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TiliK225 [7]
3 years ago
9

The length of stereocilia actually vary from 10 to 50 micrometers. Again, assuming that they behave like simple pendula, over wh

at frequency range of sound waves would they resonate. (The actual frequency range of human hearing is 20 Hz 20,000 Hz, so might there be other mechanisms involved and/or might the pendular model be rather oversimplified?)
A. About 70 Hz -160 Hz.
B. About 440 Hz - 1000 Hz.
C. About 20 Hz - 50 Hz.
D. About 0.07 to 0.16 Hz.
Physics
1 answer:
Mila [183]3 years ago
6 0

To solve this problem it is necessary to apply the concepts related to the period based on variables such as gravity, distance and frequency.

By definition, know that the Period is

T_1 = 2\pi \sqrt{\frac{L}{g}}

Where,

L = Length

g = Gravity

At the same time, frequency can be defined as,

f_1 = \frac{1}{T_1}

So using this for 10\mu m we have that,

T_1 = 2\pi \sqrt{\frac{L}{g}}

T_1 = 2\pi \sqrt{\frac{10*10^{-6}}{9.8}}

T_1 = 0.635*10^{-2}s

Then the frequency is

f_1 = \frac{1}{T_1}

f_1 = \frac{1}{0.635*10^{-2}}

f_1 = 157.6\approx 160Hz

For the second length of 50\mu m we have that

T_1 = 2\pi \sqrt{\frac{L}{g}}

T_1 = 2\pi \sqrt{\frac{5*10^{-5}}{9.8}}

T_1 = 1.4*10^{-2}s

Then the frequency is

f_1 = \frac{1}{T_1}

f_1 = \frac{1}{1.4*10^{-2}}

f_1 = 70Hz

Therefore the correct answer is A.

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Answer:

q = 8.57 10⁻⁵ mC

Explanation:

For this exercise let's use Newton's second law

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the bold are vectors, if we write the module of this expression we have

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the acceleration of the particle is centripetal

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we substitute

        qvB = m v² / r

         qBr = m v

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The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use

          v = d / t

the distance is ¼ of the circle,

          d = \frac{1}{4} \  2\pi  r

           d =\frac{\pi }{2r}

we substitute

           v =  \frac{\pi  r}{2t}

           r = \frac{2 \ t  \ v}{\pi }

           

let's calculate

           r =\frac{2 \ 2.2  \ 10^{-3} \ 88}{\pi } 2 2.2 10-3 88 /πpi

           r = 123.25 m

         

let's substitute the values

           q = \frac{ 7.2 \ 10^{-8} \ 88}{ 0.6 \ 123.25}7.2 10-8 88 / 0.6 123.25

            q = 8.57 10⁻⁸ C

Let's reduce to mC

           q = 8.57 10⁻⁸ C (10³ mC / 1C)

           q = 8.57 10⁻⁵ mC

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Answer:

The value is T_t =  2.5659 \  s    

Explanation:

From the we are told that  

         The initial speed of the object is  u =  8 \  m/s

         The greatest height it reached is  h  =  15 \  m

Generally from kinematic equation we have that

      v^2 =  u^2 + 2gH

At maximum height v  =  0 m/s

So

      0^2 =  8^2 + 2 *  - 9.8 *  H

=>    H  =  3.27 \  m

Here H is the height from the initial height to the maximum height

So the initial height is mathematically represented as  

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Generally the time taken for the object to reach maximum height is mathematically evaluated using kinematic equation as follows

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=>         t = 0.8163 \  s

Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows

       h  =  ut_1 + \frac{1}{2}  gt_1^2

Here the initial velocity is  0 m/s given that its the velocity at maximum height

Also  g is positive because we are moving in the direction of gravity  

So

       15  =  0* t  +  4.9 t^2

=>      t_1  =  1.7496

Generally the total time taken is mathematically represented as

          T_t =  t + t_1

=>        T_t =  0.8163  +   1.7496

=>        T_t =  2.5659 \  s            

 

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