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Airida [17]
3 years ago
9

An electromagnetic radio wave is received by a transmitter before it is converted to a sound wave. The radio wave has a waveleng

th of 23,076 m and a frequency of 13,000 Hz. What velocity is the wave traveling at before the transmitter converts it to a sound wave? 1. 76 m/s 0. 56 m/s 3. 31 × 102 m/s 3. 00 × 108 m/s.
Physics
1 answer:
snow_tiger [21]3 years ago
8 0

The velocity the wave was traveling at before the transmitter converts it to a sound wave is 3.0 x 10⁸ m/s.

<h3>What is Velocity of waves?</h3>

The velocity of waves is the rate of change of the wave's displacement with time.

The velocity of waves is determined by taking the product of the wave's frequency and wavelength.

v = fλ

v = 13,000 x 23,076

v = 3.0 x 10⁸ m/s.

Thus, the velocity the wave was traveling at before the transmitter converts it to a sound wave is 3.0 x 10⁸ m/s.

Learn more about velocity of waves here:  brainly.com/question/13867834

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a)E= U + K = \frac{1}{2}kx² +  \frac{1}{2}mv²

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The amplitude of the motion compares to the most extreme displacement of the mass-spring system. The displacement of the system, x(t), at time t, for a simple harmonic oscillator is given by,

x= Asin(ωt+∅)

where

amplitude  is 'A'

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c)v_{max}=\omega A<u></u>

When the elastic potential energy is zero, the maximum speed of the system occurs i.e U=0 and the kinetic energy is maximum, so:

U=0

E=\frac{1}{2}mv_{max}^2

According to the law of conservation of the mechanical energy, this energy must be equal to the energy of the system at its maximum displacement (1), so we can write

\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2

and solving for v_{max}we find an expression for the maximum speed:

v_{max}=\sqrt{\frac{kA^2}{m}}=\sqrt{\frac{k}{m}}A=\omega A

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