Answer:
The angle of incidence is greater than the angle of refraction
Explanation:
Refraction occurs when a light wave passes through the boundary between two mediums.
When a ray of light is refracted, it changes speed and direction, according to Snell's Law:
where
:
is the index of refraction of the 1st medium
is the index of refraction of the 2nd medium
is the angle of incidence (the angle between the incident ray and the normal to the boundary)
is the angle of refraction (the angle between the refracted ray and the normal to the boundary)
In this problem, we have a ray of light passing from air into clear plastic. We have:
(index of refraction of air)
approx. (index of refraction in clear plastic)
Snell's Law can be rewritten as
And since 
, we have
And so
Which means that
The angle of incidence is greater than the angle of refraction
A camera flash or lighting bolt because stored separated positive and negative charges --> caused them to do work by briefly lighting a bulb as the separated charges moved back together
The Answer Is A
Hope This Helps !
The statement about pointwise convergence follows because C is a complete metric space. If fn → f uniformly on S, then |fn(z) − fm(z)| ≤ |fn(z) − f(z)| + |f(z) − fm(z)|, hence {fn} is uniformly Cauchy. Conversely, if {fn} is uniformly Cauchy, it is pointwise Cauchy and therefore converges pointwise to a limit function f. If |fn(z)−fm(z)| ≤ ε for all n,m ≥ N and all z ∈ S, let m → ∞ to show that |fn(z)−f(z)|≤εforn≥N andallz∈S. Thusfn →f uniformlyonS.
2. This is immediate from (2.2.7).
3. We have f′(x) = (2/x3)e−1/x2 for x ̸= 0, and f′(0) = limh→0(1/h)e−1/h2 = 0. Since f(n)(x) is of the form pn(1/x)e−1/x2 for x ̸= 0, where pn is a polynomial, an induction argument shows that f(n)(0) = 0 for all n. If g is analytic on D(0,r) and g = f on (−r,r), then by (2.2.16), g(z) =
Answer:
66.6 or it could be 66.66 one of those
Explanation:
