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2 years ago

The brain receives messages in signals called nerve impulses. Which part of the ear first generates these nerve impulses?

Physics

2 answers:

liberstina [14]2 years ago

6 0

That's the job of the tiny "hair cells", located in the <em>inner ear.</em>

If you're a sound wave, this is how you reach the hair cells:

-- go into the big funnel of skin on the outside of the head, that thing we call the "ear"

-- go about an inch or two, down through a skinny dark tunnel inside the skull

-- at the end of the tunnel, hit a dead end, made of a wall of thin skin like a drum, called the "ear drum"; sound waves hit the ear drum and make it vibrate

-- on the other side of the ear drum, inside, is the chamber called the "middle ear". In there are the three smallest bones in the body; the ear drum touches the first one and makes it vibrate; the first one touches the second one and makes it vibrate; the second one touches the third one and makes it vibrate; then the third one touches another dead end made of thin skin.

-- the region on the other side of this wall of thin skin is the "inner ear"; it's a long skinny chamber, called the "cochlea", wound up in a spiral and filled with liquid; the walls of the cochlea are lined with millions of tiny hairs, sticking out into the liquid; the vibrations make waves in the liquid, and the waves make the tiny hairs wave back and forth; each tiny hair is the end of a nerve that goes into the brain; when that hair wiggles, it sends a nerve "message" into the brain.

-- there are two complete copies of this whole structure ... one on each side of your head.

boyakko [2]2 years ago

4 0

Answer:

B. cochlea

i just took the test

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An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t

gtnhenbr [62]

1. $tan \theta = \frac{v^2}{Rg}$

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

$N cos \theta = mg$ (1)

$N sin \theta = m \frac{v^2}{R}$ (2)

where in the second equation, the term $m\frac{v^2}{R}$ represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the following expression:

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2. $F=\frac{m}{R}(w^2-v^2)$

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This means that the centripetal force term

$m\frac{v^2}{R}$

is now larger than before, and therefore, the horizontal component of the normal reaction, $N sin \theta$ , is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

$N sin \theta + F = m\frac{w^2}{R}$