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mario62 [17]
3 years ago
15

(PLEASE HELP) Identify the following as electromagnetic (E) or mechanical (M) waves.

Physics
2 answers:
tatyana61 [14]3 years ago
3 0
Sound waves (m)

water waves (m)

radio waves (e)

ultraviolet (e)

waves in a wheat field (m)
EastWind [94]3 years ago
3 0
<h3><u>Answer;</u></h3>

<u>Mechanical waves; </u>

  • sound waves
  • waves in a wheat field
  • water waves

<u> Electromagnetic waves</u>

  • Radio waves
  • Ultraviolet
<h3><u> Explanation;</u></h3>
  • A wave is a transmission of a disturbance from one place, the source to another.
  • Waves can be classified as mechanical waves or electromagnetic waves depending on the requirement of a material medium.
  • <em><u>Mechanical waves </u></em><em><u>are those waves that require a material medium for transmission, they include, sound waves, water waves, etc. The transmission of these waves is by vibration of particles in the medium.</u></em>
  • <em><u>Electromagnetic waves</u></em><em><u> are those waves that do not require a material medium for transmission.They include, radio waves, microwaves, and all the waves in the electromagnetic spectrum.</u></em>

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<u>Explanation:</u>

<u>Given:</u>

speed of sound - 343 m/s

You detect a frequency that is 0.959 times as small as the frequency emitted by the car when it is stationary. So, it can be written as,

f^{\prime}=0.959 f

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If there is relative movement between an observer and source, the frequency heard by an observer differs from the actual frequency of the source. This changed frequency is called the apparent frequency. This variation in frequency of sound wave due to motion is called the Doppler shift (Doppler effect). In general,

                     f^{\prime}=\left(\frac{v+v_{0}}{v-v_{s}}\right) \times f

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f^' - Observed frequency

f – Actual frequency

v – Velocity of sound waves

v_0 – Velocity of observer

v_s - velocity of source

When source moves away from an observer at rest (v_{0} = 0), the equation would be

                        f^{\prime}=\left(\frac{v}{v-\left(-v_{s}\right)}\right) \times f

                        f^{\prime}=\left(\frac{v}{v+v_{s}}\right) \times f

                        \frac{f^{\prime}}{f}=\left(\frac{v}{v+v_{s}}\right)

By substituting the known values, we get

            0.959=\left(\frac{343}{343+v_{s}}\right)

           0.959\left(343+v_{s}\right)=343

           0.959(343)+0.959\left(v_{s}\right)=343

           328.937+0.959 v_{s}=343

           0.959 v_{s}=343-328.937=14.063

           v_{s}=\frac{14.063}{0.959}=14.66 \mathrm{m} / \mathrm{s}

Approximately 15 m/s is the speed of the car.

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