All categories

2 years ago

What effect do sound waves have on air particles

Physics

1 answer:

valkas [14]2 years ago

7 0

Answer:

Sound waves travel at 343 m/s through the air and faster through liquids and solids. The waves transfer energy from the source of the sound, e.g. a drum, to its surroundings. Your ear detects sound waves when vibrating air particles cause your ear drum to vibrate. The bigger the vibrations the louder the sound.

Explanation:

You might be interested in

Is the term used to describe an object with enough orbital energy to escape earth completely?

Komok [63]

Escape velocity describes its speed. It becomes a satellite instead of falling back to earth because as it falls under Earth’s gravity the Earth has moved so the satellite falls around the planet instead of down to the ground.

5 0

3 years ago

The frequency of violet light is 7.5x10 to the 14th power hertz. What is the wavelength?

Verizon [17]

Wavelength times frequency = speed of light
7.5E14 x wavelength = 300,000 m/s

Wavelength in meters = 300,000 divided by 7.5E14

8 0

2 years ago

Three 1.5V cells are connected in series in a circuit. What is the total potential difference?

lorasvet [3.4K]

Answer:

4.5V

Explanation:

1.5x3= 4.5

4 0

2 years ago

At what distance of separation, r, must two 3.20 x 10-9 Coulomb charges be positioned in order for the repulsive force between t

STatiana [176]

I'm not 100% sure but I think the answer is 60.4, because you multiply 3.20 by 10, to get 32 - 9 = 23, then subtract 23 from 83.4. Hope this helps you, and good luck!!!

4 0

2 years ago

Read 2 more answers

A baseball pitcher throws a ball horizontally at a speed of 34.0 m/s. A catcher is 18.6 m away from the pitcher. Find the magnit

Sidana [21]

To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

$y = \frac{1}{2} at^2+v_0t+y_0$

Where,

a = acceleration

t = time

$v_0$ = Initial velocity

$y_0$ = initial position

In addition to this we know that speed, speed is the change of position in relation to time. So

$v = \frac{x}{t}$

x = Displacement

t = time

With the data we have we can find the time as well

$v = \frac{x}{t}$

$t = \frac{x}{v}$

$t = \frac{18.6}{34}$

$t = 0.547s$

With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,

$y = \frac{1}{2} at^2+v_0t+y_0$

$y = \frac{1}{2} gt^2+0+0$

$y = \frac{1}{2} gt^2$

$y = \frac{1}{2} 9.8*0.547^2$

$y = 1.46m$

Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.

6 0

2 years ago