All categories

2 years ago

How does the amplitude of the wave change as you get farther from the speaker?

Physics

1 answer:

KatRina [158]2 years ago

3 0

Answer:

The amplitude decreases

Explanation:

From the inverse square law, we know that;

I ∝ 1/d²

Where;

I is intensity

d is distance

Also, we know that;

A² ∝ I

Where A is amplitude

Thus, we can say that;

A ∝ 1/d²

Thus, amplitude is inversely proportional to the distance.

So the larger the distance, the lesser the amplitude and the lesser the distance, the higher the amplitude.

Thus, as you move farther from the speaker the distance increases and therefore the amplitude decreases

You might be interested in

If zak's speed is 3.00 m/s when he starts to slide, what distance d will he slide before stopping? express your answer in meters

OverLord2011 [107]

That will depend on the coefficient of friction between the sliding surfaces, and also on Zak's weight. We don't have any of that information.

8 0

2 years ago

Temperature and moisture are key factors in determining where each of the seven occur. A) biomes B) biospheres

Sindrei [870]

It's when each of the seven BIOMES occur

6 0

3 years ago

An auto, moving too fast on a horizontal stretch of mountain road, slides off the road, falling into deep snow 43.9 m below the

allsm [11]

Ok so it is 42.8 because if you - 43.9 by 1.1 by the way don’t listen to any of the I just want point

7 0

2 years ago

For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103

Alona [7]

Answer:

a) P = 44850 N

b) $\delta l =0.254\ mm$

Explanation:

Given:

Cross-section area of the specimen, A = 130 mm² = 0.00013 m²

stress, σ = 345 MPa = 345 × 10⁶ Pa

Modulus of elasticity, E = 103 GPa = 103 × 10⁹ Pa

Initial length, L = 76 mm = 0.076 m

a) The stress is given as:

$\sigma=\frac{\textup{Load}}{\textup{Area}}$

on substituting the values, we get

$345\times10^6=\frac{\textup{Load}}{0.00013}$

Load, P = 44850 N

Hence<u> the maximum load that can be applied is 44850 N = 44.85 KN</u>

b)The deformation ( $\delta l$ ) due to an axial load is given as:

$\delta l =\frac{PL}{AE}$

on substituting the values, we get

$\delta l =\frac{44850\times0.076}{0.00013\times103\times 10^9}$

$\delta l =0.254\ mm$

3 0

2 years ago

Two sources of light of wavelength 720 nm are 10 m away from a pinhole of diameter 1.2 mm. How far apart must the sources be for

julia-pushkina [17]

Answer:

The value is $y = 0.00732 \ m$

Explanation:

From the question we are told that

The wavelength of each source is $\lambda = 720 \ nm = 720 *10^{-9} \ m$

The distance from the pinhole $D = 10 \ m$

The diameter is $d = 1.2 mm = \frac{1.2}{1000} = 0.0012 \ m$

Generally from Rayleigh's criterion we have that the distance between the sources of light for their diffraction patterns is mathematically represented as '

$y = \frac{ 1.22 * \lambda * D}{d}$

=> $y = \frac{ 1.22 * 720 *10^{-9}* 10 }{ 0.0012}$

=> $y = 0.00732 \ m$

6 0

3 years ago