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2 years ago

Lucy has three sources of sound that produce pure tones with wavelengths of 60cm, 100cm, and 124cm.

Physics

1 answer:

denis23 [38]2 years ago

8 0

Answer:

a) We see that the tubes of lengths 15, 45 and 75 resonate with this wavelength

b) There is resonance for the lengths 25 and 75 cm

c) Resonance occurs for tubes with length 31 and 93 cm

Explanation:

To find the length of the tube that has resonance we must find the natural frequencies of the tubes, for this at the point that the tube is closed we have a node and the open point we have a belly; in this case the fundamental wave is

λ = 4L

The next resonance called first harmonic λ₃ = 4L / 3

The next fifth harmonic resonance λ₅ = 4L / 5,

WE see that the general form is λ ₙ= 4L / n n = 1, 3, 5 ...

Let's use these expressions for our problem

Let's start with the shortest wavelength.

a) Lam = 60 cm

Let's look for the tube length that this harmonica gives

L = λ n / 4

To find the shortest tube length n = 1

L = 60 1/4

L = 15 cm

For n = 3

L = 60 3/4

L = 45 cm

For n = 5

L = 60 5/4

L = 75 cm

For n = 7

L = 60 7/4

L = 105cm

We see that the tubes of lengths 15, 45 and 75 resonate with this wavelength, in different harmonics 1, 3 and 5

.b) λ = 100 cm

For n = 1

L = 100 1/4

L = 25 cm

For n = 3

L = 100 3/4

L = 75 cm

For n = 5

L = 100 5/4

L = 125 cm

There is resonance for the lengths 25 and 75 cm in the fundamental and third ammonium frequency

c) λ= 124 cm

L = 124 1/4

L = 31 cm

For the second resonance

L = 124 3/4

L = 93 cm

Resonance occurs for tubes with length 31 and 93 cm in the fundamental harmonics and third harmonics

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A certain part of a flat screen TV has a thickness of 150 nanometers. How<br> many meters is this?

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Answer:

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2 years ago

An older camera has a lens with a focal length of 60mm and uses 34-mm-wide film to record its images. Using this camera, a photo

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Answer:

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Explanation:

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and

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Similarly, magnification of second lens

$M_2=-\frac{d'_2}{d_1}$

and

$M_2=\frac{h'_2}{h_1}$

From the above equations we get

$\frac{M_1}{M_2}=\frac{d'_1}{d_2'}$

and

$\frac{M_1}{M_2}=\frac{h'_1}{h_2'}$

which means,

$\frac{d'_1}{d_2'}=\frac{h'_1}{h_2'}$

and

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So, we get

$\frac{f_1}{f_2}=\frac{h'_1}{h_2'}\\\Rightarrow f_2=f_1\times\frac{h_2'}{h'_1}\\\Rightarrow f_2=60\times\frac{14}{34}=24.71\ mm$

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2 years ago

The position of a particle moving along the x axis may be determined from the expression x(t) = btu + ctv, where x will be in me

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now we can say it should be meter on right side also

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1 year ago

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Answer:

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Case 2:

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