All categories

3 years ago

Explain, using your own words, how noise cancelling headphones work, (physics)

Physics

1 answer:

Afina-wow [57]3 years ago

4 0

Answer:

They use noise control, creating a wave that negates outside or ambient sound and replaces it with the desired sound that listeners request.

Explanation:

I hope this helped

You might be interested in

A ball is thrown vertically upward with a speed of 12.0 m/s. (<br> a. How high does it rise?

sdas [7]

It rises till all of its Kinetic energy is converted into potential energy.

so, mgh=(1/2)m(v^2)

so, h=(v^2)/2g = 12*12/(2*9.81)=7.34 m

5 0

3 years ago

What is the worlds most common acid and also the worlds most commin base

mrs_skeptik [129]

Answer: Water is the world's most common acid and base.

3 0

2 years ago

Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note

Sophie [7]

(a) 2 Hz

The frequency of the nth-harmonic is given by

$f_n = n f_1$

where

$f_1$ is the fundamental frequency

Therefore, the frequency of the third harmonic of the A ( $f_1 = 440 Hz$ ) is

$f_3 = 3 \cdot f_1 = 3 \cdot 440 Hz =1320 Hz$

while the frequency of the second harmonic of the E ( $f_1 = 659 Hz$ ) is

$f_2 = 2 \cdot f_1 = 2 \cdot 659 Hz =1318 Hz$

So the frequency difference is

$\Delta f = 1320 Hz - 1318 Hz = 2 Hz$

(b) 2 Hz

The beat frequency between two harmonics of different frequencies f, f' is given by

$f_B = |f'-f|$

In this case, when the strings are properly tuned, we have

- Frequency of the 3rd harmonic of A-note: 1320 Hz

- Frequency of the 2nd harmonic of E-note: 1318 Hz

So, the beat frequency should be (if the strings are properly tuned)

$f_B = |1320 Hz - 1318 Hz|=2 Hz$

(c) 1324 Hz

The fundamental frequency on a string is proportional to the square root of the tension in the string:

$f_1 \propto \sqrt{T}$

this means that by tightening the string (increasing the tension), will increase the fundamental frequency also*, and therefore will increase also the frequency of the 2nd harmonic.

In this situation, the beat frequency is 4 Hz (four beats per second):

$f_B = 4 Hz$

And since the beat frequency is equal to the absolute value of the difference between the 3rd harmonic of the A-note and the 2nd harmonic of the E-note,

$f_B = |f_3-f_2|$

and $f_3 = 1320 Hz$ , we have two possible solutions for $f_2$ :

$f_2 = f_3 - f_B = 1320 Hz - 4 Hz = 1316 Hz\\f_2 = f_3 + f_B = 1320 Hz + 4 Hz = 1324 Hz$

However, we said that increasing the tension will increase also the frequency of the harmonics (*), therefore the correct frequency in this case will be

1324 Hz

8 0

3 years ago

Which of the following waves travel at the greatest speed

NeTakaya

HEY mate here your answer

p waves travel at the greatest speed and these waves can travel at solid and liquid also. p waves means primary waves.

4 0

3 years ago

Imagine you have been asked to create an experimental design to test the hypothesis that talking on a cell phone impairs driving

joja [24]

To start the control would be import to see how drivers would act with full focus on the road. The subjects should (in order for the only difference to be the phone) is to use the same vehicle same road same weather condition, just obviously one has the phone the other does not. I hope this helps

7 0

3 years ago

Read 2 more answers