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3 years ago

How is sound detected by the brain

2 answers:

ivann1987 [24]3 years ago

3 0

begin with vibrations of the eardrum. Sound waves cause the eardrum to vibrate. Cells in the cochlea detect the vibrations and send a message to the brain.

siniylev [52]3 years ago

3 0

Answer:

Sound must have a means by which it can be transported, for example air. That is why there is no sound in space.

The way our brain perceives sound is through the ears. The tympanum that the human being has is connected to three bones and when the air vibrates, the tympanum also vibrates, causing these vibrations to pass through those bones towards the inner part of the ears and from there to our brain.

That is the way the brain perceives sound.

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A pitched ball is hit by a batter at a 45degrees angle and just clears the outfield fence, 98m away. Assume that the fence is at

Alex17521 [72]

The range of a projectile can be found directly using:
R = (v²sin2∅) / g
v = √((98 x 9.81)/(sin(90)))
v = 31.0 m/s

8 0

3 years ago

An athlete of mass 70.0 kg applies a force of 500 N to a 30.0 kg luge, which is initially at rest, over a period of 5.00 s befor

ira [324]

Force = (mass) x (acceleration)

Acceleration = (force) / (mass)

Acceleration of the luge = (500 N) / (30 kg) = 16 and 2/3 m/s²

After accelerating at that rate for 5 sec, its speed is (16-2/3) x (5) = 83-1/3 m/s .

(I pause slightly at this point, to reflect that this thing is now moving
at about 186 miles per hour. I question that, and I check my work.
I reassure myself with two thoughts: 1). Maybe those things really
do move at that kind of speeds. I don't know. 2). I was given the
numbers, and I didn't make them up, so I'm only responsible for the
math, not for the plausibility of the solution.)

So the luge is moving at 83-1/3 m/s when he jumps on. In order to maintain
that force against it for 5 seconds, he had to accelerate himself to almost-
if-not-totally the same speed ... necessary, no matter how implausible.

So, although it hasn't been mentioned, the pusher is also doing an enormous
amount of other work just to accelerate himself, and when he jumps aboard,
his own velocity already matches that of his luge. I'm going to say that
after the jump, they continue on, together, coupled as one, at the same
speed as just before the jump.

Their speed together is <em>83-1/3 m/s</em> .

We can't state their <em>velocity</em>, because no information is given regarding
the direction of the track.

6 0

3 years ago

- 4x4 – 13x3 + 8x2 – 21x + 18 is divided by 2 – 3?

Alecsey [184]

Answer:

x = 3/4 or x = ((-1)^(1/3) (127 - 18 sqrt(43))^(2/3) - 13 (-1)^(2/3))/(3 (127 - 18 sqrt(43))^(1/3)) - 4/3 or x = 1/3 (13 (-1/(127 - 18 sqrt(43)))^(1/3) - (-1)^(1/3) (18 sqrt(43) - 127)^(1/3)) - 4/3 or x = (-(127 - 18 sqrt(43))^(2/3) - 13)/(3 (127 - 18 sqrt(43))^(1/3)) - 4/3

Explanation:

Solve for x over the real numbers:

4 x^4 + 13 x^3 - 8 x^2 + 21 x - 18 = 0

The left hand side factors into a product with two terms:

(4 x - 3) (x^3 + 4 x^2 + x + 6) = 0

Split into two equations:

4 x - 3 = 0 or x^3 + 4 x^2 + x + 6 = 0

Add 3 to both sides:

4 x = 3 or x^3 + 4 x^2 + x + 6 = 0

Divide both sides by 4:

x = 3/4 or x^3 + 4 x^2 + x + 6 = 0

Eliminate the quadratic term by substituting y = x + 4/3:

x = 3/4 or 14/3 + 4 (y - 4/3)^2 + (y - 4/3)^3 + y = 0

Expand out terms of the left hand side:

x = 3/4 or y^3 - (13 y)/3 + 254/27 = 0

Change coordinates by substituting y = z + λ/z, where λ is a constant value that will be determined later:

x = 3/4 or 254/27 - 13/3 (z + λ/z) + (z + λ/z)^3 = 0

Multiply both sides by z^3 and collect in terms of z:

x = 3/4 or z^6 + z^4 (3 λ - 13/3) + (254 z^3)/27 + z^2 (3 λ^2 - (13 λ)/3) + λ^3 = 0

