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telo118 [61]
3 years ago
9

One of the dispersive components of an optical instrument is a 0.750-meter focal length monochromator equipped with the same gra

ting used in part (a). The grating measures 80 mm x 80 mm. Using the line/mm number calculated (a) find the first-order reciprocal linear dispersion, D-1, (in nm/mm) for this monochromator.
Engineering
1 answer:
SVETLANKA909090 [29]3 years ago
7 0

Answer:  (a) 1.11 nm/mm

Explanation:

Monochromator is a device which is used to transmits a delectable band of wavelengths of light wavelengths available at the input.

Assuming focal length monochromator  equipped with a                                    1200-groove/mm grating

The formula for the first-order reciprocal linear dispersion is

D^{-1}  = d/n*F where F is the focal length and n belongs to order of the first spectra.

D^{-1} = \frac{(1/1200) mm * 10^6 (nm/mm)/}{1 * 0.75 m * (103 mm/m) } = 1.11 nm/mm

Make sure you have used the conversion factors correctly as it will have a major impact on the calculation of the answer.

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Turpentine flows through a 12-nominal schedule 40 pipe. What is the flow rate that corresponds to a Reynolds number of 2000?
r-ruslan [8.4K]

Answer:

flow rate is 8.0385 × x^{-4} m³/s or 12.741 gpm

Explanation:

given data

12-nominal schedule 40 pipe

Reynolds number = 2000

to find out

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solution

we know the diameter of 12-nominal schedule 40 pipe is

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D = 0.32385 m

and

dynamic viscosity of Turpentine is = 0.001375 Pa-s

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so

Reynolds number is express as

Re = \frac{\rho*V*D}{\mu}

here ρ is density and D is diameter and V is velocity and µ is viscosity

so put here all value

2000 = \frac{870*V*0.3238}{0.001375}

V = 9.7619 × x^{-3} m/s

and

flow rate is

Q = V  × A

here A is area and Q is flow rate

Q = 9.7619 × x^{-3}  ×  \frac{\pi }{4} * 0.3238^2

Q = 8.0385 × x^{-4} m³/s

so flow rate is 8.0385 × x^{-4} m³/s or 12.741 gpm

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4 years ago
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Explanation:

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Answer:

Numerals

Dimensions

Extension Lines

Arrowheads

Dimension Figures

Isometric Dimensioning

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