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Misha Larkins [42]
3 years ago
15

A bicycle gear wheel is a disc with 50 ‘teeth' equally spaced around its edge, as shown. The gear wheel is rotated 10 times each

second. A springy strip of metal is vibrated by the rotating ‘teeth'. The metal strip produces a sound of frequency that is equal to the frequency of vibration of the strip
Physics
1 answer:
Soloha48 [4]3 years ago
5 0

Answer:

<h2>0.66m</h2>

Explanation:

<em>The question is not complete, here is the complete question, also see attached the image.</em>

<em>"A bicycle gear wheel is a disc with 50 ‘teeth’ equally spaced around its edge, as shown. The gear  wheel is rotated 10 times each second. A springy strip of metal is vibrated by the rotating ‘teeth’.  The metal strip produces a sound of frequency that is equal to the frequency of vibration of the  strip. The speed of sound in air is 330ms–1.  What is the wavelength of the emitted sound?"</em>

given data

number of teeth= 50        

we are told that the gear is rotated 10 times in 1 sec,  

hence the frequency of the rotation past the strip is

=50*10  

=500Hz    

speed of sound= 330m/s

we know that

  v_w = f \Lambda-----------1

where Vw is the speed of sound,

f is its frequency, and

λ is its wavelength.

λ=Vw/f

λ=330/500

λ=0.66m

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Answer:

Explanation:

Given

When m_1 mass is attached to the spring its frequency is f_1=12\ Hz

when another mass m_2 is attached to m_1 , frequency changes to f_2=4\ Hz

frequency of spring mass system is given by

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for m_1

f_1=\frac{1}{2\pi }\sqrt{\frac{k}{m_1}}

12=\frac{1}{2\pi }\sqrt{\frac{k}{m_1}}-----1

for m_1 and m_2

f_2=\frac{1}{2\pi }\sqrt{\frac{k}{m_1+m_2}}

4=\frac{1}{2\pi }\sqrt{\frac{k}{m_1+m_2}}----2

divide 1 and 2

3=\sqrt{\frac{m_1+m_2}{m_1}}

squaring

9=1+\frac{m_2}{m_1}

\frac{m_2}{m_1}=8

   

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A nonconducting sphere is made of two layers. The innermost section has a radius of 6.0 cm and a uniform charge density of −5.0C
rosijanka [135]

Answer:

The total charge on the sphere is 46.11 x 10⁻³ C

Explanation:

Given:

charge density of innermost section, σ = −5.0C/m³

radius of the innermost section, r = 6.0cm

charge density of the outer layer ,σ = +8.0 C/m³

radius of the outer layer, r =12.0cm

volume of sphere is given as = \frac{4}{3}.\pi  r^3

Charge enclosed by the innermost section = volume x charge density

volume = \frac{4}{3}\pi  r^3 = \frac{4}{3}\pi  (0.06^3) = 9.05  X 10^{-4} m^3

Enclosed charge, q₁ = −5.0C/m³ X 9.05 x 10⁻⁴ m³

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Charge at the outer surface

Volume =  \frac{4}{3}\pi  [r_2{^3} - r_1{^3}] = \frac{4}{3}\pi  [0.12{^3} - 0.06{^3}] = 0.00633 {m^3}

Enclosed charge, q₂ = +8.0 C/m³ X 0.00633 m³

                            = 50.64 x 10⁻³ C

Total charge on the sphere; Q = q₁ + q₂

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                                                   =  46.11 x 10⁻³ C

Therefore, the total charge on the sphere is 46.11 x 10⁻³ C

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3 years ago
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