Answer: D.) increasing in distance from the earth
Explanation:
The <u>Doppler shift</u> is related to the Doppler effect and refers to the change in a wave perceived frequency (or <u>wavelength=color</u>) when the emitter of the waves, and the observer move relative to each other.
From there, it is deduced that the farther the object is, the more redshifted it is in its spectrum. For example, <u>as a galaxy moves away from the Earth, its espectrum turns towards the red and as the galaxy moves toward the Earth, its espectrum turns towards the blue.
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It should be noted that this effect bears its name in honor of the Austrian physicist Christian Andreas Doppler, who in 1842 proposed the existence of this effect for the case of light in the stars.
Write about some struggles you’ve faced and what they can do to avoid those struggles. Maybe also what you personally learned from the struggle too.
Answer:
Explanation:
a)
Using Ohms Law
Where J is the current density 
and the direction of E is the same as the direction of the current. Since J is uniform throughout the conductor 
just inside at a radius a (and anywhere else).
b)
Since we have no changing electric fields we can use Ampere’s law in it’s simplest form without displacement current
such that
and by the right hand rule, since the current is going to the right, the magnetic field is circling around the conductor such that it’s pointing out of the page at the top and into the page at the bottom.
c)
The Poynting vector is given by
and by the right hand rule it’s always pointing in towards the center of the conductor.
d)
Note: directions of these three vectors are mentioned along with their magnitudes in above 3 parts a , b and c
Answer:
ω, the angular frequency of the source equals 377 rad/s
Explanation:
From the question, V(t) = V cosωt.
Now, ω = the angular frequency of the sinusoidal wave is given by
ω = 2πf where f = the frequency of the source = 60 Hz
So, the angular frequency of the source ,ω = 2π × the frequency of the source.
So, ω = 2πf
ω = 2π × 60 Hz
ω = 120π rad/s
ω = 376.99 rad/s
ω ≅ 377 rad/s
So, ω, the angular frequency of the source equals 377 rad/s
