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Nezavi [6.7K]
3 years ago
12

Blue light, which has a wavelength of about 475 nm, is made to pass through a slit of a diffraction grating that has 425 lines p

er mm and forms a first-order bright band. What is the angle of diffraction? Round answer to the nearest whole number.
Physics
2 answers:
tatiyna3 years ago
8 0
The formula relating the variables involved is:

distance between lines * sin(angle of diffraction) = order of band * wavelength

dsin(∅) = nλ

The line spacing may be calculated using the fact that there are 425 lines per mm and 1 mm contains 0.001 m

d = 0.001 / 425 = 2.35 x 10⁻⁶ m
λ = 475 x 10⁻⁹ m

Substituting the values,

2.35 x 10⁻⁶ * sin(∅) = 1 * 475 x 10⁻⁹ 
∅ = 11.7°


The angle of the first order bright band, or maxima, is 11.7°
andreyandreev [35.5K]3 years ago
4 0

round your answer to 12

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4 years ago
Suzie (of mass 42 kg) is roller-blading down the sidewalk going 32 miles per hour. She notices a group of workers down the walkw
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Explanation:

a) We have equation of motion, v = u + at

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     Initial velocity, u = 32 mph = 14.22 m/s

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Substituting

     0 = 14.22 + a x 2.2

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Suzie’s average acceleration = -6.46 m/s²

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3 years ago
During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s
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Answer:

part a : <em>The dry unit weight is 0.0616  </em>lb/in^3<em />

part b : <em>The void ratio is 0.77</em>

part c :  <em>Degree of Saturation is 0.43</em>

part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}

Here

  • \gamma_d is the dry unit weight which is to be calculated
  • γ is the bulk unit weight given as

                                              \gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3

  • w is the moisture content in percentage, given as 12%

Substituting values

                                              \gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3

<em>The dry unit weight is 0.0616  </em>lb/in^3<em />

Part b

Void Ratio

The void ratio is given as

                                                e=\frac{G_s \gamma_w}{\gamma_d} -1

Here

  • e is the void ratio which is to be calculated
  • \gamma_d is the dry unit weight which is calculated in part a
  • \gamma_w is the water unit weight which is 62.4 lb/ft^3 or 0.04 lb/in^3
  • G is the specific gravity which is given as 2.72

Substituting values

                                              e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766

<em>The void ratio is 0.77</em>

Part c

Degree of Saturation

Degree of Saturation is given as

S=\frac{G w}{e}

Here

  • e is the void ratio which is calculated in part b
  • G is the specific gravity which is given as 2.72
  • w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

                                      S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261

<em>Degree of Saturation is 0.43</em>

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}

Here

  • \gamma_{zav} is  the zero air unit weight which is to be calculated
  • \gamma_w is the water unit weight which is 62.4 lb/ft^3 or 0.04 lb/in^3
  • G is the specific gravity which is given as 2.72
  • w is the moisture content in percentage, given as 12% or 0.12 in fraction

                                      \gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\

Now as the volume is known, the the overall weight is given as

weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

4 0
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Answer:

see below :)

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