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Zigmanuir [339]
3 years ago
7

you stand on a straight desert road at night and observe a vehicle approaching. this vehicle is equipped with two small headligh

ts that are 0.679 m apart. at what distance, in kilometers, are you marginally able to discern that there are two headlights rather than a single light source? take the wavelength of the light to be 537 nm and your pupil diameter to be 4.81 mm.
Physics
1 answer:
inysia [295]3 years ago
7 0

Answer:

The distance that you marginally able to discern that there are two headlights rather than a single light source is 6.084 km

Explanation:

Given:

d = distance = 0.679 m

λ = wavelength of the light = 537 nm = 537x10⁻⁹m

dp = pupil diameter = 4.81 mm = 0.00481 m

Question: What distance, in kilometers, are you marginally able to discern that there are two headlights rather than a single light source, dx = ?

For the separation of the peak from the central maximum it is:

sin\theta =\frac{\lambda }{d_{p} } =\frac{537x10^{-9} }{0.00481} =1.116x10^{-4}

In this case, the two small sources of the headlights have the same angle as the images that form inside the eye

d_{x} =\frac{d}{sin\theta } =\frac{0.679}{1.116x10^{-4} } =6.084x10^{3} m=6.084km

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Explanation:

the area given by the exercise is

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Replacing values:

ε = (7*(38 - 14) * (200x10^-4))/8x10^-3 = 0.42 V

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<h2>A  or B</h2>

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A boy in a wheelchair (total mass 54.5 kg) has speed 1.40 m/s at the crest of a slope 2.10 m high and 12.4 m long. At the bottom
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Answer:

630.75 j

Explanation:

from the question we have the following

total mass (m) = 54.5 kg

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final speed (Vf) = 6.6 m/s

frictional force (FF) = 41 N

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work done (wd) = ?

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wd + mgh = (0.5 mVf^2) - (0.5 mVi^2) + (FF x  d)

wd = (0.5 mVf^2) - (0.5 mVi^2) + (FF x  d) - (mgh)

where wd = work done

m = mass

h = height

g = acceleration due to gravity

FF = frictional force

d = distance

Vf and Vi = final and initial velocity

wd =  (0.5 x 54.5 x 6.9^2) - (0.5 x 54.5 x 1.4^2) + (41 x 12.4) - (54.5 X 9.8 X 2.1)            

wd = 630.75 j

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3 years ago
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