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Yuri [45]
3 years ago
9

A light ray travels in the +x direction and strikes a slanted surface with an angle of 62° between its normal and the ty axis. T

he surface is a clear solid with an index of refraction 1.33. What is the x component of the direction of the resulting reflected light ray?
Physics
1 answer:
Elena-2011 [213]3 years ago
6 0

Answer:

- 0.6

Explanation:

Given that angle between normal y axis is 62° so angle between  normal

and x axis will be 90- 62 = 28 °. Since incident ray is along x axis , 28 ° will be the angle between incident ray and normal ie it will be angle of incidence

Angle of incidence = 28 °

angle of reflection = 28°

Angle between incident ray and reflected ray = 28 + 28 = 56 °

Angle between x axis and reflected ray = 56 °

x component of reflected ray

= - cos 56 ( it will be towards - ve x axis. )

- 0.6

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You are driving your 1700 kg car at 21 m/s down a hill with a 5.0∘ slope when a deer suddenly jumps out onto the roadway. You sl
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A single insulated duct flow experiment using air operating at steady-state is performed in a lab. One measurement location (Sta
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Answer:

a) -0.0934 kJ/kg. K

b) The direction of flow is from right to left.

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A free flow diagram of the horizontal insulated duct is as shown below.

NOW,

Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:

\frac{\sigma_{cv}}{m}= (s_2-s_1) \geq  0 \ \ \ -------> \ \ \ 1

Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.

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Then, the flow is said to be  in the opposite direction

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s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \  ------->  \ \ \ 2

where;

s_1, s_2 , s^0(T_2), s^0(T1) are specific entropies

R = universal gas constant

P_1 = pressure at location 1

P_2 = pressure at location 2

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From the table A- 22

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s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )

s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97}  \ In (\frac{0.8}{0.95} )

s_2-s_1 = 1.68515 -1.8279+ 0.0493

s_2-s_1 =-0.0934 \  kJ/kg.K

Equating our result to equation (1)

s_2-s_1 \geq 0\\-0.0934 \leq 0

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

We therefore conclude that the direction of flow is from right to left.

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3 years ago
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