Ask question

Ask question

All categories

3 years ago

13

Your brother is insisting that you’ll never hear a sound produced behind a barrier wall at the end of your yard you notice that

the door within the wall is open and quickly explain to him that the sound will be able to reach you due to
A.Reverberation
B.Diffusion
C.Focusing
D.Diffraction

2 answers:

Tresset [83]3 years ago

8 0

Answer

D.Diffraction

Explanation

Diffraction is a property that is experienced by waves when they come across a barrier when they are in motion.

The ways tends to curve behind the barrier. This is called diffraction of waves.

Now, sound is a wave and it also experience diffraction. . So the brother will be able to hear the sound due to diffraction

oksian1 [2.3K]3 years ago

7 0

Answer: D. Diffraction

Explanation:

You might be interested in

Arrange the distances between Earth and various celestial objects in order from least to greatest. Use the conversion table to h

Kaylis [27]

distance to the star Betelgeuse: 640 ly

As we know that

$1 ly = 63000 AU$

also we know that

$1AU = 1.5 \times 10^8 km$

$1 ly = 63000 (1.5 \times 10^8) = 9.45 \times 10^{12} km$

So the distance of Betelgeuse = 640 ly

$d_1 = 640 \times 9.45 \times 10^{12} = 6.05 \times 10^{15} m$

distance to the star VY Canis Majoris: $3.09 × 10^8 AU$

$d_2 = 3.09\times 10^8 \times 1.5 \times 10^8 km$

$d_2 = 4.64 \times 10^{16} km$

distance to the galaxy Large Magellanic Cloud: 49976 pc

$1 pc = 3.262 ly = 3.262 \times 9.45 \times 10^{12} km$

$1pc = 3.08 \times 10^{13} km$

now we have

$d_3 = 49976 \times 3.08 \times 10^{13}$

$d_3 = 1.54 \times 10^{18} km$

distance to Neptune at the farthest: 4.7 billion km

$d_4 = 4.7 \times 10^9 km$

now the order of distance from least to greatest is as following

1. distance to Neptune at the farthest

2. distance of Betelgeuse

3. distance to the star VY Canis Majoris

4. distance to the galaxy Large Magellanic Cloud

6 0

3 years ago

A train reaches a speed of 35.0 m/s after accelerating at a rate of 5.00 m/s2 over a distance of 40.0 m. What was the train’s in

MatroZZZ [7]

Answer:

Initial velocity, U = 28.73m/s

Explanation:

Given the following data;

Final velocity, V = 35m/s

Acceleration, a = 5m/s²

Distance, S = 40m

To find the initial velocity (U), we would use the third equation of motion.

V² = U² + 2aS

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

S represents the displacement measured in meters.

Substituting into the equation, we have;

35² = U + 2*5*40

1225 = U² + 400

U² = 1225 - 400

U² = 825

Taking the square root of both sides, we have;

Initial velocity, U = 28.73m/s

5 0

3 years ago

Does every light source emit only one type of light?

blsea [12.9K]

Yes it is possible. Spectrum of emitted light depends upon the chemical composition of the source. and the way of its excitation. a clear example to us is that of sun.

3 0

2 years ago

In any problems involving circular motion, which way does the tangential speed vector point?

Anton [14]

In what may be one of the most remarkable coincidences in
all of physical science, the tangential component of circular
motion points along the tangent to the circle at every point.

The object on a circular path is moving in that exact direction
at the instant when it is located at that point in the circle. The
centripetal force ... pointing toward the center of the circle ...
is the force that bends the path of the object away from a straight
line, toward the next point on the circle. If the centripetal force
were to suddenly disappear, the object would continue moving
from that point in a straight line, along the tangent and away from
the circle.

4 0

3 years ago

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio

DanielleElmas [232]

Answer:

$\alpha =-2.2669642\times^{-10}rad/s^2$

Explanation:

Angular acceleration is defined by $\alpha =\frac{\Delta \omega}{\Delta t}=\frac{\omega_f-\omega_i}{\Delta t}$

Angular velocity is related to the period by $\omega=\frac{2\pi}{T}$

Putting all together:

$\alpha =\frac{\frac{2\pi}{T_f}-\frac{2\pi}{T_i}}{\Delta t}=\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})$

Taking our initial (i) point now and our final (f) point one year later, we would have:

$\Delta t=1\ year=(365)(24)(60)(60)s=31536000
s$

$T_i=0.0786s$

$T_f=0.0786s+7.03\times10^{-6}s$

So for our values we have:

$\alpha =\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})=\frac{2\pi}{31536000s}(\frac{1}{0.0786s+7.03\times10^{-6}s}-\frac{1}{0.0786s})=-2.2669642\times^{-10}rad/s^2$

Where the minus sign indicates it is decelerating.

8 0

3 years ago