Which transformation will map figure L onto figure L'? (1 point) Two congruent triangles Figure L and Figure L prime are drawn o
1 answer:
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Answer:
is reflected back into the region of higher index
Explanation:
Total internal reflection is a phenomenon that occurs when all the light passing from a region of higher index of refraction to a region of lower index is reflected back into the region of higher index.
According to Snell's law, refraction of ligth is described by the equation
where
n1 is the refractive index of the first medium
n2 is the refractive index of the second medium
is the angle of incidence (in the first medium)
is the angle of refraction (in the second medium)
Let's now consider a situation in which
so light is moving from a medium with higher index to a medium with lower index. We can re-write the equation as
Where is a number greater than 1. This means that above a certain value of the angle of incidence
, the term on the right can become greater than 1. So this would mean
But this is not possible (the sine cannot be larger than 1), so no refraction occurs in this case, and all the light is reflected back into the initial medium (total internal reflection). The value of the angle of incidence above which this phenomen occurs is called critical angle, and it is given by
Answer:
The answer is below
Explanation:
Given that:
x(t) = at – bt2+c
a) x(t) = at – bt2+c
Substituting a = 1.4 m/s, b = 0.06 m/s2 and c =50 m gives:
x(t) = 1.4t - 0.06t² + 50
At t = 5, x(5) = 1.4(5) - 0.06(5)² + 50 = 55.5 m
At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m
The average velocity (v) is given as:
b) x(t) = 1.4t - 0.06t² + 50
At t = 10, x(10) = 1.4(10) - 0.06(10)² + 50 = 58 m
At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m
The average velocity (v) is given as:
c) x(t) = 1.4t - 0.06t² + 50
At t = 15, x(5) = 1.4(15) - 0.06(15)² + 50 = 57.5 m
At t = 10, x(10) = 1.4(10) - 0.06(10)² + 50 = 58 m
The average velocity (v) is given as: