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Gnom [1K]
3 years ago
6

As the distance of a sound wave from its source quadruples, how is the intensity changed?

Physics
2 answers:
timurjin [86]3 years ago
8 0
Through the 1/r law, it was concluded that for the sound intensity or pressure is directly proportional to distance or radius. That is,
                                 p = k/r 
where k is the proportionality constant. If the distance of a sound wave is quadrupled then, the intensity of the sound is decreased to 1/4 of its original value. 
Ainat [17]3 years ago
6 0

Answer: Hello mate!

Here we have a sound wave, which expands itself like a sphere that starts in the source.

the model that describes the intensity of the wave with respect of the distance to the source is:

A(r) = A₀/(4πr^2)

where A₀ is the initial intensity, π is 3.14159265 and r is the distance to the source.

then if you quadriply the distance, you now have:

A(4r) = A₀/(4π(4r)^2) = A₀/(4π16r^2) = A(r)/16

this means that if you quadruply the distance to the source, the intensity of the sound wave is 16 times smaller, this is because the intensity decreases with the distance squared.

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3 years ago
A bungee jumper with mass 65.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting
ololo11 [35]

Explanation:

It is given that,

Mass of a bungee jumper is 65 kg

The time period of the oscillation is 38 s, hitting a low point eight more times.It means its time period is

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After many oscillations, he finally comes to rest 25.0 m below the level of the bridge.

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5 0
3 years ago
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Answer:

After 9 seconds the object reaches ground.

Step-by-step explanation:

We equation of motion given as h = -16t²+128t+144,

We need to find in how many seconds will the object hit the ground,

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Negative time is not possible, hence after 9 seconds the object reaches ground.

8 0
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