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.
Answer:
x = 8.3 cm
y = 5.4 cm
Explanation:
x position
200(0) + 100(12) + 350(12) / (200 + 100 + 350) = 8.307692...
y position
200(0) + 100(0) + 350(10) / (200 + 100 + 350) = 5.3846153...
At 20 seconds it will be 12.6 because at 10 seconds it was as at approximately 6.3 so we times it by 2 to get the 20s
Answer:
q = 0.0392 / V, for V= 0.1V q = 0.392 C
Explanation:
For this exercise we can assume that the power energy of the drops is transformed into kinetic energy, therefore we use the conservation of energy
starting point
Em₀ = U = q V
final point
Em_f = K = ½ m v²
Em₀ = Em_f
q V = ½ m v²
q =
let's calculate
q = ½ 0.40 10⁻³ 14² / V
q = 0.0392 / V
The object to be painted is connected to ground therefore its potential is dro, but the gun where it is painted has a given potential, suppose it is
V = 0.1 V
q = 0.0392 / 0.1
q = 0.392 C
Applying Concepts: What happens to the volume of a balloon that is taken outside on a cold winter day? ... When the air particles inside the balloon become cooler, they slow down and do not hit the inside of the balloon as often, so the balloon's volume decreases. this is google answer sorry i didnt know how to word it