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Tatiana [17]
3 years ago
8

Can anyone please help me with this question? : A target consists of two concentric circles. The larger circle has diameter 1m.

When a complete novice fires at the target, the probability of hitting the blue circle is the same as the probability of hitting the green region. Find the diameter of the blue circle.

Mathematics
1 answer:
11111nata11111 [884]3 years ago
5 0
We know that
if  <span>the probability of hitting the blue circle is the same as the probability of hitting the green region
then 
the area of the blue circle is equal to the area of the green region

Let
x----> diameter of the blue circle

area of the blue circle=pi*(x/2)</span>²----> (pi/4)*x² m²-----> equation 1

area of the green region=area of the larger circle-area of the blue circle
area of the green region=pi*(1/2)²-(pi/4)*x²
=(pi/4)-(pi/4)*x² m²----> equation 2

equate equation 1 and equation 2
(pi/4)*x²=(pi/4)-(pi/4)*x² -----> divide by (pi/4)---> x²=1-x²
2x²=1-----> x²=1/2----> x=1/√2-----> x=√2/2 m

the diameter of the blue circle is √2/2 m


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Answer:

56.39 nm

Step-by-step explanation:

In order to have constructive interference total optical path difference should be an integral number of wavelengths (crest and crest should be interfered). Therefore the constructive interference condition for soap film can be written as,

2t=(m+\frac{1}{2} ).\frac{\lambda}{n}

where λ is the wavelength of light and n is the refractive index of soap film, t is the thickness of the film, and m=0,1,2 ...

Please note that here we include an additional 1/2λ phase shift due to reflection from air-soap interface, because refractive index of latter is higher.

In order to have its longest constructive reflection at the red end (700 nm)

t_1=(m+\frac{1}{2} ).\frac{\lambda}{2.n}\\ \\ t_1=\frac{1}{2} .\frac{700}{(2)*(1.33)}\\ \\ t_1=131.58\ nm

Here we take m=0.

Similarly for the constructive reflection at the blue end (400 nm)

t_2=(m+\frac{1}{2} ).\frac{\lambda}{2.n}\\ \\ t_2=\frac{1}{2} .\frac{400}{(2)*(1.33)}\\ \\ t_2=75.19\ nm

Hence the thickness difference should be

t_1-t_2=131.58-75.19=56.39 \ nm

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Answer:

The solution of the differential equation is y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)

Step-by-step explanation:

The differential equation is given by: y" + y = Sin(2t)

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y(t)_{h}=c_{1}[Sin(t)+iCos(t)]+c_{2}[Sin(t)-iCos(t)]

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y''(t)_{p}=-4ASin(2t)-4BCos(2t)

So we use these derivatives in the differential equation:

-4ASin(2t)-4BCos(2t)+ASin(2t)+BCos(2t)=Sin(2t)

-3ASin(2t)-3BCos(2t)=Sin(2t)

As there is not a term for Cos(2t), B is equal to 0.

So the value A=-1/3

The solution is the sum of the particular function and the homogeneous function:

y(t)= - \frac{1}{3} Sin(2t) + C_{1} Sin(t) + C_{2} Cos(t)

Using the initial conditions we can check that C1=5/3 and C2=2

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To solve the differential equation we use the Laplace transformation in both members:

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By using the Table of Laplace Transform we get:

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We replace the previous data in the equation:

s²·Y(s) -2s-1+Y(s) =\frac{2}{(s^{2}+4)}

(s²+1)·Y(s)-2s-1=\frac{2}{(s^{2}+4)}

(s²+1)·Y(s)=\frac{2}{(s^{2}+4)}+2s+1=\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)}

Y(s)=\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)(s^{2}+1)}

Y(s)=\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}

Using partial franction method:

\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}=\frac{As+B}{s^{2}+4} +\frac{Cs+D}{s^{2}+1}

2s^{3}+s^{2}+8s+6=(As+B)(s²+1)+(Cs+D)(s²+4)

2s^{3}+s^{2}+8s+6=s³(A+C)+s²(B+D)+s(A+4C)+(B+4D)

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A+C=2

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