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GenaCL600 [577]
3 years ago
6

What is the approximate wavelength of a light whose first-order bright band forms a diffraction angle of 45.0° when it passes th

rough a diffraction grating that has 500.0 lines per mm?
236 nm
353 nm
943 nm
1414 nm
Physics
2 answers:
max2010maxim [7]3 years ago
7 0

Answer:

its D

Explanation:

edg2020

Bond [772]3 years ago
6 0

Answer:

D) 1414 nm

Explanation:

This is correct on Edge.

Using the equation dsin(angle)=n(wavelength), we can solve for wavelength.

First we must convert the 500 lines per mm to nm. We do this by 1/500, giving you 0.002. Then move the decimal over six places to the right, resulting in 2000.

Then by plugging in the other values, we have 2000sin(45)=1(wavelength).

N is one because we are just solving for a first-order band.

So 2000sin(45)=wavelength

By using a calculator, we can see that the wavelength equals approximately 1414nm.

I hope this helped. If it did, I would really appreciate a Brainliest!!

Have a great day:)

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The 1.18-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in
sattari [20]

Answer:

 k = 11,564 N / m,   w = 6.06 rad / s

Explanation:

In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;

 let's apply the equilibrium condition at this point

                 

Axis y

          W_{y} - Fr = 0

          Fr = k y

let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal

             sin 46 = W_{y} / W

             W_{y} = W sin 46

     

 we substitute

           mg sin 46 = k y

           k = mg / y sin 46

If the length of the bar is L

          sin 46 = y / L

           y = L sin46

 

we substitute

           k = mg / L sin 46 sin 46

           k = mg / L

for an explicit calculation the length of the bar must be known, for example L = 1 m

           k = 1.18 9.8 / 1

           k = 11,564 N / m

With this value we look for the angular velocity for the point tea = 30º

let's use the conservation of mechanical energy

starting point, higher

          Em₀ = U = mgy

end point. Point at 30º

         Em_{f} = K -Ke = ½ I w² - ½ k y²

          em₀ = Em_{f}

          mgy = ½ I w² - ½ k y²

          w = √ (mgy + ½ ky²) 2 / I

the height by 30º

           sin 30 = y / L

           y = L sin 30

           y = 0.5 m

the moment of inertia of a bar that rotates at one end is

          I = ⅓ mL 2

          I = ½ 1.18 12

          I = 0.3933 kg m²

let's calculate

          w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)

          w = 6.06 rad / s

7 0
3 years ago
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