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maks197457 [2]
4 years ago
8

A 40-cmcm-long tube has a 40-cmcm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. A

s the insert is slowly pulled out, the sound from the tuning fork creates standing waves in the tube when the total length LL is 42.5 cmcm , 58.5 cmcm
What is the frequency of the tuning fork? Assume velocity of sound = 343 m/s
Physics
1 answer:
Licemer1 [7]4 years ago
7 0

Answer:

1070 Hz

Explanation:

First, I should point out there might be a typo in the question or the question has inconsistent values. If the tube is 40 cm long, standing waves cannot be produced at 42.5 cm and 58.5 cm lengths. I assume the length is more than the value in the question then. Under this assumption, we proceed as below:

The insert in the tube creates a closed pipe with one end open and the other closed. For a closed pipe, the difference between successive resonances is a half wavelength \frac{\lambda}{2}.

Hence, we have

\dfrac{\lambda}{2}=58.5-42.5=16 \text{ cm}

\lambda=32\text{ cm}=0.32 \text{ m}.

The speed of a wave is the product of its wavelength and its frequency.

v=f\lambda

f=\dfrac{v}{\lambda}

f=\dfrac{343}{0.32}=1070 \text{ Hz}

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3 years ago
Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00
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Answer:

2.62898\times 10^{-6}\ C/m^3

1979.99974\ N/C

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

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r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C

Volume charge density is given by

\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3

The volume charge density for the sphere is 2.62898\times 10^{-6}\ C/m^3

E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C

The magnitude of the electric field is 1979.99974\ N/C

8 0
3 years ago
When the wedges are removed, the cars will move. Predict which direction they will move and when they will stop
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Answer: hello your question lacks the required diagram attached below is the required diagram

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Explanation:

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The cars will later come to a stop due to frictional forces between the cars and the surface.

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3 years ago
Compute the expected shell-model quadrupole moment of 209Bi () and compare with the experimental value, - 0.37 b
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Answer:

0.22 b

Explanation:

Quadrupole moment of the nucleon is,

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R^{2}=R^{2} _{0}A^{\frac{2}{3} }

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Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2} _{0}A^{\frac{2}{3} }

For Bismuth j=\frac{9}{2} and A is 209.

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Therefore, the expected value of quadrupole is 0.22 b which is quite related with experimental value which is 0.37 b

3 0
3 years ago
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