The frequency of the light observed from the Earth is [tex]4.945\cdot 10^{14}Hz[/tex]
Explanation:
First of all, we start by noticing that the galaxy is receding from Earth (moving away): this means that according to the Doppler effect, the frequency of the light as seen from the Earth must be shorter than the real frequency of the light emitted by the galaxy.
Furthermore, we can quantify the change in frequency of the light using the following equation:
[tex]\frac{\Delta f}{f}=\frac{v}{c}[/tex]
where
[tex]\Delta f[/tex] is the change in frequency
f is the real frequency
v is the velocity of recession of the galaxy (negative if the galaxy is moving away)
c is the speed of light
In this problem, we have:
[tex]f=5.00\cdot 10^{14} Hz[/tex]
[tex]v=-3325 km/s = -3.325\cdot 10^6 m/s[/tex]
[tex]c=3\cdot 10^8 m/s[/tex]
Substituting and solving for [tex]\Delta f[/tex], we find
[tex]\Delta f = \frac{v}{c}f=\frac{-3.325\cdot 10^6}{3\cdot 10^8}(5.00\cdot 10^{14})=-5.542\cdot 10^{12} Hz[/tex]
And therefore, the frequency of the light observed from the Earth is
[tex]f'=f+\Delta f = 5.00\cdot 10^{14} +(-5.52\cdot 10^{12})=4.945\cdot 10^{14}Hz[/tex]
Learn more about frequency and waves:
brainly.com/question/5354733
brainly.com/question/9077368
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