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sesenic [268]
4 years ago
7

Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitt

ed by each speaker is 172 Hz. You are 8.00 m from speaker A. Take the speed of sound in air to be 344 m/s.What is the closest you can be to speaker B and be at a point of perfectly destructive interference?
Physics
2 answers:
Tamiku [17]4 years ago
8 0

Answer:

7 m .

Explanation:

For destructive interference

Path difference = odd multiple of λ /2

Wave length of sound from each of  A and B.

= speed / frequency

λ = 334 / 172 = 2 m

λ/2 = 1 m

If I am  1 m away from B , the path difference will be

8 - 1 = 7 m  which is  odd multiple of 1 or λ /2

So path difference becomes odd multiple of  λ /2.

This is the condition of destructive interference.

So one meter is the closest distance which I can remain at so that i can hear destructive interference.

Zolol [24]4 years ago
3 0

Answer:

1m.

Explanation:

Frequency = f = 172 Hz

Speed = V = 344 m/s

For the destructive interference,

Wavelength = λ = V/f = 344/172 = 2m

Path difference = d2 – d1 = n λ/2

Where “n” is an odd integer.

If you are 1m away from the loudspeaker B then the path difference will be,

Path difference = d2 - d1 = 8 – 1 = 7m  

Which is odd multiple of λ/2 and also condition for destructive interference. Hence, to hear the perfect destructive interference you have to stay 1m close to loudspeaker B.  

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3 years ago
From a vantage point very close to the track at a stock car race, you hear the sound emitted by a moving car. You detect a frequ
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Answer:

55.8 m/s

Explanation:

f = Actual frequency of sound emitted by car

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3 years ago
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antiseptic1488 [7]

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From a window that is 20 m from the ground a stone with a speed of 10m / s is thrown vertically upwards. Calculate:
Oduvanchick [21]

a)

consider the motion in upward direction as positive and down direction as negative

Y₀ = initial position of the stone = 20 m

v₀ = initial velocity of the stone = 10 m/s

a = acceleration = - 9.8 m/s²

Y = final position of the stone when it reach the maximum height

v = final velocity at the maximum height = 0 m/s

t = time taken to reach the maximum height

Using the equation

v² = v₀² + 2 a (Y - Y₀)

0² = 10² + 2 (- 9.8) (Y - 20)

Y = 25.1 m


also using the equation

v = v₀ + a t

inserting the values

0 = 10 + (- 9.8) t

t = 1.02 sec


b)

consider the motion in upward direction as positive and down direction as negative

Y₀ = initial position of the stone = 20 m

v₀ = initial velocity of the stone = 10 m/s

a = acceleration = - 9.8 m/s²

Y = final position of the stone when it reach the ground = 0 m

t = time taken to reach the ground

Using the equation

Y = Y₀ + v₀ t + (0.5) a t²

0 = 20 + 10 t + (0.5) (- 9.8) t²

t = 3.3 sec

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3 years ago
In some circumstances, it is useful to look at the linear velocity of a point on the blade. The linear velocity of a point in un
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Answer:

v=wr

Explanation:

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In the uniform circular motion, an object describes the same angles in the same times. If \theta is the angle formed by the trajectory of the object in a time t, then its angular velocity is

\displaystyle w=\frac{\theta}{t}

if \theta is expressed in radians and t in seconds the units of w is rad/s. If the circular motion is uniform, the object forms an angle 2\theta in 2t, or 3\theta in 3t, etc. Thus the angular velocity is constant.

The magnitude of the tangent or linear velocity is computed as the ratio between the arc length and the time taken to travel that distance:

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Replacing the formula for w, we have

\boxed{ v=wr}

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