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RUDIKE [14]
3 years ago
5

A light beam is incident on a piece of fluorite (n = 1.434) at the Brewster's angle.

Physics
1 answer:
trasher [3.6K]3 years ago
5 0
The Brewster angle is the angle can be found through this equation
θ=arctan(n2/n1).
This came from Snell's Law:
n1sinθ=n2sin(90-θ)
          =n2cosθ
n2/n1=tanθ --- θ=arctan(n2/n1)=arctan(1.434)=55.11 degrees

Next, the angle of refraction can be found by using Snell's Law
n1sinθ=n2sinθ'
sin(55.11)=1.434sinθ'
sinθ'=sin(55.11)/1.434=0.572 --- θ'=arcsin(0.572)=34.89 degrees
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What is the equivalent resistance of a circuit that contains two 50.00
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Answer:

A

Explanation:

Resistors in series add. There is only one path the current can take. That's why Christmas Tree lights sometimes give a lot of trouble. If a bulb burns out, it could be any one of them and time is needed to find the burned out bulb.

That being the case R = R1 + R2

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Answer A

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Find the electric flux and the disp at t=0.50ns 
<span>Given: </span>
<span>Resistor R = 160 Ω </span>
<span>Voltage ε = 22.0 V </span>
<span>Capacitor C = 3.10 pF = 3.10 * 10^-12 F </span>
<span>time t = 0.5 ns = 0.5 * 10^-9 s </span>
<span>ε0 = 8.85 * 10^-12 </span>
<span>Solution: </span>
<span>ELECTRIC FLUX: </span>
<span>Φ = Q/ε0 </span>
<span>we have ε0, we need to find Q the charge </span>
<span>STEP 1: FIND Q </span>
<span>Q = C ε ( 1 - e^(-t/RC) ) </span>
<span>Q = { 3.10 * 10^-12 } { 22.0 } { 1 - e^(- 0.5 * 10^-9 / 160 *3.10 * 10^-12 ) } </span>
<span>Q = { 3.10 * 10^-12 } { 22.0 } { 1 - 0.365 } </span>
<span>Q = { 3.10 * 10^-12 } { 22.0 } { 0.635 } </span>
<span>Q = 43.31 * 10^-12 C </span>
<span>STEP 2: WE HAVE Q AND ε0 > >>> SOLVE FOR ELECTRIC FLUX >>> </span>
<span>Φ = Q/ε0 </span>
<span>Φ = { 43.31 * 10^-12 C } / { ε0 = 8.85 * 10^-12 } </span>
<span>Φ = 4.8937 = 4.9 V.m </span>
<span>DISPLACEMENT CURRENT </span>
<span>we use the following equation: </span>
<span>I = { ε / R } { e^(-t/RC) } </span>
<span>I = { 22 / 160 } { e^(- 0.5 * 10^-9 / 160 *3.10 * 10^-12 ) } </span>
<span>I = { 0.1375 } { 0.365 } </span>
<span>I = 0.0502 A = 0.05 A </span>
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