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Nitella [24]
3 years ago
5

How does reverberation explain why your voice sounds different in a gym than it does in your living room

Physics
1 answer:
TEA [102]3 years ago
6 0
Sound can reach the inner ear by way of two separate paths, and those paths in turn affect what we perceive. Air-conducted sound is transmitted from the surrounding environment through the external auditory canal, eardrum and middle ear to the cochlea, the fluid-filled spiral in the inner ear. Bone-conducted sound reaches the cochlea directly through the tissues of the head.
When you speak, sound energy spreads in the air around you and reaches your cochlea through your external ear by air conduction. Sound also travels from your vocal cords and other structures directly to the cochlea, but the mechanical properties of your head enhance its deeper, lower-frequency vibrations. The voice you hear when you speak is the combination of sound carried along both paths. When you listen to a recording of yourself speaking, the bone-conducted pathway that you consider part of your “normal” voice is eliminated, and you hear only the air-conducted component in unfamiliar isolation. You can experience the reverse effect by putting in earplugs so you hear only bone-conducted vibrations.
Some people have abnormalities of the inner ear that enhance their sensitivity to this component so much that the sound of their own breathing becomes overwhelming, and they may even hear their eyeballs moving in their sockets.
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7 0
4 years ago
Part A
cupoosta [38]

Answer:

A) U₀ = ϵ₀AV²/2d

B) U₁ = (ϵ₀AV²)/6d

This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.

C) U₂ = (kϵ₀AV²)/2d

Explanation:

A) The energy stored in a capacitor is given by (1/2) (CV²)

Energy in the capacitor initially

U₀ = CV²/2

V = voltage across the plates of the capacitor

C = capacitance of the capacitor

But the capacitance of a capacitor depends on the geometry of the capacitor is given by

C = ϵA/d

ϵ = Absolute permissivity of the dielectric material

ϵ = kϵ₀

where k = dielectric constant

ϵ₀ = permissivity of free space/air/vacuum

A = Cross sectional Area of the capacitor

d = separation between the capacitor

If air/vacuum/free space are the dielectric constants,

So, k = 1 and ϵ = ϵ₀

U₀ = CV²/2

Substituting for C

U₀ = ϵ₀AV²/2d

B) Now, for U₁, the new distance between plates, d₁ = 3d

U₁ = ϵ₀AV²/2d₁

U₁ = ϵ₀AV²/(2(3d))

U₁ = (ϵ₀AV²)/6d

This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.

C) U₂ = CV²/2

Substituting for C

U₂ = ϵAV²/2d

The dielectric material has a dielectric constant of k

ϵ = kϵ₀

U₂ = (kϵ₀AV²)/2d

3 0
4 years ago
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