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-BARSIC- [3]
2 years ago
11

A lab technician uses laser light with a wavelength of 670 nm to test a diffraction grating. When the grating is 40.0 cm from th

e screen, the first-order maxima appear 6.00 cm from the center of the pattern. How many lines per millimeter does this grating have?
Physics
1 answer:
Juliette [100K]2 years ago
8 0

Answer:

N = 221.4 lines / mm

Explanation:

Given:

- The wavelength of the source λ = 670 nm

- Distance of the grating from screen B = 40.0 cm

- The distance of first bright fringe from central order P = 6.0 cm

Find:

How many lines per millimeter does this grating have?

Solution:

- The derived results from Young's experiment that relates the order of bright fringes about the central order is given by:

                                          sin (Q) = n*λ*N

Where,

n is the order number 0, 1 , 2, 3 , ....

λ  is the wavelength of the light source

Q is the angle of sweep respective fringe from central order

N is the number of lines/mm the grating has

- We will first compute the length along which the light travels for the first bright fringe:

                                            L^2 = P^2 + B^2

                                            L^2 = 40^2 + 6^2

                                            L^2 = 1636

                                            L = 40.45 cm  

- Now calculate the sin(Q) that the fringe makes with the central order:

                                            sin (Q) = P / L

                                            sin (Q) = 6 / 40.45

- Now we will use the derived results:

                                           N = sin(Q) / n*λ

  Where, n = 1 - First order

  Plug values in                N = (6 / 40.45) / (670 *10^-6)

                                          N = 221.4 lines / mm

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A charge of 5.67 x 10-18 C is placed 3.5 x 10 m away from another charge of - 3.79 x 10 "C
miskamm [114]

Answer:

1. 579 x 10 ^-22N

Explanation:

F = kq1q2/r^2

   = 9.0 x 10^9 x 5.67 x 10^-18 x 3.79 x 10^-18/ (3.5 x 10^-2)^2

    = 1. 579 x 10 ^-22N

6 0
1 year ago
A stone is thrown horizontally from 2.4m above the ground at 35m/s. The wall is 14m away and 1m high.At what height the stone wi
KIM [24]

The stone reaches the wall at a height of <u>1.62 m</u>.

The stone lands at a point <u>24.5 m</u> from the point of projection.

The stone is projected horizontally with a velocity u at a height <em>h</em> from the ground. The wall is located at a distance <em>x</em> from the point of projection. The stone takes a time <em>t</em> to reach the wall and in the same time the stone falls a vertical distance <em>y</em>.

The horizontal distance <em>x</em> is traveled with a constant velocity <em>u</em>.

x=ut

Calculate the time taken <em>t</em>.

t=\frac{x}{u} \\ =\frac{14m}{35 m/s} \\ =0.40s

The stone's initial vertical velocity is zero. It falls through a distance <em>y</em> in the time <em>t</em> under the action of acceleration due to gravity <em>g</em>.

y=\frac{1}{2} gt^2\\ \frac{1}{2} (9.81m/s^2)(0.40s)^2\\ =0.784m

The height  <em>h₁ </em>of the stone above the ground when it reaches the wall  is given by,

h_1=h-y\\ =(2.4m)-(0.784m)\\ =1.616m=1.62m

When the stone reaches the wall, its height from the ground is <u>1.62m.</u>

The stone thus crosses over the wall, since the height of the wall is 1 m. It reaches the ground at a distance <em>R</em> from the point of projection. If the time taken by the stone to reach the ground is <em>t₁, </em>then,

h=\frac{1}{2} gt_1^2

Calculate the time taken by the stone to reach the ground.

t_1=\sqrt{\frac{2h}{g} } \\=\sqrt{\frac{2(2.4m)}{9.81m/s^2} } \\ =0.699 s

The horizontal distance traveled by the stone is given by,

R=ut_1 \\ =(35m/s)(0.699s)\\ =24.5m

The stone lands at point 24.5 m from the point of projection and 10.5 m from the wall.

3 0
1 year ago
Now suppose the initial velocity of the train is 4 m/s and the hill is 4 meters tall. If the train has a mass of 30000 kg, what
hoa [83]

Answer:

<h2>187,500N/m</h2>

Explanation:

From the question, the kinectic energy of the train will be equal to the energy stored in the spring.

Kinetic energy = 1/2 mv² and energy stored in a spring E = 1/2 ke².

Equating both we will have;

1/2 mv² = 1/2ke²

mv² = ke²

m is the mass of the train

v is the velocity of then train

k is the spring constant

e is the extension caused by the spring.

Given m = 30000kg, v = 4 m/s, e = 4 - 2.4 = 1.6m

Substituting this values into the formula will give;

30000*4² =  k*1.6²

k = \frac{30,000*16}{1.6^2}\\ \\k = \frac{480,000}{2.56}\\ \\k = 187,500Nm^{-1}

The value of the spring constant is 187,500N/m

7 0
2 years ago
Given a die, would it be more likely to get a single 6 in six rolls, at least two 6s in twelve rolls, or at least one-hundred 6s
vichka [17]

Answer:

Explanation:

In first case we are interested in one time 6 in six rolls

Thus probability = number of chances required/Total chances

= 1/6

Similarly in the second case probability = 2/12 = 1/6

In the same way in last case probability = 100/600 = 1/6

The probability is the same . Thus all the cases has equal chances  

4 0
2 years ago
A careful photographic survey of Jupiter’s moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to
Y_Kistochka [10]

Answer:

529.15 m/s

Explanation:

h = Maximum height = 70000 m

g = Acceleration due to gravity = 2 m/s²

m = Mass of sulfur

As the potential and kinetic energies are conserved

mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 2\times 70000}\\\Rightarrow v=529.15\ m/s

The speed with which the liquid sulfur left the volcano is 529.15 m/s

7 0
2 years ago
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