequency f₁₀=v/2l₀ .......... (2)

Where f₁₀ = fundamental frequency of a closed pipe, v= velocity of sound and l₀=length of the resulting open pipe.

from the question, the fundamental mode of the resulting open pipe = The fifth harmonic of the original stopped pipe.

∴ f₅=f₁₀

⇒5v/4L = v/2l₀

Equating v from both side of the equation,

⇒ 5/4L = 1/2l₀

Cross multiplying the equation,

5×2l₀ = 4L× 1

10l₀ = 4L

Dividing both side of the equation by the coefficient of l₀ i.e 10

10l₀/10 = 4L/10

∴ l₀ = 4/10(L)

∴ 4/10 of L must be cut off

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Driving your Ferrari through the Italian countryside at a speedy 88 m/s, you approach an opera diva singing a high C (1,046 Hz).
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Answer:

You will hear the note E₆

Explanation:

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The Doppler effect says that:

f' = \frac{v \pm v0 }{v \mp vs}*f

Where:

f = original frequency

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v = velocity of the sound wave

v0 = your velocity

vs = velocity of the source, in this case, the source is the diva, we assume that she does not move, so vs = 0.

Replacing the values that we know in the equation we have:

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This frequency is close to the note E₆ (1,318.5 Hz)

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2 years ago
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2 years ago
When you urinate, you increase pressure in your bladder to produce the flow. For an elephant, gravity does the work. An elephant
weqwewe [10]

Answer:

a) v =  1.19 m / s , b)   P₁ = 0.922 10⁵ Pa

Explanation:

1) Let's use the fluid continuity equation

       Q = A v

The area of ​​a circle is

      A = π r2 = π d²/4

     

     v = Q / A = Q 4 / pi d²

     v = 0.006 4/π 0.08²

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2) write Bernoulli's equation, where point 1 is the bladder and point 2 is the urine exit point

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The exercise tell us

P₂ = 1.0013 105 Pa

v₁ = 0

y₁ = 1 m

y₂=0  

Rho (water) = 1000 kg / m³

      P₁ + rho y₁ = P₂ + ½ rho v₂²

      P₁ = P₂ + ½ rho v₂² - rho g y₁

      P₁ = 1.013 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1

      P₁ = 1.013 10⁵ +708.5  - 9800

      P₁ =  92208.5Pa

      P₁ = 0.922 10⁵ Pa

8 0
1 year ago
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