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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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Vinil7 [7]

Explanation:

1. Mass of the proton, m_p=1.67\times 10^{-27}\ kg

Wavelength, \lambda_p=4.23\times 10^{-12}\ m

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

\lambda=\dfrac{h}{\sqrt{2m_pq_pV}}

V=\dfrac{h^2}{2q_pm_p\lambda^2}

V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 1.67\times 10^{-27}\times (4.23\times 10^{-12})^2}

V = 45.83 volts

2. Mass of the electron, m_p=9.1\times 10^{-31}\ kg

Wavelength, \lambda_p=4.23\times 10^{-12}\ m

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

\lambda=\dfrac{h}{\sqrt{2m_eq_eV}}

V=\dfrac{h^2}{2q_em_e\lambda^2}

V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 9.1\times 10^{-31}\times (4.23\times 10^{-12})^2}

V=6.92\times 10^{34}\ V

V = 84109.27 volt

Hence, this is the required solution.

7 0
3 years ago
6. A golf ball is hit a distance of 300 yards in 10 sec. What is the speed of the golf ballo
katen-ka-za [31]

The speed of the ball is 27.4 m/s

Explanation:

The speed of an object is given by:

speed=\frac{d}{t}

where

d is the distance covered

t is the time taken

In this problem, we have:

d = 300 yards is the distance covered by the golf ball

t = 10 s is the time taken

Keeping in mind that

1 yard = 0.914 m

We can convert the distance from yards to meters:

d = 300 \cdot 0.914 = 274.2 m

And substituting into the equation, we find the speed of the ball:

speed=\frac{274.2}{10}=27.4 m/s

Learn more about speed:

brainly.com/question/8893949

#LearnwithBrainly

3 0
3 years ago
Which activity is a model of thermal energy transfer by conduction? A. Freezing ice cubes in a freezer B. Using a heat lamp to w
Ne4ueva [31]

Answer:

Hot water rises and cold water sinks is a model of thermal energy transfer by conduction.

3 0
3 years ago
Read 2 more answers
Nucleus A decays into the stable nucleus B with a half-life of 22.07 s. At t=0 s there are 1,293 A nuclei and no B nuclei. At wh
Alexeev081 [22]

Answer:

29.38 seconds

Explanation:

Half life, T = 22.07 s

No = 1293

Let N be the number of atoms left after time t

N = 1293 - 779 = 514

By the use of law of radioactivity

N=N_{0}e^{-\lambda t}

Where, λ is the decay constant

λ = 0.6931 / T = 0.6931 / 22.07 = 0.0314 decay per second

so,

514=1293e^{-0.0314t}

2.5155=e^{0.0314t}

take natural log on both the sides

0.9225 = 0.0314 t

t = 29.38 seconds

5 0
3 years ago
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NemiM [27]

Answer:

I think the 1st statement is right.

Explanation:

Wind patterns doesn't stay the same.

Waves don't follow the same patterns.

Waves move further up the shore.

I didn't hear about "waves adding" before..so i guess 1st statement is right.

3 0
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