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oksano4ka [1.4K]
2 years ago
7

You are standing at the midpoint between two speakers, a distance D away from each. The speakers are playing the exact same soun

d of frequency 170 Hz in phase with each other. Describe what happens when you walk 1.5 m directly toward one of the speakers. Assume the speed of sound is 340 m/s. Also assume you can't hear any sounds except those produced by the speakers.
A. You begin by hearing a loud sound that gets quieter until you can't hear it anymore. Then it gets louder again until it is as loud as where you started. Then it gets quieter again until you can't hear it anymore.
B. You begin by hearing nothing, but the sound gets louder.
C. You begin by hearing nothing, but the sound gets louder. Then it gets quieter again until you can't hear it anymore.
D. You begin by hearing a loud sound that gets quieter until you can't hear it anymore.
E.You begin by hearing a loud sound that gets quieter until you can't hear it anymore. Then it gets louder again until it is as loud as where you started.
F. You begin by hearing nothing, but the sound gets louder. Then it gets quieter again until you can't hear it anymore. Then it gets louder again.
Physics
1 answer:
Rzqust [24]2 years ago
3 0

Answer:

Explanation:

wave length of sound waves = velocity / frequency

= 340 / 170

λ = 2 m.

When the position of man is exactly at the meddle point between the speakers , sound waves from the speakers reaching man are in same phases ( path difference is zero. ) so intensity of sound is maximum .

Now , the man starts moving towards one of the speakers , his distance from one speaker becomes closer than the other creating path difference for the sound waves reaching his ears.

If he walks a distance of .5 m towards one speaker , path difference created

= .5 x 2 = 1 m

So , path difference = λ /2 ,

there will be destructive interference so minimum sound will be heard there.

When he walks a distance of 1 m , path difference created = 2m

path difference =  λ

so there is constructive interference and maximum  sound will be heard there.

Again we he walks a distance of 1.5 m , path difference created = 3 m

path difference = 3 λ /2

So there will be destructive interference so minimum sound will be heard there.

In this way we see that man starts  from a point of maximum sound intensity , reaches a point of minimum sound intensity , then reaches a point of maximum sound intensity . At last he reaches a position of minimum sound intensity.

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Answer:

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Explanation:

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So

v  =  0   +  13.5 *  3.10

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The is distance it covers at this time is

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=> s =  +  \frac{1}{2} * 13.5 *  3.10^2

=> s =64.87

Now when sled stops its the final velocity is v_f =  0 m/s while the initial velocity will be the velocity after its acceleration i.e v  =  41.85 \ m/s

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Here  a_1 =  - 4.65, the negative sign shows that it is deceleration

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3 0
2 years ago
A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At
GuDViN [60]

Answer:

E/4

Explanation:

The formula for electric field of a very large (essentially infinitely large) plane of charge is given by:

E = σ/(2ε₀)

Where;

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ε₀ is the electric constant.

Formula to calculate σ is;

σ = Q/A

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We are told the elastic sheet is a square with a side length as d, thus ;

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Putting Q/d² for σ in the electric field equation to obtain;

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E ∝ 1/d²

The electric field at P has some magnitude E. We now double the side length of the sheet to 2L while keeping the same amount of charge Q distributed over the sheet.

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Answer:

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Part A. This is because when the man pulls on the suit upwards, he exerts a force in the upward direction. This takes part of the force of weight of the suitcase and decreases the force the suitcase is exerting on the ground. Thus, the normal force (force exerted by suitcase on the ground) also decreases by the same force as the pull.

Part B. The statements for this part were not given in the question, but the answer reflects what is going to happen in that scenario. Since the man sits on the suitcase, the total weight acting on the ground through the suitcase is that of the suitcase plus the man. Since this force (acting on the ground) is normal force, the statement given in the answer is correct.

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