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3 years ago

11

the velocity of a sound on a particular day outside is 331 meters/second. what is the frequency of a tone if it has a wavelength

of 0.6 meters?

1 answer:

hichkok12 [17]3 years ago

3 0

Wavelength*frequency=velocity
(331m/s)/(.6m)
Frequency = 551.666 1/s

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Answer:

t=7.33 s

Explanation:

According to Newton's second law:

$\sum F=m*a$

because we don't want the box to slide, the acceleration has to be zero.

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we know that:

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Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly

loris [4]

Answer: $36.86\°$

Explanation:

According to the described situation we have the following data:

Horizontal distance between lily pads: $d=2.4 m$

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Time it takes a jump: $t=0.6 s$

We need to find the angle $\theta$ at which Ferdinand jumps.

In order to do this, we first have to find the <u>horizontal component (or x-component)</u> of this initial velocity. Since we are dealing with parabolic movement, where velocity has x-component and y-component, and in this case we will choose the x-component to find the angle:

$V_{x}=\frac{d}{t}$ (1)

$V_{x}=\frac{2.4 m}{0.6 s}$ (2)

$V_{x}=4 m/s$ (3)

On the other hand, the x-component of the velocity is expressed as:

$V_{x}=V_{o}cos\theta$ (4)

Substituting (3) in (4):

$4 m/s=5 m/s cos\theta$ (5)

Clearing $\theta$ :

$\theta=cos^{-1} (\frac{4 m/s}{5 m/s})$

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Near the surface of reflection, reflected wave may interfere with incident wave leading to production of constructive and as well as destructive interference. This in turn, can result to resonance as well as enhancement of the sound intensity as the waves of reflection adds to incident wave. Therefore, the girl would higher intensity of reflected waves as compared to incident waves.

Therefore, statement A is correct.

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Let's use the mirror equation to solve the problem:
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Answer:

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