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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.

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Hydrogen fluoride gas (HF) and sodium hydroxide (NaOH) react in a test tube. They form water and sodium fluoride (NaF). Which ty

solmaris [256]

The type of reaction that occurred in the test tube is ACID BASE REACTION.
Acid base reaction is a type of chemical reaction in which an acid and a base react together to form salt and water molecules. It is also known as neutralization reaction. For the question given above, the applicable chemical reaction is as follows:
HF + NaOH = NaF + H2O
From this chemical reaction, it can be seen that, hydrogen fluoride, which is a weak acid react with sodium hydroxide which is a strong base to produce sodium fluoride, which is a salt and a molecule of water.

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3 years ago

How many atoms of each element are in the chemical formula NH4NO3?

Igoryamba

<span>2 Nitrogen, 4 Hydrogen, 3 Oxygen

9 atoms per molecule.

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4 0

3 years ago

Read 2 more answers

At the moment t = 0, a 24.0-v battery is connected to a 5.00-mh coil and a 6.00-ω resistor. (a) immediately thereafter, how does

agasfer [191]

At the moment the answer is Yeet

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3 years ago

Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ

bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; $H_0 = 51km/s/Mly$

Age of the universe; $t = \ ?$

We know that, the reciprocal of the Hubble's constant ( $H_0$ ) gives an estimate of the age of the universe ( $t$ ). It is expressed as:

$Age\ of\ Universe; t = \frac{1}{H_0}$

Now,

Hubble's constant; $H_0 = 51km/s/Mly$

We know that;

$1\ light\ years = 9.46*10^{15}m$

$1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m$

Therefore;

$H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\$

Now, we input this Hubble's constant value into our equation;

$Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years$

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0

3 years ago

Calculate the induced electric field (in V/m) in a 40-turn coil with a diameter of 11 cm that is placed in a spatially uniform m

Naddik [55]

Answer:

Explanation:

Given that,

Number of turn N = 40

Diameter of the coil d= 11cm = 0.11m

Then, radius = d/2 = 0.11/2 =0.055m

r = 0.055m

Then, the area is given as

A =πr²

A = π × 0.055²

A = 9.503 × 10^-3 m²

Magnetic Field B = 0.35T

Magnetic field reduce to zero in 0.1s, t = 0.1s

so we want to find induce electric field. To find the electric field,(E) we need to find the electric potential (V).

E.M.F is given as

ε = —N • dΦ/dt

Where magnetic flux is given as

Φ = BA

Then, ε = —N • dΦ/dt

ε = —N • dBA/dt

ε = —NBA/t

Then, its magnitude is

ε = NBA/t

Inserting the values of N, B, A and t

ε = 40×0.35×9.503×10^-3/0.1

ε = 1.33 V

Then, using the relationship between Electric field and electric potential

V = Ed

ε = E•d

E = ε/d

E = 1.33/0.11

E = 12.09 V/m

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3 years ago