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Trava [24]
3 years ago
12

Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light

by total internal reflection at the critical angle for the interface between the core (ncore = 1.482 ) and the cladding (ncladding = 1.44).
Suppose you wanted the largest angle at which total internal reflection occurred to be θmax = 5 degrees. What index of refraction does the cladding need if the core is unchanged?
Physics
1 answer:
motikmotik3 years ago
7 0

Answer:

n_cladding = 1.4764

Explanation:

We are told that θ_max = 5 °

Thus;

θ_max + θ_c = 90°

θ_c = 90° - θ_max

θ_c = 90° - 5°

θ_c = 85°

Now, critical angle is given by;

θ_c = sin^(-1) (n_cladding/n_core)

sin θ_c = (n_cladding/n_core)

n_cladding = (n_core) × sin θ_c

Plugging in the relevant values, we have;

n_cladding = 1.482 × sin 85

n_cladding = 1.4764

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Robin Hood wishes to split an arrow already in the bull's-eye of a target 40 m away.
tamaranim1 [39]

Answer:

5.843 m

Explanation:

suppose that the arrow leave the bow with a horizontal speed , towards he bull's eye.

lets consider that horizontal motion

distance = speed * time

time = 40/ 37 = 1.081 s

arrow doesnot have a initial vertical velocity component. but it has a vertical motion due to gravity , which may cause a miss of the target.

applying motion equation

(assume g = 10 m/s²)

s=ut+\frac{1}{2}*gt^{2}  \\= 0+\frac{1}{2}*10*1.081^{2}\\= 5.843 m

Arrow misses the target by 5.843m ig the arrow us split horizontally

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3 years ago
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77julia77 [94]
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umka2103 [35]

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6 0
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An enconomy car has a mass of 800 kg. What is the weight of the car?
Maksim231197 [3]

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A child of mass m is standing at the edge of a carousel. Both the carousel and the child are initially stationary. The carousel
suter [353]

Answer:

the angular velocity of the carousel after the child has started running =

\frac{2F}{mR} \delta t

Explanation:

Given that

the mass of the child = m

The radius of the disc = R

moment of inertia I = \frac{1}{2} mR^2

change in time = \delta \ t

By using the torque around the inertia ; we have:

T = I×∝

where

R×F = I × ∝

R×F = \frac{1}{2} mR^2∝

F = \frac{1}{2} mR∝

∝ = \frac{2F}{mR}           ( expression for angular  angular acceleration)

The first equation of motion of rotating wheel can be expressed as :

\omega = \omega_0  + \alpha  \delta t

where ;

∝ = \frac{2F}{mR}    

Then;

\omega = 0+ \frac{2F}{mR} \delta t

\omega =  \frac{2F}{mR} \delta t

 

∴ the angular velocity of the carousel after the child has started running =

\frac{2F}{mR} \delta t

7 0
3 years ago
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