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White raven [17]
3 years ago
5

21.13 The index of refraction of corundum (Al2O3) is anisotropic. Suppose that visible light is passing from one grain to anothe

r of different crystal-lographic orientation and at normal incidence to the grain boundary. Calculate the reflectivity at the boundary if the indices of refraction for the two grains are 1.757 and 1.779 in the direction of light propagation.
Engineering
1 answer:
d1i1m1o1n [39]3 years ago
6 0

Answer:

3.87 x 10⁻⁵

Explanation:

Given parameters

normal incidence

refraction index of grain1, n₁ = 1.757

refraction index of grain2, n₂ = 1.779

to calculate reflectivity index between the two grains of different orientation an at normal incidence, we use the relation

R = [\frac{n2 - n1}{n2 + n1} ]^{2}

since the incidence is normal where R, is the index of relativity

R = [\frac{1.779 - 1.757}{1.779 + 1.757} ]^{2}

R = 3.87 x 10⁻⁵

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Answer:

The overflow rate is 4.24×10^-4 m/s.

The detention time is 7069.5 s

Explanation:

Overflow rate is given as volumetric flow rate ÷ area

volumetric flow rate = 0.3 m^3/s

area = πd^2/4 = 3.142×30^2/4 = 706.95 m^2

Overflow rate = 0.3 m^3/s ÷ 706.95 m^2 = 4.24×10^-4 m/s

Detention time = volume ÷ volumetric flow rate

volume = area × depth = 706.95 m^2 × 3 m = 2120.85 m^3

Detention time = 2120.85 m^3 ÷ 0.3 m^3/s = 7069.5 s

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3 years ago
A skull and crossbones pictogram indicates this kind of information about a chemical.
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Answer:

This pictogram indicates that the chemical is toxic and dangerous.

Explanation:

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2 years ago
A venturi meter is to be installed in a 63 mm bore section of a piping system to measure the flow rate of water in it. From spac
Alex Ar [27]

Answer:

Throat diameter d_2=28.60 mm

Explanation:

 Bore diameter d_1=63mm  ⇒A_1=3.09\times 10^{-3} m^2

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Flow rate Q=240 Lt/min⇒ Q=.004\frac{m^3}{s}

Coefficient of discharge C_d=0.8

We know that discharge through venturi meter

 Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}

h=x(\dfrac{S_m}{S_w}-1)

S_m=13.6 for Hg and S_w=1 for water.

h=0.235(\dfrac{13.6}{1}-1)

h=2.961 m

Now by putting the all value in

Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}

0.004=0.8\times \dfrac{3.09\times 10^{-3} A_2\sqrt{2\times 9.81\times 2.961}}{\sqrt{(3.09\times 10^{-3})^2-A_2^2}}

A_2=6.42\times 10^{-4} m^2

 ⇒d_2=28.60 mm

So throat diameter d_2=28.60 mm

     

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3 years ago
What do you do if building doesn't have enough water pressure for sprinklers? a. Increase pipe size b. Adjust budget to accommod
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Explanation:

the condition of education adda

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2 years ago
An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate (ϵ˙) is found to be
cupoosta [38]

Answer:

Activation energy for creep in this temperature range is Q = 252.2 kJ/mol

Explanation:

To calculate the creep rate at a particular temperature

creep rate, \zeta_{\theta} = C \exp(\frac{-Q}{R \theta} )

Creep rate at 800⁰C, \zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )

\zeta_{800} = C \exp(\frac{-Q}{1073R} )\\\zeta_{800} = 1 \% per hour =0.01\\

0.01 = C \exp(\frac{-Q}{1073R} ).........................(1)

Creep rate at 700⁰C

\zeta_{700} = C \exp(\frac{-Q}{R (700+273)} )

\zeta_{800} = C \exp(\frac{-Q}{973R} )\\\zeta_{800} = 5.5 * 10^{-2}  \% per hour =5.5 * 10^{-4}

5.5 * 10^{-4}  = C \exp(\frac{-Q}{1073R} ).................(2)

Divide equation (1) by equation (2)

\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\

Take the natural log of both sides

ln 18.182= 0.0000115Q\\2.9004 = 0.0000115Q\\Q = 2.9004/0.0000115\\Q = 252211.49 J/mol\\Q = 252.2 kJ/mol

3 0
3 years ago
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