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lukranit [14]
4 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 0.691 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 539 nm and your pupil diameter as 5.11 mm. ______________km
Physics
1 answer:
spayn [35]4 years ago
4 0

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

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

Here,

D is diameter of the eye

D = \frac{1.22 (539nm)}{5.11 mm}

D= 1.287*10^{-4}m

The angle that relates the distance between the lights and the distance to the lamp is given by,

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

For small angle, sin\theta = \theta

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

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle sin \theta = \theta

Therefore,

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

L = \frac{0.691m}{1.287*10^{-4}}

L = 5367m

Therefore the distance is 5.367km.

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A small object with mass m, charge q, and initial speed v0 = 5.00 * 103 m>s is projected into a uniform electric field betwee
kotykmax [81]

Answer:

q/m = 2177.4 C/kg

Explanation:

We are given;

Initial speed v_o = 5 × 10³ m/s = 5000 m/s

Now, time of travel in electric field is given by;

t_1 = D_1/v_o

Also, deflection down is given by;

d_1 = ½at²

Now,we know that in electric field;

F = ma = qE

Thus, a = qE/m

So;

d_1 = ½ × (qE/m) × (D_1/v_o)²

Velocity gained is;

V_y = (a × t_1) = (qE/m) × (D_1/v_o)

Now, time of flight out of field is given by;

t_2 = D_2/v_o

The deflection due to this is;

d_2 = V_y × t_2

Thus, d_2 = (qE/m) × (D_1/v_o) × (D_2/v_o)

d_2 = (qE/m) × (D_1•D_2/(v_o)²)

Total deflection down is;

d = d_1 + d_2

d = [½ × (qE/m) × (D_1/v_o)²] + [(qE/m) × (D_1•D_2/(v_o)²)]

d = (qE/m•v_o²)[½(D_1)² + D_1•D_2]

Making q/m the subject, we have;

q/m = (d•v_o²)/[E((D_1²/2) + (D_1•D_2))]

We have;

E = 800 N/C

d = 1.25 cm = 0.0125 m

D_1 = 26.0 cm = 0.26 m

D_2 = 56 cm = 0.56 m

Thus;

q/m = (0.0125 × 5000²)/[800((0.26²/2) + (0.26 × 0.56))]

q/m = 312500/143.52

q/m = 2177.4 C/kg

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3 years ago
Distinguish between average speed and average<br>velocity​
Butoxors [25]

Answer:

The average speed of an object is defined as the distance traveled divided by the time elapsed. Velocity is a vector quantity, and average velocity can be defined as the displacement divided by the time.

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3 years ago
Which material is likely to slow the flow of electric charges the most ? explain.
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Metal as it is a conductor of electricity
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3 years ago
Read 2 more answers
Pls can anyone help me ​
Mamont248 [21]

Explanation:

y = y₀ + v₀ t + ½ at²

For the first ball:

0 = h + v₀ t − 4.9t²

For the second ball:

0 = h − 4.9 (t−1)²

a) If h = 20.0, find v₀.

0 = 20 − 4.9 (t−1)²

t = 3.02 s

0 = 20 + v₀ (3.02) − 4.9 (3.02)²

v₀ = 8.18 m/s

Graph:

desmos.com/calculator/uk1wzkxybt

If v₀ is given, find h.

First, find t in terms of v₀:

h + v₀ t − 4.9t² = h − 4.9 (t−1)²

v₀ t − 4.9t² = -4.9 (t−1)²

v₀ t − 4.9t² = -4.9 (t² − 2t + 1)

v₀ t − 4.9t² = -4.9t² + 9.8t − 4.9

v₀ t = 9.8t − 4.9

(9.8 − v₀) t = 4.9

t = 4.9 / (9.8 − v₀)

Therefore:

h = 4.9 (4.9 / (9.8 − v₀) − 1)²

bi) If v₀ = 6.0 m/s:

h = 4.9 (1 / (9.8 − 6.0) − 1)²

h = 2.66 m

bii) If v₀ = 9.5 m/s:

h = 4.9 (1 / (9.8 − 9.5) − 1)²

h = 26.7 m

c) As found in part a, the time it takes for the first ball to land is:

t = 4.9 / (9.8 − v₀)

If v₀ is greater than 9.8 m/s, the time becomes negative, which isn't possible.  Therefore, vmax = 9.8 m/s.  At this speed, the ball would reach its highest point after 1 second, the same time that the second ball is dropped.  Two balls dropped at the same time from different heights cannot land at the same time.

d) If v₀ is less than 4.9 m/s, the time for the first ball to land becomes less than 1 second.  Which means it will have already landed before the second ball is dropped.  Therefore, vmin = 4.9 m/s.

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Answer:

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