All categories

3 years ago

How are sound waves similar to light waves?

Physics

2 answers:

arlik [135]3 years ago

7 0

Answer:

When light moves from one medium (like air) to another medium (like water) it will change directions. This is a "wave-like" behavior and is called refraction. In this way light behaves like other waves such as sound waves. The speed of the light wave also changes when it moves from medium to medium.

Explanation:

Tom [10]3 years ago

6 0

Answer:

Sound and light are similar in that both are forms of energy that travel in waves.

Explanation:

They both have properties of wavelength, freqency and amplitude. ... Sound can only travel through a medium (substance) while light can travel through empty space. Sound is a form of mechanical energy caused by vibrations of matter.

You might be interested in

A block of mass 0.490 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. T

Dennis_Churaev [7]

Answer:

The value of x is $x = 0.1955\ m$

The value of the velocity at the top is $v_{top} = 2.25 \ m/s$

Explanation:

The diagram of this process is on the first uploaded image

From the question we are told that

The mass of the block is $m_b = 0.490kg$

The distance of compression is $x$

The force constant is $k = 450 N/m$

The radius of the circular track is $R =1.00m$

The speed at the bottom of the track is $v_A = 13.4 \ m/s$

The frictional force experienced is $F_f = 7.00 \ N$

Now looking at this process we see that the potential energy of the spring is been transformed into the kinetic energy of the block . So,

$PE \ of \ spring = KE \ of \ block$

mathematically i.e

$\frac{1}{2}k x^2 = \frac{1}{2} m_bv^2_A$

$0.5 * 450 * x^2 = 0.5 * 0.490 * 13.4^2$

Making x the subject of the formula

$x = \sqrt{\frac{0.5 * 0.490 * 13.4^2}{ 0.5 * 450} }$

$= 0.1955\ m$

The average workdone by friction is $W_f = F_f * \pi$

Here $\pi is \ the\ net\ displacement$

$W_f = 7 * 3.142$

$= 21.99 J$

The kinetic energy at the bottom is

$KE = \frac{1}{2} * 0.490 * 13.4^2$

$= 43.99 \ N$

The potential energy gained at the top of the circle is

$PE = m_bgh$

Here h is the height which is equal to d(diameter) = 2r = 2 × 1 = 2 m and

g is acceleration due to gravity $= 9.8m/s^2$

Now substituting values

$PE = 0.490 * 9.8 *2$

$=9.604 J$

Since energy is can not be created nor destroyed but transformed according to first law of thermodynamics

$KE_{at \bottom} = PE _ {\ at \ top } + W_f + KE_{\ at \ top}$

$43.99 = 9.604 +21.99 + [\frac{1}{2} m_b * v_{top} ^2]$

$12.369= [2.45* v_{top} ^2]$

$v_{top} =\sqrt{\frac{12.369}{2.45} }$

$= 2.25 \ m/s$

8 0

3 years ago

A 3 kg block is released from rest to slide down a ramp with friction. The length that the block slides is 2 meters and the angl

Arte-miy333 [17]

Answer:

(a). The acceleration is 3.20 m/s².

(b). The amount of work done by each force is 19.2 J.

(c). The total work done on the block is 19.2 J.

(d). The final speed of the block is 6.26 m/s.

Explanation:

Given that,

Mass of block = 3 kg

Distance = 2 m

Angle = 30°

Coefficient of kinetic friction = 0.20

(a). We need to calculate the acceleration

Using balance equation of force

$N = mg\co\theta$

$mg\sin\theta-f_{k}=ma$

$a = g\sin\theta-\mu g\cos30$

Put the value into the formula

$a=g(\sin30-0.20\cos30)$

$a=9.8(\dfrac{1}{2}-0.20\times\dfrac{\sqrt{3}}{2})$

$a=3.20\ m/s^2$

The acceleration is 3.20 m/s².

(b). We need to calculate the amount of work done by each force

Using formula of work done

Normal force is

$N = mg\cos30$

So due to this the net force is zero then the no work done by reaction force.

By another force,

$W= F\times d$

$W=ma\times d$

Put the value into the formula

$W= 3\times3.20\times2$

$W=19.2\ J$

The amount of work done by each force is 19.2 J.

(c). We need to calculate the total work done on the block

The total work done on the block is 19.2 J.

(d). We need to calculate the final speed of the block

Using equation of motion

$v^2=u^2+2gs$

Put the value into the formula

$v^2=0+2\times9.8\times2$

$v=\sqrt{2\times9.8\times2}$

$v=6.26\ m/s$

The final speed of the block is 6.26 m/s.

Hence, This is the required solution.

4 0

3 years ago

What is defect of vision

Anuta_ua [19.1K]

Answer:

The vision becomes blurred due to the refractive defects of the eye. There are mainly three common refractive defects of vision. These are (i) myopia or near-sightedness, (ii) Hypermetropia or far – sightedness, and (iii) Presbyopia. These defects can be corrected by the use of suitable spherical lenses.

4 0

3 years ago

Convection currents occur when _________ energy transfers between two parts of a fluid

valentinak56 [21]

Answer:

heat

Explanation:

7 0

3 years ago

A 66.0 kg diver is 3.10 m above the water, falling at speed of 6.10 m/s. Calculate her kinetic energy as she hits the water. (Ne

castortr0y [4]

Kinetic energy as she hits the water is 3300 joule.

To find the answer, we need to know about the Newton's equation of motion.

<h3>What's the Newton's equation of motion to determine the final velocity?</h3>

The final velocity is determined as

V²=U²+2aS

V= final velocity, U= initial velocity, a= acceleration and S= distance

<h3>What's the final velocity of the driver falling from 3.10m with initial velocity of 6.10m/s?</h3>

Here, a= 9.8m/s², U= 6.10m/s and S= 3.10m
So, V²= 6.1²+2×9.8×3.10= 98
V= √98= 10m/s

<h3>What's the kinetic energy of the driver when touches the water?</h3>

Kinetic energy= 1/2×mass×velocity²

= 1/2 × 66 × 10²

= 3300J

Thus, we can conclude that the kinetic energy of the driver is 3300 Joule.

Learn more about the kinetic energy here:

brainly.com/question/25959744

#SPJ4

3 0

2 years ago