A light string is wrapped around the edge of the smaller disk, and a 1.50 kg block is suspended from the free end of the string.
1 answer:
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When light moves from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle above which there is no refracted ray, and it is given by:
(2)
where
is the refractive index of the second medium and 
is the refractive index of the first medium.
We can find the ratio
by using Snell's law:
(1)
where
is the angle of incidence
is the angle of refraction
By using the data of the problem and re-arranging (1), we find
and if we use eq.(2) we can now find the value of the critical angle:
where
We can find the ratio
where
By using the data of the problem and re-arranging (1), we find
and if we use eq.(2) we can now find the value of the critical angle:
Answer:
kg m/s
Explanation:
e = Charge = C
V = Voltage =
c = Speed of light = m/s
Momentum is given by
The unit of MeV/c in SI fundamental units is kg m/s