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nordsb [41]
2 years ago
7

A fiber-optic rod consists of a central strand of material surrounded by an outer coating. the interior portion of the rod has a

n index of refraction of 1.55. if all rays striking the interior walls of the rod with incident angles greater than 59.5° are subject to total internal reflection, what is the index of refraction of the coating?
Physics
1 answer:
algol132 years ago
5 0

Answer:

the index of refraction of the coating is 1.33  

Explanation:

Given the data in the question;

refraction index of  interior portion of the rod η_{interior = 1.55

angle of incidence θ_i = 59.5°

From Snell's law, we know that;

η_{interior × sinθ_i  = η_{coating × sinθ_r

where η_{interior  is the index of refraction of the rod ( material 1 )

θ_i  is the angle of incidence

η_{coating is the index of refraction in outer coating ( material 2 )

θ_r is the angle of refraction

so we substitute our values into the equation;

η_{interior × sinθ_i  = η_{coating × sinθ_r

1.55 × sin( 59.5° )  = η_{coating × sin( 90° )

1.55 × 0.861629  = η_{coating × 1

1.3355 = η_{coating × 1

η_{coating = 1.33   { 2 decimal places }

Therefore, the index of refraction of the coating is 1.33  

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Two protons (each with q = 1.60 x 10-19)
otez555 [7]

Answer:

230.4 N

Explanation:

From the question given above, the following data were obtained:

Charge (q) of each protons = 1.6×10¯¹⁹ C

Distance apart (r) = 1×10¯¹⁵ m

Force (F) =?

NOTE: Electric constant (K) = 9×10⁹ Nm²/C²

The force exerted can be obtained as follow:

F = Kq₁q₂ / r²

F = 9×10⁹ × (1.6×10¯¹⁹)² / (1×10¯¹⁵)²

F = 9×10⁹ × 2.56×10¯³⁸ / 1×10¯³⁰

F = 2.304×10¯²⁸ / 1×10¯³⁰

F = 230.4 N

Therefore, the force exerted is 230.4 N

5 0
2 years ago
When temperature increases the kinetic energy of the particles in a material, true or false?
saul85 [17]

Answer:

True

Explanation:

kinetic energy is proportional to temperature

4 0
3 years ago
A 250 kg flatcar 25 m long is moving with a speed of 3.0 m/s along horizontal frictionless rails. A 61 kg worker starts walking
Anna11 [10]

Answer:

x=31.09m

Explanation:

p1=p2

The momentum of flatcar and the momentum of the worker so

The velocity of the worker is:

m_{f}*v_{f}=m_{w}*v_{w}\\\\v_{f}=\frac{m_{f}*v_{f}}{m_{w}}\\v_{f}=\frac{61kg*3.0\frac{m}{s}}{250kg}\\v_{f}=0.732\frac{m}{s}

The total motion has a total velocity and is

Vt=v_{w}+v_{f}\\Vt=0.732\frac{m}{s}+3.0\frac{m}{s}\\Vt=3.732\frac{m}{s}

The time the worker take walking is

t=\frac{x}{v_{w}}\\t=\frac{25m}{3\frac{m}{s}}=8.33s

Now the total time and the total velocity determinate the motion of tha flatcar how far has moved

x=t*Vt\\x=8.33s*3.732\frac{m}{s} \\x=31.09m

5 0
2 years ago
two electrons are an angstrom (1x10^-10m) apart. What electrostatic force do they exert on one another?
Irina-Kira [14]

Answer:

2.30 × 10⁻⁸ N if the two electrons are in a vacuum.

Explanation:

The Coulomb's Law gives the size of the electrostatic force F between two charged objects:

\displaystyle F = -\frac{k\cdot q_1 \cdot q_2}{r^{2}},

where

  • k is coulomb's constant. k = 8.99\times 10^{8}\;\text{N}\cdot\text{m}^{2}\cdot\text{C}^{-2} in vacuum.
  • q_1 and q_2 are the signed charge of the objects.
  • r is the distance between the two objects.

For the two electrons:

  • q_1 = q_2 = 1.60\times 10^{-19}\;\text{C}.
  • r = 1\times 10^{-10}\;\text{m}.
  • \displaystyle F = -\frac{k\cdot q_1 \cdot q_2}{r^{2}} = -\frac{8.99\times 10^{8}\times (1.60\times 10^{-19})^{2}}{(1\times 10^{-10})^{2}} = 2.30\times 10^{-8}\;\text{N}.

The sign of F is negative. In other words, the two electrons repel each other since the signs of their charges are the same.

8 0
2 years ago
Paragraph about how the constellations were used by ancient civilizations.
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Answer:

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Explanation:

8 0
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