Question 1 of 10

How do noise canceling headphones work?

postnew [5]

Answer:

B.) by interfering with sound waves

Explanation:

As we know that the interference of sound waves is of two types

1). constructive interference

2). destructive interference

now we know that constructive interference means the resultant intensity will be more than the intensity of interfering waves as here two waves are in same phase.

In destructive interference the resultant of two waves is given by the minimum resultant of the intensity as here the phase of two waves are opposite to each other.

So we will say that

$I = \sqrt{(I_1 + I_2 + 2\sqrt{I_1I_2}cos\theta)}$

here in case of noise cancelling headphones we know that the phase of noise is always made in opposite phase with the sound which is used to cancelled the noise.

This will reduce the noise and we will get a clear sound

5 0

2 years ago

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What is the magnitude of the sum of the two vectors A = 36 units at 53 degrees, and B =47 units at 157 degrees.

Thepotemich [5.8K]

Answer:

51.82

Explanation:

First of all, let's convert both vectors to cartesian coordinates:

Va = 36 < 53° = (36*cos(53), 36*sin(53))

Va = (21.67, 28.75)

Vb = 47 < 157° = (47*cos(157), 47*sin(157))

Vb = (-43.26, 18.36)

The sum of both vectors will be:

Va+Vb = (-21.59, 47.11) Now we will calculate the module of this vector:

$|Va+Vb| = \sqrt{(-21.59)^2+(47.11)^2}=51.82$

4 0

3 years ago

The loudness of a sound is the waves amplitude True or false

kakasveta [241]

Answer:

true i think

Explanation:

The amplitude of a sound wave determines its loudness or volume. A larger amplitude means a louder sound, and a smaller amplitude means a softer sound. In Figure 10.2 sound C is louder than sound B. The vibration of a source sets the amplitude of a wave.

4 0

2 years ago

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Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$