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Alexxandr [17]
3 years ago
10

A speaker generates a continuous tone of 440 Hz. In the drawing, sound travels into a tube that splits into two segments, one lo

nger than the other. The sound waves recombine before being detected by a microphone. The speed of sound in air is 339 m/s. What is the minimum difference in the lengths of the two paths for sound travel if the waves arrive in phase at the microphone?
Physics
1 answer:
chubhunter [2.5K]3 years ago
7 0

Answer:

The minimum difference between the lengths of the two tubes should be 0.385 meters.

Explanation:

As we known that for any two waves to arrive in phase at any point the difference in the path traveled by the waves should be an integral multiple of the wavelength of the wave.

Mathematically we can write:

\Delta x=n\frac{\lambda }{2}

For the given wave we have

\lambda =\frac{v}{\nu }

Applying values we get

\lambda =\frac{339}{440 }=0.77m

Thus the minimum difference in the lengths of the tubes can be obtained by putting the value of n = 1

\therefore \Delta x=1\times \frac{0.77}{2}=0.385m

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Answer:

2000 to 2010

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A rock is thrown upward with a velocity of 22 meters per second from the top of a 25 meter high cliff, and it misses the cliff o
Mice21 [21]

Answer:

The rock will reach 9 m from the ground at eaxactly 5.06 s after it was initially thrown upwards.

Explanation:

We will use the equations of motion for this.

u = initial velocity of the rock = 22 m/s

g = acceleration due to gravity = -9.8 m/s²

y = vertical position of the rock at a time t = 9 m

y₀ = initial height of the rock = 25 m

t = time it takes for the rock to reach height of 9 m.

(y-y₀) = ut + 0.5gt²

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- 14 = 22t - 4.9t²

4.9t² - 22t - 14 = 0

solving this quadratic equation,

t = 5.055 s or - 0.565 s

Since time cannot be negative,

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Hope this Helps!!!

6 0
3 years ago
Read 2 more answers
When you whisper you produce a 10-dB sound.
masha68 [24]
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A. How far does a 100-newton force have to move to do 1,000 joules
Aloiza [94]

Work done by a force is given as the product of force and the distance moved by the force.

<h3>What is work done?</h3>

Work done is the product of force and the distance moved by the the force.

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Thus, distance required by the 100 N force is given as:

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Distance = 1000/100 = 10 m

Distance to be moved is 10 m.

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Work done by the athlete after one push-up = 250 × 0.25 m

Work done by the athlete = 62.5 J

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Work done = 500 × 0 = 0 N

Therefore, for work to be done, force has to move a distance.

Learn more about work done at: brainly.com/question/25573309

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