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AnnZ [28]
3 years ago
15

In which part of the ear is the sound wave converted into electrical impulse

Physics
2 answers:
Elenna [48]3 years ago
7 0

Answer:

The correct answer would be cochlea.

The cochlea is a part of an inner ear and serves the function of hearing in an organism.

The perilymph (watery liquid) present in the cochlea moves in response to the vibrations made by the sound waves.

The movement of cochlea leads to the generation of movement in the organ of Corti and basilar membrane.

Then stereocilia present in the hair cells help in mechanosensing these movements and convert these into electrical signals.

These signals are then passed to the brain through auditory nerves for further processing.

Kryger [21]3 years ago
3 0
The part of the ear where the sound wave converted into electrical impulse would be the cochlea. This part is the  auditory portion of the inner ear which produces nerve impulses in response to sound vibrations. Hope this answers the question. Have a nice day.
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One of your classmates, in a fit of unrestrained ego, jumps onto a lab table:
Anettt [7]

The equilibrium condition allows finding the results for the forces of the system are

      a) The free body diagram is in the attachment

      b) The normal force is N = 737 N

      c) The mass of the table is  10.2 kg

Newton's second law indicates that the net force is proportional to the product of the mass and the acceleration of the bodies, in the special case that the acceleration is zero, it is called the equilibrium condition

          ∑ F = 0

Where the bold letters indicate vectors, F is the external forces

a) A free body diagram is a scheme of the forces without the details of the bodies, in the attachmentt we see a free body diagram of the system.

b) The reaction force of the ground is applied in each of the legs of the table, in general this force has the same magnitude in each leg, therefore in Newton's second law we can place it as a single force

             N = N₁ + N₂ + N₃ + N₄₄

Let's apply the equilibrium condition

                N -  W_m -w_{table} = 0

                N =  W_m +w_{table}

                N = M_m g + w_{table}  

They indicate the pose of the boy is 65 kg, for the weight of the table of a laboratory table is approximately 100 N

                N = 65 9.8 + 100

                N = 737 N

c) To calculate the mass of the table we use the relation

                W = m_{table} g

                m_{table} = \frac{w_{table}}{g}

                m_{tabble}= \frac{100}{9.8}  

               m_{table}e = 10.2 kg

In conclusion using the equilibrium condition we can find the results for the forces are

      a) The free body diagram is in the attachment

      b) The normal force is N = 737 N

      c) The mass of the table is  10.2 kg

Learn more here:  brainly.com/question/19860811

7 0
2 years ago
There are 5510 lines per centimeter in a grating that is used with light whose wavelegth is 467 nm. A flat observation screen is
Mademuasel [1]

Answer:

1.696 nm

Explanation:

For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,

dsinθ = mλ = (1)λ = λ

dsinθ = λ

sinθ = λ/d.

Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m

From trig ratios 1 + cot²θ = cosec²θ

1 + (1/tan²θ) = 1/(sin²θ)

substituting the values of sinθ and tanθ we have

1 + (D/w)² = (d/λ)²

(D/w)² = (d/λ)² - 1

(w/D)² = 1/[(d/λ)² - 1]

(w/D) = 1/√[(d/λ)² - 1]

w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹  = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.

w is also the distance from the center to the other principal maximum on the other side.

So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm

So, the minimum width of the screen must be 1.696 nm

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3 years ago
Which type of energy is released when a bond between atoms is broken
scoray [572]

Answer:

D

Explanation:

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A boulder with a weight of 780 N is resting at the edge of a cliff that rises 123 m above the ground. What is the gravitational
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Potential energy U = mgh

Given h = 123 m,
mg = F = 780 N

Then
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