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Dmitriy789 [7]
2 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]2 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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All valid equations in physics have consistent units. Are all equations that have consistent units valid?
sweet [91]

Answer:

c. No. An equation may have consistent units but still be numerically invaid.

Explanation:

For an equation to be corrected, it should have consistent units and also be numerically correct.

Most equation are of the form;

(Actual quantity) = (dimensionless constant) × (dimensionally correct quantity)

From the above, without the dimensionless constant the equation would be numerically wrong.

For example; Kinetic energy equation.

KE = 0.5(mv^2)

Without the dimensionless constant '0.5' the equation would be dimensionally correct but numerically wrong.

8 0
3 years ago
A strontium vapor laser beam is reflected from the surface of a CD onto a wall. The brightest spot is the reflected beam at an a
sladkih [1.3K]

Answer:

d=1.29*10^{-6}m

Explanation:

From the question we are told that:

Distance of wall from CD D=1.4

Second bright fringe y_2= 0.803 m

Let

Strontium vapor laser has a wavelength \lambda= 431 nm=>431 *10^{-9}m

Generally the equation for Interference is mathematically given by

y=frac{n*\lambda*D}{d}

Where

d=\frac{n*\lambda*D}{y}

d=\frac{2*431 *10^{-9}m*1.4}{0.803}

d=1.29*10^{-6}m

8 0
2 years ago
What is water that flows across eatlrtus surface
ira [324]

Water that flows across the surface is called a;

Runoff

That's when rain has saturated the ground to the point it cant hold anymore and it runs over the surface.

4 0
3 years ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
3 years ago
What happens when charged object is brought near uncharged object?<br> Attract or Repel ?
Makovka662 [10]
When a charged object is brought near to but does not touch a neutral object, it causes the side of the neutral object that the charged object is near to become the other charge. It causes charge migration within the neutral object so the two charges (positive and negative) move to opposite sides of the object. Because the two objects do not touch, they do not repel each other, but rather have a slight attraction because of charge migration. If the two object were to touch then they would repel.
5 0
2 years ago
Read 2 more answers
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