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2 years ago

Find Lm= Help me please thanks

Mathematics

1 answer:

Ymorist [56]2 years ago

4 0

A triangle is a three-edged polygon with three vertices. The length of LM is 3√3 cms.

<h3>What is a triangle?</h3>

A triangle is a three-edged polygon with three vertices. It is a fundamental form in geometry. The sum of all the angles of a triangle is always equal to 180°.

The triangle formed by MN, MO, and NO is an equilateral triangle, with the length of each side of the triangle equal to the radius of the triangle. Therefore, the length of MN will be,

MN = r

Now, the length of arc MN can be written as,

2 × π × r × (60° /360°) = π

r = 360 / (60 × 2)

r = 3 cms

Further, the radius of the circle can be written as,

LO = ON = OM = MN = Radius = 3 cms

Since, the triangle formed in a semicircle is a right-angle triangle, therefore, using the Pythagorean theorem we can write,

LM² + MN² = LN²

LM² + MN² = (LO + ON)²

LM² + r² = (r+r)²

LM² + r² = 4r²

LM² = 3r²

LM² = 3(3²)

LM = √27 cms

LM = 3√3 cms

Hence, the length of LM is 3√3 cms.

Learn more about Triangle:

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3 years ago

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

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3 years ago

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Answer:

Never

Step-by-step explanation:

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300/180 +2 = 11/3 or 3.666. Polygons cannot have decimal sides therefore your answer is never.

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3 years ago