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Butoxors [25]
2 years ago
4

Suppose ocean waves are hitting a shore at a frequency of 20 waves per minute. Two swimmers are in the water. One swimmer says t

he frequency is 25 waves per minute and the other says the frequency is 15 waves per minute. How can the Doppler effect explain this apparent difference?
Physics
1 answer:
ratelena [41]2 years ago
8 0
<h3><u>Answer and explanation</u>;</h3>
  • One swimmer is swimming out to sea and is moving against the direction the waves are traveling, which causes the swimmer to encounter a wave 25 times per minute.
  • The other swimmer is swimming toward shore in the same direction the waves are traveling, so that swimmer encounters waves less frequently.
  • Therefore; swimmer who sees 25 waves a minute is swimming toward the waves, while the other swimmer is swimming with the waves. As you move toward a wave, the Doppler effect makes the frequency seen faster, and visa versa.
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A large mass collides with a stationary, smaller mass. How will the masses behave if the collision is inelastic?
Mamont248 [21]
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7 0
2 years ago
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Explain whether or not there is any difference between a light ray emitted by a candle flame and one reflected off the cover of
nexus9112 [7]

Answer:

the difference between the two is that the candle forms an emission spectrum and the book an absorption spectrum.

the book it is observed in all directions so that its reflection has to be diffused

Explanation:

The ray of light emitted by a candle is the light generated by the temperature of the flame, which is made up of the emissions of a black body at this temperature plus the emissions of the chemical elements that make up the candle.

The Light reflected from the cover of a book is the same incident light spectrum minus the wavelengths that create transitions in the elements of the cover, these wavelengths will be seen as dark areas.

As a consequence of the above, the difference between the two is that the candle forms an emission spectrum and the book an absorption spectrum.

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6 0
2 years ago
What is the threshold frequency for sodium metal if a photon with frequency 6.66 × 1014 s−1 ejects a photon with 7.74 × 10−20 J
FrozenT [24]

Answer:

5.5 × 10^14 Hz or s^-1

no orange light has less frequency so no photoelectric effect

Explanation:

hf = hf0 + K.E

HERE h is Planck 's constant having value 6.63 × 10 ^-34 J s

f is frequency of incident photon and f0 is threshold frequency

hf0 = hf- k.E

6.63 × 10 ^-34 × f0 = 6.63 × 10 ^-34× 6.66 × 10^14 - 7.74× 10^-20

6.63 × 10 ^-34 × f0 = 3.64158×10^-19

                           f0 = 3.64158×10^-19/ 6.63 × 10 ^-34

                           f0 = 5.4925 × 10^14

                            f0 =5.5 × 10^14 Hz or s^-1

frequency of orange light is 4.82 × 10^14 Hz which is less than threshold frequency hence photo electric effect will not be observed for orange light

8 0
2 years ago
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because he has more weight and is moving faster

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