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maxonik [38]
2 years ago
6

You are standing 10 meters from a light source. Then, you back away from the light source until you are 20 meters away from it.

From your perspective, what has happened to the intensity of the wave?
A. It has decreased by a factor of two.
B. It has increased by a factor of two.
C. It has decreased by a factor of four.
D. It has increased by a factor of four.
Physics
1 answer:
bogdanovich [222]2 years ago
5 0
Inverse Square Law of Light states that light intensity falls off rapidly with distance from its source. T<span>he intensity varies with the square of the flash-to-subject distance.
 intensity at distance 1/intensity at distance 2=distance2^2/distance1^2
In our case, distance1=10m and distance2=20m, 20^2/10^2= 400/100=4, which means that t</span>he intensity of the wave<span> has increased by a factor of four.
Answer:D</span>
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Answer:

a. 8.33 x 10 ⁻⁶ Pa

b. 8.19 x 10 ⁻¹¹ atm

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Explanation:

Given:

a.

I = 2500 W / m² , us = 3.0 x 10 ⁸ m /s

P rad = I / us

P rad  = 2500 W / m² / 3.0 x 10 ⁸ m/s

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b.

P rad = 8.33 x 10 ⁻⁶ Pa *[  9.8 x 10 ⁻⁶ atm / 1 Pa ]

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c.

P rad = 2 * I / us = ( 2 * 2500 w / m²) / [ 3.0 x 10 ⁸ m /s ]

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P rad = 1.67 x 10 ⁻⁵ Pa / 1.013 x 10 ⁵ Pa /atm = 1.65 x 10 ⁻¹⁰ atm

d.

P rad  = I / us

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Answer:

7.9 m/s

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Distance, <em>s</em> = 4.0 m - 1.2 m = 2.8 m

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Motion of plastic ball

This has two components: a vertical and a horizontal.

The vertical motion is under gravity.

Considering the vertical motion,

Initial velocity, <em>u </em>= ?

Distance, <em>s</em> = 1.2 m

Acceleration, <em>a</em> = -<em>g                   </em> (It is going up)

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s = ut+\frac{1}{2}at^2

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1.2\text{ m} = u\sqrt{\dfrac{5.6}{g}}-\dfrac{1}{2}\times g\times\dfrac{5.6}{g}

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horizontal component of the initial velocity is horizontal distance ÷ time.

u_h = \dfrac{4.4\text{ m}}{0.756\text{ s}} = 5.82\text{ m/s}

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