All categories

3 years ago

In which part of the ear are the vibrations occurring in a gas, liquid, and solid?

Physics

1 answer:

UkoKoshka [18]3 years ago

5 0

Answer:

eardrum is the correct answer

hope this helps ❤️

Explanation:

Sound waves enter the ear canal and cause the eardrum to vibrate. Three small bones transmit these vibrations to the cochlea. This produces electrical signals which pass through the auditory nerve to the brain, where they are interpreted as sound.

You might be interested in

[]Answer the question below[]

marysya [2.9K]

Answer:

The answer is D. density.

8 0

2 years ago

Which feature of electromagnets makes them more useful than permanent

mezya [45]

I think it’s D................

4 0

3 years ago

A proton is projected in the positive x direction into a region of a uniform electric field E S 5 126.00 3 105 2 i^ N/C at t 5 0

Dafna11 [192]

Answer:

(a). The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$

(b). The initial peed of the protion is $2.83\times10^{6}\ m/s$

(c). The time is $0.493\times10^{-7}\ sec$ .

Explanation:

Given that,

Electric field $E=-6.00\times10^{5}i\ N/C$

Time = 5.0 sec

Distance 7.00 cm

(a). We need to calculate the acceleration

Using formula of force

$F=F_{e}$

$ma=qE$

$a=\dfrac{Eq}{m}$

Where, E = electric field

m = mass of proton

q = charge of proton

Put the value into the formula

$a=\dfrac{-6.00\times10^{5}\times1.6\times10^{-19}}{1.67\times10^{-27}}$

$a=-5.74\times10^{13}\ m/s62$

The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$

(b). We need to calculate the initial peed

Using equation of motion

$v^2-u^2=2as$

Where, s = distance

Put the value into the formula

$0-u^2=2\times(-5.74\times10^{13})\times7.00\times10^{-2}$

$u=\sqrt{2\times5.74\times10^{13}\times7.00\times10^{-2}}$

$u=2834783.94$

$u=2.83\times10^{6}\ m/s$

The initial peed of the protion is $2.83\times10^{6}\ m/s$

(c). We need to calculate the time interval over which the proton comes to rest

Using formula

$t=\dfrac{u}{a}$

Where, u = initial velocity

a = acceleration

Put the value into the formula

$t=\dfrac{2.83\times10^{6}}{5.74\times10^{13}}$

$t=0.493\times10^{-7}\ sec$

Hence, (a). The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$

(b). The initial peed of the protion is $2.83\times10^{6}\ m/s$

(c). The time is $0.493\times10^{-7}\ sec$ .

5 0

3 years ago

What is the relationship between gravity and pressure in a nebula?

zheka24 [161]

The relationship between gravity and pressure in a nebula is that pressure balances gravity. The pressure exerted by a static fluid depends only upon the depth of the fluid, the density of the fluid, and the acceleration of gravity. The answer is B.

8 0

3 years ago

Read 2 more answers

Use the drop-down menus to identify the variables for the second hypothesis.

Debora [2.8K]

Answer: 1) independent

2) dependent

3) voltage

Explanation:

8 0

3 years ago

Read 2 more answers