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jeka57 [31]
4 years ago
9

Which part of the ear receives the signal from the cochlea and transmits it to the brain?

Physics
2 answers:
iogann1982 [59]4 years ago
8 0
Its b cochlea is the part of the ear is used for hearing
bazaltina [42]4 years ago
6 0

Im pretty sure it would be A) auditory nerve

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The less internal heat a jovian planet emits the more it stirs up its clouds.
g100num [7]
The correct answer would be false. The less internal heat a jovian planet emits the lesser it stirs up its clouds making the atmosphere hotter. All of the four Jovian planets have unique atmospheres. They have more or less the same structures but they differ in their average temperature. As the distance of the planet is closer to the sun the atmosphere would be cooler. These planets are Jupiter, Saturn, Uranus and Neptune. They do not have a solid structure instead they are primarily composed of helium and hydrogen which makes them gas giants or ice planets. They are larger in size than the remaining planets in the solar system. 
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Is acceleration directly proportional to mass, or is it inversely proportional to mass?
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Answer:

The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object

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NEED ASAP!! A box of mass 10 kg requires 20 N to slide it across a surface. What is the weight of the box? What is the coefficie
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jenyasd209 [6]

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4 0
3 years ago
Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d
Tasya [4]

Answer:

I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

I(x)  = k × ({6}^{6}/6) × (  {\pi /4} - sin\pi /4)

I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

3 0
3 years ago
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