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Flauer [41]
3 years ago
6

Check my answer(there’s one above) I really just chose my best guess on this.for the ones below, can you provide a simple explan

ation and the answer? Please help me...
What is the length of a stopped pipe that has a fundamental frequency of 153 Hz when the speed of sound is 340 m/s?
28 cm
43 cm
56 cm
2.1 cm
17 cm
For an open pipe of length L, what are the possible wavelengths of a standing wave? (Let n = a time positive integer)
2L/3n
Ln
L/n
2L/n
A standing wave of the third harmonic is induced in a stopped pipe. The speed of sound through the air of the pipe is 340 m/s, and the frequency of the wave is 235 Hz. What is the length of the pipe?
133 cm
189 cm
109 cm
124 cm
WARNING IF YOU ANSWER JUST FOR FREE POINTS, I WILL REPORT YOU AND THEY WILL BE TAKEN AWAY.

Physics
1 answer:
almond37 [142]3 years ago
3 0

#1

For closed pipe the fundamental frequency will be given as

f = \frac{v}{4L}

153 = \frac{340}{4L}

L = 0.56 m

L = 55 cm

#2

For open pipe the wavelength for Nth harmonic will be given as

n \times \frac{\lambda}{2} = L

\lambda = \frac{2L}{n}

#3

for a closed pipe the frequency is given as

f = \frac{nv}{4L}

here for third harmonic n = 3

f = \frac{3v}{4L}

now plug in all values

235 = \frac{3\times 340}{4L}

L = 1.09 m

L = 109 cm

#10

since in the given figure there are three complete loops

so in this case the <em>number of harmonics are THREE</em>

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Ishani and John now try a problem involving a charging capacitor. An uncharged capacitor with C = 6.81 μF and a resistor with R
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Answer:

Q=81.72\times10^{-6}C

I=2.1\times10^{-5}A

Explanation:

The maximum charge on the capacitor will be, at the end of the process, given by the formula (and for our values):

Q=CV=(6.81\times10^{-6}F)(12V)=81.72\times10^{-6}C

The maximum current on the resistor will be, at the beginning of the process, given by the formula (and for our values):

I=\frac{V}{R}=\frac{12V}{5.8\times10^{5}\Omega}=2.1\times10^{-5}A

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2 years ago
Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou
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Answer:

a. The station is rotating at 1.496 \frac{rev}{min}

b. the rotation needed is 2.8502 \frac{rev}{min}

Explanation:

We know that the centripetal acceleration is

a_{c}= \omega ^2 r

where \omega is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is <u>NOT INERTIAL,</u> the centrifugal force is a fictitious force, the real force is the centripetal).

<h3>a. </h3>

The rotational speed  is :

2.7 \frac{m}{s^2} = \omega ^2 * 110  \ m

\omega ^2 = \frac{2.7 \frac{m}{s^2}} {110 \ m}

\omega  = \sqrt{ 0.02454 \frac{rad^2}{s^2} }

\omega  = 0.1567 \frac{rad}{s}

Knowing that there are 2\pi \ rad in a revolution and 60 seconds in a minute.

\omega  = 0.1567 \frac{rad}{s}  \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}

\omega  = 1.496 \frac{rev}{min}

<h3>b. </h3>

The rotational speed needed is :

9.8 \frac{m}{s^2} = \omega ^2 * 110  \ m

\omega ^2 = \frac{9.8 \frac{m}{s^2}} {110 \ m}

\omega  = \sqrt{ 0.08909 \frac{rad^2}{s^2} }

\omega  = 0.2985 \frac{rad}{s}

Knowing that there are 2\pi \ rad in a revolution and 60 seconds in a minute.

\omega  = 0.2985 \frac{rev}{min}  \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}

\omega  = 2.8502 \frac{rev}{min}

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