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2 years ago

Sounds are reflected best when bounced off [answer] hard surfaces.

Physics

1 answer:

yuradex [85]2 years ago

3 0

Sounds are reflected best when bounced off the flat surfaces.

<h3>What is a sound wave?</h3>

A sound wave is produced when a medium begins to vibrate. When an entity vibrates, a pressure wave is formed, which causes sound.

Since sound waves cannot easily pass through hard surfaces, they bounce off them instead, reflecting back to the source which they first came from.

When sound is reflected off of flat surfaces, it does so most effectively.

Hence the flat surface is the correct answer.

To learn more about the sound wave refer to;

brainly.com/question/11797560

#SPJ1

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(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5

Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Given that,

Kinetic energy = 6.2 MeV

Radius = 0.500 m

We need to calculate the acceleration

Using formula of acceleration

$a=\dfrac{v^2}{r}$

Put the value into the formula

$a=\dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}mr}$

Put the value into the formula

$a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}$

$a=2.32\times10^{15}\ m/s^2$

We need to calculate the rate at which it emits energy because of its acceleration is

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Put the value into the formula

$\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}$

$\dfrac{dE}{dt}=3.00\times10^{-23}\ J/s$

The energy in ev/s

$\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s$

$\dfrac{dE}{dt}=1.875\times10^{-4}\ ev/s$

We need to calculate the fraction of its energy that it radiates every second

$\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}$

$\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}$

Hence, The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

5 0

3 years ago

The field inside a charged parallel-plate capacitor is __________.

Bess [88]

<span>Uniform. A parallel plate capacitor is a simple arrangement of electrodes and dielectric to form a capacitor. Here two parallel conductive plates are used as electrodes with a medium or dielectric in between them. Charge separation in a parallel-plate capacitor causes an internal electric field, which is uniform.</span>

5 0

4 years ago

Read 2 more answers

A truck, initially at rest, rolls down a frictionless hill and attains a speed of 20 m/s at the bottom. To achieve a speed of 40

Crank

Answer:

To achieve the velocity of 40 m/sec height will become 4 times

Explanation:

We have given initially truck is at rest and attains a speed of 20 m/sec

Let the mass of the truck is m

At the top of the hill potential energy is mgh and kinetic energy is $\frac{1}{2}mv^2$