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lukranit [14]
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 0.691 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]3 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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Select the correct answer
DiKsa [7]

Answer:

C or 3

Explanation:

A: no they are called sources of sound. A is incorrect.

B: It does. Many people attest to this. But this is not a property of physics.

C: A media is required is the correct answer.

D: Dogs might. In general we don't.

I would pick C

7 0
2 years ago
A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.3 m/s rel
Sonja [21]

consider east-west direction along X-axis  and north-south direction along Y-axis

V_{ra} = velocity of migrating robin relative to air = 12 j m/s

(where "j" is unit vector in Y-direction)

V_{ag} = velocity of air relative to ground = 6.3 i m/s

(where "i" is unit vector in X-direction)

V_{rg} = velocity of migrating robin relative to ground = ?

using the equation

V_{rg} = V_{ra} + V_{ag}

V_{rg} = 12 j + 6.3 i

V_{rg} = 6.3 i + 12 j

magnitude : sqrt((6.3)² + (12)²) = 13.6 m/s

direction : tan⁻¹(12/6.3) = 62.3 deg north of east

4 0
3 years ago
How can parents help children to gain friends?
Andre45 [30]

Answer:

You could try finding a familiar peer to join the activity with your child. Or ask your child who their friends are at school, or what they look for in a friend at school.

7 0
3 years ago
Read 2 more answers
Wat would happen if a feather and a ball were released from the same height at the same time? Gravity Experiments for Kids – Sci
solniwko [45]

Answer:

They would land at the same time

Explanation:

They would land at the same exact time.

As weird, impossible and unbelievable as it appears. When in a vacuum, every weight, body and material when released from the same height would land on the ground at the same time. This also means that like in the question, a feather and a ball would land at the same time. And just for illustrations as well, a feather and a car would land at the same time as well.

6 0
3 years ago
You have a 3.00-liter container filled with N₂ at 25°C and 4.45 atm pressure connected to a 2.00-liter container filled with Ar
LuckyWell [14K]

Answer : The final pressure in the two containers is, 2.62 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

Thus, the expression for final pressure in the two containers will be:

PV=P_1V_1+P_2V_2

P=\frac{P_1V_1+P_2V_2}{V}

where,

P_1 = pressure of N₂ gas = 4.45 atm

P_2 = pressure of Ar gas = 2.75 atm

V_1 = volume of N₂ gas = 3.00 L

V_2 = volume of Ar gas = 2.00 L

P = final pressure of gas = ?

V = final volume of gas = (4.45 + 2.75) L = 7.2 L

Now put all the given values in the above equation, we get:

P=\frac{(4.45atm)\times (3.00L)+(2.75atm)\times (2.00L)}{7.2L}

P=2.62atm

Thus, the final pressure in the two containers is, 2.62 atm

8 0
3 years ago
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