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neonofarm [45]
3 years ago
12

For 589-nm light, calculate the critical angle for the following materials surrounded by air. (a) fluorite (n = 1.434) ° (b) cro

wn glass (n = 1.52) ° (c) ice (n = 1.309)
Physics
1 answer:
Simora [160]3 years ago
4 0

Answer:

a) θc = 44.21⁰.

b) θc = 41.14⁰

c)  θc = 40.36⁰

Explanation:

sin θ = \dfrac{n_2}{n_1}

For air,

n₂ = 1

so,

sin θ =  \dfrac{1}{n_1}

(a) fluorite : n₁ = 1.434⁰

sin θ = 0.697                   θc = 44.21⁰.

(b) crown glass: n₁ = 1.52⁰  

sin θ = 0.657                   θc = 41.14⁰.

(c) sodium chloride : n₁= 1.544⁰

sin θ = 0.647                   θc = 40.36⁰.

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andrezito [222]

Let's check the relationship

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4 0
2 years ago
A particular car engine operates between temperatures of 440°C (inside the cylinders of the engine) and 20°C (the temperature of
Step2247 [10]

One of the concepts to be used to solve this problem is that of thermal efficiency, that is, that coefficient or dimensionless ratio calculated as the ratio of the energy produced and the energy supplied to the machine.

From the temperature the value is given as

\eta = 1-\frac{T_L}{T_H}

Where,

T_L = Cold focus temperature

T_H = Hot spot temperature

Our values are given as,

T_L = 20\° C = (20+273) K = 293 K

T_H = 440\° C = (440+273) K = 713 K

Replacing we have,

\eta = 1-\frac{T_L}{T_H}

\eta = 1-\frac{293}{713}

\eta = 0.589

Therefore the maximum possible efficiency the car can have is 58.9%

4 0
3 years ago
A tube of mercury with resistivity 9.84 × 10 -7 Ω ∙ m has an electric field inside the column of mercury of magnitude 23 N/C tha
slava [35]

Answer:

The current through the tube is 73.39A.

Explanation:

The relationship between the resistivity \rho, the electric field E, and the current density J is given by

\rho = \dfrac{E}{J}

This equation can be solved for J to get:

J = \dfrac{E}{\rho}

Since the current is I = J\cdot A

I= J\cdot A  = \dfrac{E}{\rho} \cdot A

Now, for the tube of mercury \rho = 9.84*10^{-7}\: \Omega \cdot m, E = 23N/C, and the area is A = \pi r^2 = \pi (1.0*10^{-3}m)^2 = 3.14*10^{-6}m^2; therefore,

I= \dfrac{23N/C}{9.84*10^{-7}\Omega\cdot m } *3.14*10^{-6}m^2

\boxed{I = 73.39A.}

Hence, the current through the mercury tube is 73.39A.

5 0
3 years ago
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Tpy6a [65]
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7 0
3 years ago
What frequency corresponds to a period of 4.31s.<br> T =1/f = 1/4.31s = 0.232hz correct?
Korvikt [17]

Answer:correct

Explanation: Period T is the reciprocal of frequency (i.e T=1/f)

Frequency is the reciprocal of period (i.e F= 1/T)

Therefore if T=4.31s

Frequency F= 1/4.31s=0.232hz

8 0
3 years ago
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