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4 years ago

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Numerous engineering and scientific applications require finding solutions to a set of equations. Ex: 8x + 7y = 38 and 3x - 5y =

-1 have a solution x = 3, y = 2. Given integer coefficients of two linear equations with variables x and y, use brute force to find an integer solution for x and y in the range -10 to 10.

2 answers:

Irina-Kira [14]4 years ago

6 0

Answer:

we do not have the integer coeficients, so the solutions may not be in the range -10 to 10 (it does not happen for all coeficients) so let's do other thing:

Ok, suppose you have the equations:

1) A*x + B*y = C

2) a*x + b*y = c

where A, B, C, a, b and d are knowed.

Let's solve them in the most general way, and you always can use those solutions in the future, just need to replace the values of the constants up there.

First, we isolate x or y in a equation, let's do it with x in equation 1:

A*x = C - B*y

x = (C - B*y)/A

Now we can replace this in the equation 2, and solve it for y:

a*x + b*y = c

a*(C - B*y)/A + b*y = c

y*(b - a*B/A) = c - a*C/A

y = (c - a*C/A)/(b - a*B/A)

Now you have a solution for y, and we you know the value of y, you can put it in the eqution:

x = (C - B*y)/A

and find the value of x

mote1985 [20]4 years ago

5 0

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a) Total power output: $3.845\cdot 10^{26} W$

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is $540 W/m^2$

Explanation:

a)

The intensity of electromagnetic radiation is given by

$I=\frac{P}{A}$

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P is the power output

A is the surface area considered

In this problem, we have

$I=1360 W/m^2$ is the intensity of the solar radiation at the Earth

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Therefore

$A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2$

And now, using the first equation, we can find the total power output of the Sun:

$P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W$

b)

The energy of the solar radiation is directly proportional to its frequency, given the relationship

$E=hf$

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

$P=\frac{E}{t}$

where t is the time.

This means that the power output is proportional to the frequency:

$P\propto f$

Here the frequency increases by 1 MHz: the original frequency was

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so the relative percentage change in frequency is

$\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%$

And therefore, the power also increases by 1.67 %.

c)

In this second case, we have to calculate the new power output of the Sun:

$P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W$

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

$r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m$

Now we can find the radiation intensity with the equation

$I=\frac{P}{A}$

Where the area is

$A=4\pi r'^2 = 4\pi(2.4\cdot 10^{11})^2=7.24\cdot 10^{23} m^2$

And substituting,

$I=\frac{3.910\cdot 10^{26}}{7.24\cdot 10^{23}}=540 W/m^2$

Learn more about electromagnetic radiation:

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3 years ago

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Answer:

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The general formula for power of a lens is:

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Therefore, for the top half of the lens:

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Answer: The unpolarized light's intensity is reduced by the factor of two when it passes through the polaroid and becomes linearly polarized in the plane of the Polaroid. When the polarized light passes through the polaroid with the plane of polarization at an angle $\theta$ with respect to the polarization plane of the incoming light, the light's intensity is reduced by the factor of $\cos^2\theta$ (this is the Law of Malus).

Explanation: Let us say we have a beam of unpolarized light of intensity $I_0$ that passes through two parallel Polaroid discs with the angle of $\theta$ between their planes of polarization. We are asked to find $\theta$ such that the intensity of the outgoing beam is $I_2$ . To solve this we follow the steps below:

Step 1. It is known that when the unpolarized light passes through a polaroid its intensity is reduced by the factor of two, meaning that the intensity of the beam passing through the first polaroid is

$I_1=\frac{I_0}{2}.$

This beam also becomes polarized in the plane of the first polaroid.

Step 2. Now the polarized beam hits the surface of the second polaroid whose polarization plane is at an angle $\theta$ with respect to the plane of the polarization of the beam. After passing through the polaroid, the beam remains polarized but in the plane of the second polaroid and its intensity is reduced, according to the Law of Malus, by the factor of $\cos^2\theta.$ This yields $I_2=I_1\cos^2\theta$ . Substituting from the previous step we get

$I_2=\frac{I_0}{2}\cos^2\theta$

yielding

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and finally,

$\theta=\arccos\sqrt{\frac{2I_2}{I_0}}$

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3 years ago