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hichkok12 [17]
4 years ago
11

In the first-order spectrum, maxima for two different wavelengths are separated on the screen by 3.40 mm . what is the differenc

e between these wavelengths?
Physics
1 answer:
Strike441 [17]4 years ago
6 0

The solution would be like this for this specific problem:

 

Given:  

diffraction grating slits = 900 slits per centimeter

interference pattern that is observed on a screen from the grating = 2.38m

maxima for two different wavelengths = 3.40mm

 

slit separation .. d = 1/900cm = 1.11^-3cm = 1.111^-5 m <span>

Whenas n = 1, maxima (grating equation) sinθ = λ/d 
Grant distance of each maxima from centre = y .. 
<span>As sinθ ≈ y/D  y/D = λ/d λ = yd / D </span>

∆λ = (λ2 - λ1) = y2.d/D - y1.d/D 
∆λ = (d/D) [y2 -y1] 

<span>∆λ = 1.111^-5m x [3.40^-3m] / 2.38m .. .. ►∆λ = 1.587^-8 m</span></span>

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A certain airport runway of length L allows planes to
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The length of the new runway In terms of (L) is 4L.

<em />

<em>Your question is incomplete, it seems your question is missing the following information below;</em>

"In terms of (L) , what must be the length of the new runway?"

The given parameters;

  • length of the runway, = L
  • let the take off speed = V₁

For the newly designed planes;

the take off speed is double of the original = 2V₁

the acceleration for both take is constant = a

The take off speed for the given constant acceleration is given as;

v^2 = u^2 + 2as\\\\v^2 = 0 + 2as\\\\v^2 = 2as\\\\a = \frac{v^2}{2s} \\\\

The length of the new runway In terms of (L) is calculated as follows;

a =  \frac{v_1^2}{2s_1}  = \frac{v_2^2}{2s_2}  \\\\substitute \ the \ corresponding \ values \ for \ initial \ acceleration \ and \ new \\acceleration\\\\\frac{v_1^2}{2L}  = \frac{(2v_1)^2}{2s_2}\\\\2s_2v_1^2 = 2L(4v_1^2)\\\\s_2 = \frac{2L(4v_1^2)}{2v_1^2} = \frac{8Lv_1^2}{2v_1^2} = 4L

Thus, the length of the new runway In terms of (L) is 4L.

Learn more here: brainly.com/question/15206976

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3 years ago
A 59 kg man has a total mechanical energy of 150,023. J. If he is swinging downward and is currently 2.6 m above the ground, wha
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Answer:

v = 70.95 \ m/s.

Explanation:

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Mass of the man, m = 59 \ kg

Total mechanical energy, E_{i} = 150,023 \ \rm J

Height, h = 2.6 \ m

Suppose there is no external force acting on the man. In this situation, the total mechanical energy (kinetic + potential) will remain steady.

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E_{i} = E_{f}

E_{i} = \frac{1}{2}mv^{2} + mgh

150023 = 0.5 \times 59 \times v^{2} + 59 \times 9.80 \times 2.6

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