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weqwewe [10]
3 years ago
8

Parallel light waves hit the surface of a still lake and reflect in the same direction. Which interaction of light and matter do

es this illustrate? scattering diffuse reflection refraction regular reflection
Physics
1 answer:
vampirchik [111]3 years ago
6 0

Answer:

Regular reflection

Explanation:

- Reflection is the phenomenon that occurs when a light wave hits the interface between two different mediums and it bounces off back into the same medium. The angle of reflection (measured between the reflected ray and the perpendicular to the interface) is equal to the angle of incidence (measured between the incident ray and the perpendicular to the interface).

There are two different types of reflection:

- Regular reflection: this occurs when the interface between the two mediums is smooth (such as in the case of the still lake), so all the parallel light waves (which have same angle of incidence) are reflected exactly with the same angle of reflection (so, they come out all with same direction)

- Diffuse reflection: this occurs when the interface between the two mediums is not smooth, so each light ray is reflected with a different angle because it hits the interface with a different angle of incidence.

Therefore, in the case of the still lake, the correct answer is regular reflection.

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How much work is done when a 214 newton force pushes a sleeping cow 37m across a field.
nata0808 [166]
Hello!

Answer: 
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Explanation:

We are assuming that the floor (field) is completely horizontal since there's no information about that in the statement. 

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For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo
pickupchik [31]
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height. 

<span>In that particular situation, you can prove it like this: </span>

<span>initial velocity is Vo </span>
<span>launch angle is α </span>

<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>

<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>

<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
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<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>

<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>

<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
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<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>

<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
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