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svetoff [14.1K]
3 years ago
9

How many wavelengths are seen in this image of a sound wave? three four six ten

Physics
2 answers:
Anuta_ua [19.1K]3 years ago
7 0
Four water levels are shown
qwelly [4]3 years ago
3 0

There are four wavelengths seen in the image of the sound wave.

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If you were on a ship at sea, and a tsunami passed under your ship, what would probably be your reaction? explain.
swat32
<span>Extremely powerful single waves have no effect on ships at sea since the depth of water allows the energy to be distributed over hundreds and thousands of feet. In deep water, the bigger the wave, the faster it moves and the slower the surface changes height. As the wave gets into shallow waters, it slows down and can start to pile up to large heights.</span>
6 0
2 years ago
What is the wavelength (in air) of a 13.0 khz vlf radio wave?
bonufazy [111]
Since,
Speed = Frequency * WaveLength

=> WaveLength = Speed / Frequency --- (A)

Frequency = 13.0 kHz.
As the radio waves are electromagnetic waves, their speed is equals to the speed of light. Therefore,
Speed = C = 3 *  10^{8}  m/ s^{2}

Plug in the values in equation(A):

A => WaveLength = \frac{3 * 10^{8}}{13*10^{3}}

Ans: Wavelength = 23.077 kilometers.
-i
5 0
3 years ago
What kind of wave is this?
torisob [31]

Answer:

I think its transverse

Explanation:

6 0
3 years ago
:What will be the value of the refractive index of the medium? Critical angle of that medium is 30 degree
earnstyle [38]

Answer:

Let the second medium be air (n₁=1)

The refractive index n₂ of the medium where first medium is air is found (a)

(a) n₂ = 2

Explanation:

Critical angle can be defined as the angle of incidence that provides the angle of refraction of 90°.

Refractive index of a medium can be defined as a number that describes that how fast a light will travel through that medium.

Critical angle and Refractive index are related by:

\theta_{critical}= sin^{-1}(\frac{n_1}{n_2})

sin \theta_{critical}=\frac{n_1}{n_2}

To find refractive index of medium with respect to air, substitute n₁=1 (Refractive index of air is 1)

Also θ(critical)=30°

Find n₂ :

sin30= \frac{1}{n_2}\\0.5=\frac{1}{n_2}\\n_2=\frac{1}{0.5}\\n_2=2

8 0
2 years ago
An electrical motor spins at a constant 1975.0 rpm. If the armature radius is 7.112 cm, what is the acceleration of the edge of
stealth61 [152]

Answer:

Option D is the correct answer.

Explanation:

Since value of angular acceleration is constant, the body has only centripetal acceleration.

Centripetal acceleration

               a=\frac{v^2}{r}=\frac{(r\omega )^2}{r}=r\omega ^2

We have radius = 7.112 cm = 0.07112 m

Frequency, f = 1975 rpm = 32.92 rps

Angular frequency, ω = 2πf = 2 x π x 32.92 = 206.82 rad/s

Substituting in centripetal acceleration equation,

              a=r\omega ^2=0.07112\times 206.82^2=3042.17m/s^2

Option D is the correct answer.                

3 0
3 years ago
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