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fomenos
3 years ago
8

Consider a civilization broadcasting a signal with a power of 2.0×104 watts. The Arecibo radio telescope, which is about 300 met

ers in diameter, could detect this signal if it is coming from as far away as 141 light-years. Suppose instead that the signal is being broadcast from the other side of the Milky Way Galaxy, about 70000 light-years away.
Mathematics
1 answer:
Vitek1552 [10]3 years ago
7 0

For light: Inverse Square Law 

<span>
The signal is weaken by the inverse square of the difference of distances this means that one over the square of the difference. </span>

<span>
</span>

Therefore, 70000LY / 141 LY =496.45 and that squared =246462.602 times less signal, or 1/246462.602.


 2 For a circular telescope like the Arecibo radio telescope that equals Pi times the radius squared. Sensitivity of the telescope depends on its area.<span>
The new telescope needs 246462.60 times that area to have enough sensitivity </span>

<span>
<span>(150 x 150 x Pi) = 70 685 sq m area for the Arecibo telescope and that times 246462.60
= 1.74212 x 10^10</span></span>

<span><span>
</span></span>

What is the diameter of a circle with that area?

 <span>
<span>1. Divide the area by Pi = 5.548156 x 10^9
2. Take the square root of the result to find the radius = 74485.94293</span></span>

3. Multiply by 2 to find the diameter = <span>148971.886</span>

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Simple, rewrite this equation, in y=mx+b form.

Move over the 5x, making it look like, -y=-5x+14

Then, just simply divide by the negative one in front of the y.

y=5x-14.
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3 years ago
Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f
drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

y''+3y'=0

The characteristic equation is

r^2+3r=r(r+3)=0

with roots r=0 and r=-3, giving the two solutions C_1 and C_2e^{-3t}.

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

y''+3y'=2t^4

Assume the ansatz solution,

{y_p}=at^5+bt^4+ct^3+dt^2+et

\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e

\implies {y_p}''=20at^3+12bt^2+6ct+2d

(You could include a constant term <em>f</em> here, but it would get absorbed by the first solution C_1 anyway.)

Substitute these into the ODE:

(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4

15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4

\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}

y''+3y'=t^2e^{-3t}

e^{-3t} is already accounted for, so assume an ansatz of the form

y_p=(at^3+bt^2+ct)e^{-3t}

\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}

\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}

Substitute into the ODE:

(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}

9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2

-9at^2+(6a-6b)t+2b-3c=t^2

\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}

y''+3y'=\sin(3t)

Assume an ansatz solution

y_p=a\sin(3t)+b\cos(3t)

\implies {y_p}'=3a\cos(3t)-3b\sin(3t)

\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)

Substitute into the ODE:

(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)

(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)

\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}

So, the general solution of the original ODE is

y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}

3 0
4 years ago
Please please help!!!
Rashid [163]

Answer:

  1/2

Step-by-step explanation:

cos(θ) = √(1 -sin(θ)²) = √(1 -3/4) = √(1/4)

cos(θ) = 1/2

3 0
3 years ago
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