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olasank [31]
3 years ago
5

Consider a diffraction grating of width 5cm with slits of width 0.0001 cm separated by distance of 0.0002cm 1:what is the corres

ponding grating element 2:how many orders will be observed at a wavelength of 0.000055 cm 3:calculate the dispersion in the different orders
Physics
1 answer:
sashaice [31]3 years ago
3 0

Answer:

a) d= 0.0002cm

b) m=3

Explanation:

From the question we are told that:

Width of diffraction grating W_g= 5cm

Width of silt w_s 0.0001 cm

Distance d= 0.0002cm

a)

Generally the grafting element is the distance b/w  the silts

d= 0.0002cm

Therefore the grafting element is

d= 0.0002cm

b)

Generally the equation for diffraction state is mathematically given by

 dsin\theta=m\lambda

Since

 sin\theta \leq 1

Therefore

 m \leq \frac{d}{\lambda}

 m \leq \frac{0.0002}{0.000055}

 m \leq 3.6

Therefore orders observed with \lambda is

 m=3

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Zepler [3.9K]

Answer:

The angular magnification is M = 2.808

Explanation:

From the question we are told

           The focal length is  f = 8.9cm

          The near point is n_p = 25.0cm

The angular magnification is mathematically represented as

                          M = \frac{n_p}{f}

Substituting values

                        M = \frac{25}{8.9}

                           = 2.808

4 0
3 years ago
If the humidity in a room of volume 410 m3 at 25 ∘C is 70%, what mass of water can still evaporate from an open pan? Express you
Keith_Richards [23]

Answer:

m=1864.68\ g

Explanation:

Given:

volume of air in the room, V=410\ m^3

temperature of the room, T=25+273=298\ K

<u>Saturation water vapor pressure at any temperature T K is given as:</u>

<u />p_{sw}=\frac{e^{(77.3450 + 0.0057\times T - \frac{ 7235}{T}  )}}{T^{8.2}}<u />

putting T=298 K we have

p_{sw}=3130\ Pa

<u>The no. of moles of water molecules that this volume of air can hold is:</u>

Using Ideal gas law,

P.V=n.R.T

n=\frac{P_{sw}.V}{R.T}

n=\frac{3130\times 410}{8.314\times 298}

n=518\ moles is the maximum capacity of the given volume of air to hold the moisture.

Currently we have 80% of n, so the mass of 20% of n:

m=(20\%\ of\ n)\times M}

where;

M= molecular mass of water

m=0.2\times 518\times 18

m=1864.68\ g is the mass of water that can vaporize further.

3 0
2 years ago
One of the fastest recorded pitches major-league baseball, thrown by nolan ryan in 1974, was clocked at 100.8 mi/hr. If a pitch
lawyer [7]

<u>Answer:</u>

 The ball fall vertically 2.69 ft by the time it reached home plate 60.0 ft away.

<u>Explanation:</u>

    Fastest recorded pitches major-league baseball, thrown by nolan ryan in 1974 = 100.8 mi/hr = 44.8 m/s

    The horizontal distance to home plate = 60.0 ft = 18.288 m

  We have the horizontal velocity = 44.8 m/s

    So time taken = 18.288/44.8 = 0.408 seconds.

  The distance traveled by baseball vertically is found out by equation s=ut+\frac{1}{2} at^2

  Here u =0m/s, a = 9.81 m/s^2 and t = 0.408 s

 Substituting

     S = 0*0.408+\frac{1}{2} *9.81*0.408^2 = 0.82 m\\ \\S = 0.82 m= 2.69 ft

So vertical distance traveled = 0.82 m = 2.69 ft

7 0
2 years ago
Read 2 more answers
A particle moves along the x axis. It is initially at the position 0.180 m, moving with velocity 0.060 m/s and acceleration -0.3
shusha [124]

Answer:

a

  x_2 = -2.3356

b

 v = -1.384 \ m/s

Explanation:

From the question we are told that

  The initial position of the particle is  x_1 = 0.180 \ m

  The initial  velocity of the particle is  u = 0.060  \  m/s

  The acceleration is   a = -0.380 \  m/s^2

   The time duration is  t = 3.80  \ s

Generally from kinematic equation

    v = u + at

=>  v = 0.060 + (-0.380 * 3.80)

=>  v = -1.384 \ m/s

Generally from kinematic equation

   v^2 = u^2 + 2as

Here s is the distance covered by the particle, so

   (-1.384)^2 = (0.060)^2 + 2(-0.380)* s

=>  s = -2.5156 \ m

Generally the final position of the particle is  

    x_2 = x_1 + s

=>   x_2 = 0.180 + (-2.5156)

=>   x_2 = -2.3356

6 0
3 years ago
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