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RSB [31]
4 years ago
12

A diffraction grating has 2605 lines per centimeter, and it produces a principal maximum at = 30.3°. The grating is used with li

ght that contains all wavelengths between 410 and 660 nm. What are the two wavelengths of the incident light that could have produced this maximum? Give the longer wavelength as answer (a).
Physics
1 answer:
Yuri [45]4 years ago
6 0

Answer:

so m = 3 ,4 wavelength between 410 to 660 nm  so 484.25 nm and 645.6 nm light produce principal maximum

Explanation:

given data

grating =2605 lines per centimeter

angle = 30.3°

wavelengths between 410 and 660 nm

to find out

two wavelengths of the incident light that could have produced this maximum

solution

we know diffraction grating has 2605 lines / cm

so d = ( 1/ 2605 cm )

and we know equation

d sinθ = m× λ

so λ = d sinθ / m

θ = 30.3 so sin30.3 =

λ = (1/2605) sin30.3 / m

λ = (1937 nm ) / m

here put m = 1 , 2 , 3 , 4

if m = 1

λ = (1937 nm ) / 1  = 1937 nm

if m = 2

λ = (1937 nm ) / 2  = 968.5 nm

if m = 3

λ = (1937 nm ) / 3 = 645.6 nm

if m = 4

λ = (1937 nm ) / 4 = 484.25 nm

so that m = 3 ,4 wavelength between 410 to 660 nm  so 484.25 nm and 645.6 nm light produce principal maximum

and longest wavelength is between 410 to 660 nm is 645.6 nm

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We need to write down momentum and energy conservation laws, this will give us a system of equation that we can solve to get our final answer. On the right-hand side, I will write term after the collision and on the left-hand side, I will write terms before the collision.
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Because both carts have the same mass we can cancel those out:
3v=v_A+2v_B
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Part A
We need to solve our system for v_a. We will solve second equation for v_b and then plug that in the first equation.
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We will multiply the first equation with 2 and divide by m:
3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_B=\frac{3v-v_A}{2}\\
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8 0
3 years ago
Three struts are pinned together at A, B, and C to form a triangular
ZanzabumX [31]

Answer:

a. Fab = 3000 N tension

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Fbc = 3000 N compression

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Explanation:

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∑F = ma

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∑F = ma

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Fbc = -Fab

Fbc = -H / sin 60°

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c. Draw a free body diagram of the entire triangle ABC.  Assume there is both a horizontal and vertical reaction force at C, and assume they point in the +x and +y directions.

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∑F = ma

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Fcy = V

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The magnitude of the net force is:

Fc² = Fcx² + Fcy²

Fc² = (-2600 N)² + (3000 N)²

Fc = 4000 N

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Answer it might be this

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