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skad [1K]
3 years ago
15

The intensity of light from a central source varies inversely as the square of the distance. If you lived on a planet only half

as far from the Sun as our Earth, how would the intensity compare with that on Earth?
Physics
1 answer:
sertanlavr [38]3 years ago
5 0

Answer:

the intensity will be 4 times that of the earth.

Explanation:

let us assume the following:

intensity of light on earth =J

distance of earth from sun = d

intensity of light on other planet = K

distance of other planet from sun = \frac{d}{2} (from the question, the planet is half as far from the sun as earth)

from the question the intensity is inversely proportional to the square of the distance, hence

  • intensity on earth : J = \frac{1}{d^{2} }

        Jd^{2} = 1 ... equation 1

  • intensity on other planet : K =  \frac{1}{(\frac{d}{2}) ^{2} }  (the planet is half as far from the sun as earth)

        K(\frac{d}{2}) ^{2} = 1 ....equation 2

  • equating both equation 1 and 2 we have

       Jd^{2} = K(\frac{d}{2}) ^{2}

       Jd^{2} = K\frac{d^{2}}{4}

       J = \frac{K}{4}

        K = 4J

       intensity of light on other planet (K) = 4 times intensity of light on earth (J)

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now if we move closer to some some distance the sound level is now 50 dB

now the intensity is given as

L = 10 Log\frac{I}{I_0}

50 = 10Log\frac{I}{10^{-12}}

I_2 = 10^{-7}W/m^2

now we know that

\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}

\frac{10^{-10}}{10^{-7}} = \frac{r_2^2}{15^2}

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3 years ago
In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue
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Answer:

a)  p₀ = 1.2 kg m / s,  b) p_f = 1.2 kg m / s,  c)   θ = 12.36, d)  v_{2f} = 1.278 m/s

Explanation:

a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved

a) the initial impulse is

        p₀ = m v₁₀ + 0

        p₀ = 0.6 2

        p₀ = 1.2 kg m / s

b) as the system is isolated, the moment is conserved so

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we define a reference system where the x-axis coincides with the initial movement of the cue ball

we write the final moment for each axis

X axis

        p₀ₓ = 1.2 kg m / s

        p_{fx} = m v1f cos 20 + m v2f cos θ

        p₀ = p_f

       1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ

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Y axis  

       p_{oy} = 0

       p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ

       0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ

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we write our system of equations

        0.2736 = v_{2f} sin θ

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divide to solve

        0.219 = tan θ

         θ = tan⁻¹ 0.21919

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let's look for speed

           0.2736 = v_{2f} sin θ

            v_{2f} = 0.2736 / sin 12.36

           v_{2f} = 1.278 m / s

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Answer:

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Explanation:

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