Substitute λ = 13/9 and then u = z^3, yielding a quadratic equation in the variable u:

x = 3/4 or u^2 + (254 u)/27 + 2197/729 = 0

Find the positive solution to the quadratic equation:

x = 3/4 or u = 1/27 (18 sqrt(43) - 127)

Substitute back for u = z^3:

x = 3/4 or z^3 = 1/27 (18 sqrt(43) - 127)

Taking cube roots gives 1/3 (18 sqrt(43) - 127)^(1/3) times the third roots of unity:

x = 3/4 or z = 1/3 (18 sqrt(43) - 127)^(1/3) or z = -1/3 (-1)^(1/3) (18 sqrt(43) - 127)^(1/3) or z = 1/3 (-1)^(2/3) (18 sqrt(43) - 127)^(1/3)

Substitute each value of z into y = z + 13/(9 z):

x = 3/4 or y = 1/3 (18 sqrt(43) - 127)^(1/3) - (13 (-1)^(2/3))/(3 (127 - 18 sqrt(43))^(1/3)) or y = 13/3 ((-1)/(127 - 18 sqrt(43)))^(1/3) - 1/3 (-1)^(1/3) (18 sqrt(43) - 127)^(1/3) or y = 1/3 (-1)^(2/3) (18 sqrt(43) - 127)^(1/3) - 13/(3 (127 - 18 sqrt(43))^(1/3))

Bring each solution to a common denominator and simplify:

x = 3/4 or y = ((-1)^(1/3) (127 - 18 sqrt(43))^(2/3) - 13 (-1)^(2/3))/(3 (127 - 18 sqrt(43))^(1/3)) or y = 1/3 (13 ((-1)/(127 - 18 sqrt(43)))^(1/3) - (-1)^(1/3) (18 sqrt(43) - 127)^(1/3)) or y = (-(127 - 18 sqrt(43))^(2/3) - 13)/(3 (127 - 18 sqrt(43))^(1/3))

Substitute back for x = y - 4/3:

Answer: x = 3/4 or x = ((-1)^(1/3) (127 - 18 sqrt(43))^(2/3) - 13 (-1)^(2/3))/(3 (127 - 18 sqrt(43))^(1/3)) - 4/3 or x = 1/3 (13 (-1/(127 - 18 sqrt(43)))^(1/3) - (-1)^(1/3) (18 sqrt(43) - 127)^(1/3)) - 4/3 or x = (-(127 - 18 sqrt(43))^(2/3) - 13)/(3 (127 - 18 sqrt(43))^(1/3)) - 4/3

7 0

4 years ago

Read 2 more answers

Would sound travel faster at the north pole or in the sahara desert? Explain ur answer

gtnhenbr [62]

Sound needs a medium to propagate through and this medium which is moves through is the air. The colder the temperature is, the less excited (meaning kinetic energy) the air particles will be meaning that the sound wave will travel slower. In the sahara desert, since the temperatures are warmer, and more excited, the sound will propagate faster through the air.

*Note that sound waves rely on vibration to move and if the air is hotter the molecules in the air are vibrate more than compared to lower temperatures.

6 0

3 years ago

I’m between radiant energy to mechanical and kinetic energy to mechanical

Rama09 [41]

The energy in wind CAME from radiant energy originally, but it isn't
radiant energy any more. It's the energy in the mass of moving air,
and that's KINETIC energy.

You could move the turbine blades with stones or pingpong balls,
push the blades with your hands, or shoot water at them from a hose.
In any case, you'd have moving mass with kinetic energy transfer that
energy to the turbine. A moving mass of air does the same thing.

4 0

3 years ago

Read 2 more answers