Nothing can travel faster than the speed of light. As such, perceptions of objects and time change as they approach light speed, but the laws of physics remain consistent regardless of speed. Objects will appear shortened and time will appear to slow down around an observer approaching near light speeds, but all quantities still exist as they did before and all causality is preserved, even if observers in different points or traveling at different speeds will report different things.
We've never discovered any way for wavelengths of light to
appear to get longer after they leave the source, except for
the Doppler effect ... when the observer and the source of
the light are moving apart.
Since we don't know any other way for it to happen, whenever
we see a star or a galaxy whose light is red-shifted, we conclude
that it is moving away from Earth.
If you think about this for a few seconds, there are two things
about it that should blow your mind:
#1). "Red-shifted" means that the wavelengths of the light we see
are longer than the wavelengths of the light that left the star.
How do we know what the wavelength was when it left the star ?
#2). Our whole picture of the "expanding universe" ... which is a
big part of the foundation that supports the whole picture of the
"big bang" ... rests on the red-shift of light from distant galaxies,
and our conclusion that they are all moving away from us.
If we ever discover a DIFFERENT way that light can become
red-shifted, that may be IT for the expanding universe and the
big bang, both. They may both be out the window.
<span>Noble gases are elements found in the last column of the periodic table and are characterized by their reaction rate, or lack there of. Each element in that column does not react since they do not need to gain or lose electrons. So, helium has a "full" valence electron shell.</span>
The equation for work (W) done by an electric field is:
W = qΔV
where q is the magnitude of the charge and ΔV is the potential difference. The question gives you W and q, so plug n' play to find ΔV:
10 = 2ΔV
ΔV = 5