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4 years ago

What happens to sound waves after they are received by our ears?

Physics

2 answers:

wariber [46]4 years ago

7 0

Answer:

The vibrational energy of the waves is converted into electrical signals and sent to the brain.

Explanation:

Daniel [21]4 years ago

6 0

Answer: Option C

Explanation:

The sound waves that travels to ear are send to the brain in order to hear it. The waves are in the form of vibration energy which is received by the ear. These signals are converted from the vibration into electrical signals.

These signals from the ear are sent to the brain in the form of electrical energy. These signals are processed there only and the feedback is reverted by the brain cells to the body.

Hence, the correct answer is option C

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Can someone help me label this flower??

tensa zangetsu [6.8K]

Hope this helps!!!!!!!!!!!!!

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3 years ago

A 25-foot ladder is leaning against a house. The base of the ladder is pulled away from the house at a rate of 2 feet per second

grandymaker [24]

A ladder 25 feet long is leaning against a house. The base of the ladder is pulled away at a rate of 2 ft/sec.

a.) How fast is the top of the ladder moving down the wall when the base of the ladder is 12 feet from the wall?

Answer:

dy/dt = -1.094ft/sec

Explanation:

Given that:

dz/dt = 0,

dx/dt = 2,

dy/dt = ?

Hence, we have the following

Using Pythagoras theorem

We have 25ft as the hypotenuse, y as the opposite or height of wall, and x as the base of the triangle

X² + y² = z²,

12² + y² = 25²,

y² = 25² - 12²

y = √481

Therefore, we have the following:

2x dx/dt + 2y dy/dt,

= 2z dz/dt,

= 12 (2) √481 dy/dt,

= √481 dy/dt = -24,

= dy/dt = -1.094ft/sec

Therefore, final answer is -1.094ft/sec

4 0

3 years ago

As a tractor pulls a plow across a field, it exerts a force on the plow in the forward direction. What is the "equal and opposit

Karolina [17]

Answer:

the force of the plow pulling backward on the tractor

Explanation:

7 0

3 years ago

A train rolls past a stationary observer. To him, the train is moving at a speed of 23m/s west, and a woman on the train is movi

GREYUIT [131]

Answer:

21.7 seconds.

Explanation:

Woman's velocity relative to train (23 m/s - 22.4 m/s) = 0.6 m/s

Distance woman wants to travel = 13m

To find how long she will take to move 13m relative to the train, take the distance she wants to travel divided by her velocity relative to the train.

(13m)/(0.6 m/s) = 21.6667 seconds or 21.7 seconds.

Therefore, it will take the woman 21.7 seconds to move 13m.

4 0

3 years ago

Two parallel plates are a distance apart with a potential difference between them. A point charge moves from the negatively char

IgorC [24]

Answer:

K' = 1200 J

Explanation:

To find the kinetic energy you first take into account the formula for the kinetic energy of the charge:

$K=\frac{1}{2}mv^2$ = 800J (1)

m: mass of the charge

v: final speed of the charge when it reaches the positively charged plate.

Furthermore, you have that the acceleration of the charge is obtained by using the second Newton law:

$F=ma=qE\\\\a=\frac{qE}{m}$ (2)

a: acceleration

E: electric field

q: charge

The electric field between two parallel plates is V/d, being V the potential difference and d the separation between plates. You replace E in (2) and obtain:

$a=\frac{qV}{md}$

Next, you take into account the following formula for the calculation of the final speed of the charge:

$v^2=v_o^2+2ad\\\\v_o=0m/s\\\\v=\sqrt{\frac{2qVd}{md}}=\sqrt{\frac{2qV}{m}}$

Next, you replace this value of v in (1):

$K=\frac{1}{2}mv^2=\frac{1}{2}m(\frac{2qV}{m})=qV$ = 880J (3)

If the distance between plates is tripled, and the potential difference is halved, you have for the new final speed:

$v'^2=v'_o^2+2a(3d)\\\\v_o=0m/s\\\\v'=\sqrt{6ad}=\sqrt{6(\frac{q}{md})\frac{V}{2}d}=\sqrt{\frac{3qV}{m}}$

And the kinetic energy becomes:

$K'=\frac{1}{2}mv^2=\frac{1}{2}m(\frac{3qV}{m})=\frac{3}{2}qV$ (4)

You calculate the ratio between both kinetic energies K and K', that is, you divide equations (3) and (4), in order to find the new kinetic energy:

$K=qV=800J\\\\K'=\frac{3}{2}qV\\\\\frac{K}{K'}=\frac{qV}{3/2\ qV}=\frac{2}{3}\\\\K'=\frac{3}{2}K=\frac{3}{2}(800J)=1200J$

hence, the kinetic energy of the charge incresases to 1200J

4 0

3 years ago