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Dmitriy789 [7]
3 years ago
15

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 0.673 m 0.673 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 543 nm 543 nm and your pupil diameter to be 5.37 mm.
Physics
1 answer:
g100num [7]3 years ago
5 0

To solve this problem we will apply trigonometric and optical concepts that allow us to obtain the minimum distance required. The resolution of the eye is given under the following condition,

\theta = \frac{1.22\lambda}{D}

Here,

\lambda = Wavelength

D = Diameter

With the values we have that the diameter will be,

\theta = \frac{1.22(534nm)}{5.37mm}

\theta = 1.213*10^{-4}

The relation between the distance of the lights and the distance from the eye to the lamp is given under the function,

sin\theta = \frac{d}{L}

For small angles sin\theta = \theta, then

\theta = \frac{d}{L}

Here,

d = Distance between lights

L = Distance from eye to lamp

1.213*10^{-4} = \frac{0.673m}{L}

L = \frac{0.673m}{1.213*10^{-4}}

L = 5548.22m

Therefore the distance will be 5.5km

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The visible spectrum refers to the tiny portion of the electromagnetic spectrum that we ________.
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7 0
3 years ago
A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p
777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s

Now, we can calculate the angular acceleration  (w0=0rad/s)

\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}

\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}

with this value we can compute the angular velocity

\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}

and the tangential velocity of point B, and then the acceleration of point B:

v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}

hope this helps!!

6 0
3 years ago
Read 2 more answers
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