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musickatia [10]
3 years ago
7

If a soundproof door is almost shut but still leaves a small opening, what wave phenomenon is responsible for someone being able

to hear sound from the room while standing by the wall to the side of the opening?
Physics
2 answers:
lana66690 [7]3 years ago
5 0

The phenomenon which is responsible for this effect is called diffraction.

Diffraction is the ability of a wave to propagate when it meets an obstacle or a slit. When the wave encounters the obstacle or the slit, it 'bends' around it and it continues propagate beyond it. A classical example of this phenomenon is when a sound wave propagates through a wall where there is a small aperture (as in the example of this problem)

AysviL [449]3 years ago
4 0

Answer:

diffraction

Explanation:

It is due to the diffraction.

Diffraction is the phenomenon in which the waves (light or sound) bends their path after striking to the sharp edged object or the sharp orifice.

Diffraction is possible if the wavelength of the waves has the order of width of the orifice.

You might be interested in
Electrical wire with a diameter of .5 cm is wound on a spool with a radius of 30 cm and a height of 24 cm.
kow [346]

Answer:

a)   # lap = 301.59 rad , b)   L = 90.48 m

Explanation:

a) Let's use a direct proportions rule (rule of three). If one turn of the wire covers 0.05 cm, how many turns do you need to cover 24 cm

          # turns = 1 turn (24 cm / 0.5 cm)

         # laps = 48 laps

Let's reduce to radians

        # laps = 48 laps (2 round / 1 round)

       # lap = 301.59 rad

b) Each lap gives a length equal to the length of the circle

          L₀ = 2π R

          L = # turns L₀

          L = # turns 2π R

          L = 48 2π 30

          L = 9047.79 cm

          L = 90.48 m

6 0
3 years ago
^^ help :c
ipn [44]
I need help to I have a test and I need help to
6 0
3 years ago
The intensity of light from a star varies inversely as the square of the distance. If you lived on a planet ten times farther aw
frosja888 [35]

Answer:

the intensity of the sun on the other planet is a hundredth of that of the intensity of the sun on earth.

That is,

Intensity of sun on the other planet, Iₒ = (intensity of the sun on earth, Iₑ)/100

Explanation:

Let the intensity of light be represented by I

Let the distance of the star be d

I ∝ (1/d²)

I = k/d²

For the earth,

Iₑ = k/dₑ²

k = Iₑdₑ²

For the other planet, let intensity be Iₒ and distance be dₒ

Iₒ = k/dₒ²

But dₒ = 10dₑ

Iₒ = k/(10dₑ)²

Iₒ = k/100dₑ²

But k = Iₑdₑ²

Iₒ = Iₑdₑ²/100dₑ² = Iₑ/100

Iₒ = Iₑ/100

Meaning the intensity of the sun on the other planet is a hundredth of that of the intensity on earth.

3 0
3 years ago
A cart of mass m = 0.12 kg moves with a speed v = 0.45 m/s on a frictionless air track and collides with an identical cart that
lina2011 [118]

Answer:

0.006075Joules

Explanation:

The final kinetic energy of the system is expressed as;

KE = 1/2(m1+m2)v²

m1 and m2 are the masses of the two bodies

v is the final velocity of the bodies after collision

get the final velocity using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

0.12(0.45) + 0/12(0) = (0.12+0.12)v

0.054 = 0.24v

v = 0.054/0.24

v = 0.225m/s

Get the final kinetic energy;

KE = 1/2(m1+m2)v

KE = 1/2(0.12+0.12)(0.225)²

KE = 1/2(0.24)(0.050625)

KE = 0.12*0.050625

KE = 0.006075Joules

Hence the final kinetic energy of the system is 0.006075Joules

5 0
2 years ago
А
bagirrra123 [75]

Explanation:

The reading on the scale is

W = m(g + a)

= (77 kg)(9.8 m/s^2 + 2 m/s^2)

= 908.6 N

7 0
2 years ago
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