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Sonbull [250]
3 years ago
9

Some headpieces also employ noise reduction technology to eliminate sounds not coming from the earpieces of the stethoscope. How

might such technology work?
Physics
1 answer:
sweet [91]3 years ago
8 0

Answer:

Noise cancelling headphones have a microphone which listens to the ambient noise. The headphone then processes the sound and provides a wave to produce destructive interference. This cancels the ambient noise.

Explanation:

First, let us understand what interference is. Interference is the phenomenon that occurs when two waves superpose to form a wave which has a different wavelength, frequency, amplitude than the original two waves. When the crest of one wave superposes with the other waves crest then this results in the amplitude increasing. This is known as constructive interference. When the crest of one wave superposes with the other waves trough then this results in the amplitude decreasing. This known as destructive interference.

Noise cancelling headphones have a microphone which listens to the ambient noise. The headphone then processes the sound and provides a wave to produce destructive interference. This cancels the ambient noise.

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No it's the quite opposite simple
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3 years ago
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Answer:

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Explanation:

Hope This Helps!!

7 0
2 years ago
9. A 5.0 kg block on an inclined plane is acted upon by a horizontal force of 100 N shown in the figure below. The coefficient o
Helga [31]

Answer:

A: The acceleration is 7.7 m/s up the inclined plane.

B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane

Explanation:

Let us work with variables and set

m=5kg\\\\F_H=100N\\\\\mu=0.3\\\\\theta=37^o.

As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.

Part A:

From the free body diagram we see that the total force along the x-axis is:

F_{tot}=mg*sin(\theta)+F_s-F_Hcos(\theta).

Now the force of friction is F_s=\mu*N, where N is the normal force and from the diagram it is F_y=mg*cos(\theta).

Thus F_s=\mu*N=\mu*mg*cos(\theta).

Therefore,

F_{tot}=mg*sin(\theta)+\mu*mg*cos(\theta)-F_Hcos(\theta)\\\\=mg(sin(\theta)+\mu*cos(\theta))-F_Hcos(\theta).

Substituting the value for F_H,m,\mu, and \:\theta we get:

F_{tot}= -38.63N.

Now acceleration is simply

a=\frac{F_H}{m} =\frac{-38.63N}{5kg} =-7.7m/s.

The negative sign indicates that the acceleration is directed up the incline.

Part B:

d=\frac{1}{2} at^2

Which can be rearranged to solve for t:

t=\sqrt{\frac{2d}{a} }

Substitute the value of d=0.50m and a=7.7m/s and we get:

t=0.36s.

which is our answer.

Notice that in using the formula to calculate time we used the positive value of a, because for this formula absolute value is needed.

5 0
3 years ago
The sunspot cycle is a pattern of solar activity where the average number of sunspots gradually _______________________________
zzz [600]

Answer:

option D

Explanation:

Sunspots are the spot that appears on the sun, this spot appears darker than the surrounding surface of the sun.

Sun magnetic field goes through a cycle and this cycle is called the Sunspot cycle. Every 11 years the magnetic field of the sun completely flips. This sunspot cycle affects activity on the surface of the sun.

Sunspot cycle is the pattern of solar activity where an average number of sunspot gradually increase and decrease.

Hence, the correct answer is option D

8 0
3 years ago
A long solenoid with 1.65 103 turns per meter and radius 2.00 cm carries an oscillating current I = 6.00 sin 90πt, where I is in
Leno4ka [110]

Answer:

The  electric field  is 35\cos(90\pi t)\ mV/m

Explanation:

Given that,

Radius = 2.00 cm

Number of turns per unit length n= 1.65\times10^{3}

Current I = 6.00\sin 90\pi t

We need to calculate the induced emf

\epsilon =\mu_{0}nA\dfrac{dI}{dt}

Where, n = number of turns per unit length

A = area of cross section

\dfrac{dI}{dt}=rate of current

Formula of electric field is defined as,

E=\dfrac{\epsilon}{2\pi r}

Where, r = radius

Put the value of emf in equation (I)

E=\dfrac{\mu_{0}nA\dfrac{dI}{dt}}{2\pi r}....(II)

We need to calculate the rate of current

I=6.00\sin 90\pi t....(III)

On differentiating equation (III)

\dfrac{dI}{dt}=90\pi\times6.00\cos(90\pi t)

Now, put the value of rate of current in equation (II)

E=\dfrac{4\pi\times10^{-7}\times1.65\times10^{3}\times\pi\times(2.00\times10^{-2})^2\times90\pi\times6.00\cos(90\pi t)}{2\pi\times 2.00\times10^{-2}}

E=35\cos(90\pi t)\ mV/m

Hence, The  electric field  is 35\cos(90\pi t)\ mV/m

7 0
3 years ago
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