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Flura [38]
3 years ago
15

What is the net force for this free body diagram below?

Mathematics
1 answer:
IgorC [24]3 years ago
3 0
The answer is 5N right
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The intensity of light with wavelength λ traveling through a diffraction grating with N slits at an angle θ is given by I(θ) = N
Ymorist [56]

Answer:

0.007502795

Step-by-step explanation:

We have

N = 10,000

\bf d=10^{-4}

\bf \lambda = 632.8*10^{-9}

Replacing these values in the expression for k:

\bf k=\frac{\pi Ndsin\theta}{\lambda}=\frac{\pi10^4*10^{-4}sin\theta}{632.8*10^{-9}}=\frac{\pi 10^9sin\theta}{632.8}

So, the intensity is given by the function

\bf I(\theta)=\frac{N^2sin^2(k)}{k^2}=\frac{10^8sin^2(\frac{\pi 10^9sin\theta}{632.8})}{(\frac{\pi 10^9sin\theta}{632.8})^2}

The <em>total light intensity</em> is then

\bf \int_{-10^{-6}}^{10^{-6}} I(\theta)d\theta=\int_{-10^{-6}}^{10{-6}}\frac{10^8sin^2(\frac{\pi 10^9sin\theta}{632.8})}{(\frac{\pi 10^9sin\theta}{632.8})^2}d\theta

Since \bf I(\theta) is an <em>even function</em>

\bf \int_{-10^{-6}}^{10^{-6}} I(\theta)d\theta=2\int_{0}^{10^{-6}}I(\theta)d\theta

and we only have to divide the interval \bf [0,10^{-6}] in five equal sub-intervals \bf I_1,I_2,I_3,I_4,I_5 with midpoints \bf m_1,m_2,m_3,m_4,m_5

The sub-intervals and their midpoints are

\bf I_1=[0,\frac{10^{-6}}{5}]\;,m_1=10^{-5}\\I_2=[\frac{10^{-6}}{5},2\frac{10^{-6}}{5}]\;,m_2=3*10^{-5}\\I_3=[2\frac{10^{-6}}{5},3\frac{10^{-6}}{5}]\;,m_3=5*10^{-5}\\I_4=[3\frac{10^{-6}}{5},4\frac{10^{-6}}{5}]\;,m_4=7*10^{-5}\\I_5=[4\frac{10^{-6}}{5},10^{-6}]\;,m_5=9*10^{-5}

<em>By the midpoint rule</em>

\bf \int_{0}^{10^{-6}}I(\theta)d\theta\approx\frac{10^{-6}}{5}[I(m_1)+I(m_2)+...+I(m_5)]

computing the values of I:

\bf I(m_1)=I(10^{-5})=\frac{10^8sin^2(\frac{\pi 10^9sin(10^{-5})}{632.8})}{(\frac{\pi 10^9sin(10^{-5})}{632.8})^2}=13681.31478

\bf I(m_2)=I(3*10^{-5})=\frac{10^8sin^2(\frac{\pi 10^9sin(3*10^{-5})}{632.8})}{(\frac{\pi 10^9sin(3*10^{-5})}{632.8})^2}=4144.509447

Similarly with the help of a calculator or spreadsheet we find

\bf I(m_3)=3.09562973\\I(m_4)=716.7480066\\I(m_5)=211.3187228

and we have

\bf \int_{0}^{10^{-6}}I(\theta)d\theta\approx\frac{10^{-6}}{5}[I(m_1)+I(m_2)+...+I(m_5)]=\frac{10^{-6}}{5}(18756.98654)=0.003751395

Finally the the total light intensity

would be 2*0.003751395 = 0.007502795

8 0
3 years ago
(50 POINTS) The figure is transformed as shown in the diagram. Describe the transformation. A) dilation, then reflection B) refl
Helga [31]

Answer:

Hi there I was just working on this question on UsaTestPrep and idk if the answer is really correct but here: C: reflection then rotation.

Step-by-step explanation:

Extra Info:

Dilation is when the shape changes in size, which doesn't happen.

Translation is when it just moves across. It may look like that is happening here, but I'll explain.

The triangle is first reflected along the line . Imaging putting a mirror on the x=0 line (the y axis, basically), and you'd see 1 to 2.

Then, it is rotated around the point . Get a bit of tracing paper, draw over 2, and then, holding the tracing paper down at  (works well with a pencil), rotate it round and you will find it fits perfectly over 3.

Also is this the diagram?

6 0
3 years ago
Brian works in a lab, and fills an empty cylindrical flask with 200 cubic centimetres of liquid. The flaskhas a radius of 3 cm a
Anestetic [448]
I think it's 41%,based on my calculation
8 0
3 years ago
Read 2 more answers
What set of reflections would carry trapezoid ABCD onto itself
Aloiza [94]

Answer:

all sides probably. ( I think )

7 0
3 years ago
6 ten thousands mutilply 10 = what
Oksana_A [137]

Answer:

I believe the answer is six-hundred thousand (600000). That's if I read your question right. I'm not sure what you were asking exactly.

Step-by-step explanation:

6 (10000 x 10)

6 (100000)

600000

3 0
3 years ago
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