All categories

3 years ago

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

2 answers:

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.

You might be interested in

Can density be used to identify what a substance is made of?

Korolek [52]

Possibly, if you have list of densities and you have to match it. I can't think of any other scenarios in which it would be able to.

Hope I helped! :)

7 0

3 years ago

Two balls are dropped from rest and allowed to fall. If one ball is allowed to fall for 1 s and the other for 3 s compare the di

kirill115 [55]

The second ball traveled a greater distance when compared to the first ball because the second ball spent more time in motion.

The given parameters;

time of fall of the first ball, t = 1 s
time of fall of the second ball, t = 3 s

The distance traveled by each ball is calculated using the second equation of motion as shown below.

The distance traveled by the first ball is calculated as follows;

$h = u_0t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\h = (0.5\times 9.8\times 1^2)\\\\h = 4.9 \ m$

The distance traveled by the second ball is calculated as follows;

$h = \frac{1}{2} gt^2\\\\h = (0.5\times 9.8\times 3^2)\\\\h = 44.1\ m$

Thus, the second ball traveled a greater distance because it spent more time in motion.

Learn more here:brainly.com/question/5868480

3 0

1 year ago

Changing classes, you walk 20.0 m down a hall, turn left, and then walk 10.0 m down another hall. Define a coordinate system and

worty [1.4K]

Explanation:

Calculate position vectors in a multidimensional displacement problem. Solve for the displacement in two or three dimensions. Calculate the velocity vector

7 0

2 years ago

A beam of light has a wavelength of 4.5 x10^-7 meter in a vacuum. the frequency of this light is

valkas [14]

The basic relationship between frequency and wavelength for light (which is an electromagnetic wave) is
$c= f \lambda$
where c is the speed of light, f the frequency and $\lambda$ the wavelength of the wave.
Using $\lambda=4.5 \cdot 10^{-7} m$ and $c=3 \cdot 10^8 m/s$ , we can find the value of the frequency:
$f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{4.5 \cdot 10^{-7} m}=6.7 \cdot 10^{14} Hz$

3 0

3 years ago

You hear a sound in the distance. Suddenly the sound gets deeper, decreasing in pitch. Which can you assume about the sound wave

oksian1 [2.3K]

Answer:

A. The wavelengths of the new sound waves are longer

Explanation:

This is the Doppler effect which can be best illustraded for the case of a siren of an ambulance approaching us having a greater frequency and getting lower in frequency and deeper as the ambulance passes us.

Since the wavelength is inversely proportional to the frequency it follows the wavelengths are longer when the frequency decreases lowering its pitch and getting deeper.

8 0

3 years ago