Why do cars have a push start button???

Engineering

1 answer:

QveST [7]3 years ago

8 0

To be more easy for them

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An open-open organ pipe is 73.5 cm long. An open-closed pipe has a fundamental frequency equal to the third harmonic of the open

Anon25 [30]

Answer:

The length of the open closed pipe is 12.25 cm

Explanation:

In open - open organ pipe, third harmonic has Antinode to Node, Node to Node, Node to Node and Node to Antinode.

The length of the open-open organ pipe is equal to the sum of wavelength in Antinode to Node, Node to Node, Node to Node and Node to Antinode.

L = A→N + N→N + N→N + N→A

$L =\frac{\lambda }{4} + \frac{\lambda }{2} + \frac{\lambda }{2} + \frac{\lambda }{4} \\\\L = \frac{ 3\lambda }{2}\\\\\lambda = \frac{2L}{3} \\\\F = \frac{V}{\lambda} = \frac{3V}{2L}$

In open-closed pipe, Fundamental frequency has Antinode to Node.

Thus, length of the open-closed organ pipe is equal to the wavelength in Antinode to Node.

L = A→N

$L_o = \frac{\lambda}{4} \\\\\lambda =4L_o\\\\F_o = \frac{V}{4L_o}$

From the information in the question, fundamental frequency of open-closed pipe is to the third harmonic of the open-open pipe.

F₀ = F

$\frac{V}{4L_o} =\frac{3V}{2L} \\\\\frac{1}{4L_o} = \frac{3}{2*73.5 \ cm}\\\\\frac{1}{L_o} = \frac{3*4}{2*73.5 \ cm}\\\\\frac{1}{L_o} = \frac{12}{147 \ cm}\\\\L_o = \frac{147 \ cm}{12} = 12.25 \ cm$

Therefore, the length of the open closed pipe is 12.25 cm

4 0

3 years ago

Read 2 more answers

Consider a very long, slender rod. One end of the rod is attached to a base surface maintained at Tb, while the surface of the r

9966 [12]

Answer:

(a) Calculate the rod base temperature (°C). = 299.86°C

(b) Determine the rod length (mm) for the case where the ratio of the heat transfer from a finite length fin to the heat transfer from a very long fin under the same conditions is 99 percent. = 0.4325m

Explanation:

see attached file below