If an electric circuit has 0 A of current running through it then .
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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:
First, create an illustration of the motion of the two cars as shown in the attached picture. The essential equations used are:
For constant acceleration:
a = v,final - v,initial /t
d = v,initial*t + 1/2*at²
For constant velocity:
d = constant velocity*time
The solutions is as follows:
a = v,final - v,initial /t
3.8 = (v₁ - 0)/4.6 s
v₁ = 17.48 m/s
Total distance = d1 + d2 + d3
d1 = d = v,initial*t + 1/2*at²
d2 = constant velocity*time
Total distance = 0*(4.6) + 1/2*(3.8)(4.6)² + (17.48)(9.2) + d3= 257.71
d3 = 56.69 m
For constant acceleration:
a = v,final - v,initial /t
d = v,initial*t + 1/2*at²
For constant velocity:
d = constant velocity*time
The solutions is as follows:
a = v,final - v,initial /t
3.8 = (v₁ - 0)/4.6 s
v₁ = 17.48 m/s
Total distance = d1 + d2 + d3
d1 = d = v,initial*t + 1/2*at²
d2 = constant velocity*time
Total distance = 0*(4.6) + 1/2*(3.8)(4.6)² + (17.48)(9.2) + d3= 257.71
d3 = 56.69 m