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vladimir2022 [97]
3 years ago
9

Is the refractive index constant?

Physics
2 answers:
Vlad [161]3 years ago
8 0

Answer:The refractive index (n?) of a material is defined as the ratio of velocity of light in a vacuum to that in a defined material. The refractive index of a glass is not a constant, but depends on the wavelength of the incident light, known as dispersion

GenaCL600 [577]3 years ago
4 0

Answer: No

Explanation:

The refractive index (n?) of a material is defined as the ratio of velocity of light in a vacuum to that in a defined material. The refractive index of a glass is not a constant, but depends on the wavelength of the incident light, known as dispersion.

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Two objects of the same size are both perfect blackbodies. One has a temperature of 3000 K, so its frequency of maximum emission
bija089 [108]

Answer:

a) The colder body (3000k), b) hearter body c) 12000K body

Explanation:

This exercise should know the power emitted by the objects and the distribution of this emission in the energy spectrum, for this we will use Stefan's laws and that of Wien's displacement

Stefan's Law                     P = σ A e T⁴

Wien displacement law   λ T = 2,898 10⁻³ m K

Let's calculate the power emitted for each object.

As they are perfect black bodies e = 1, they also indicate that they have the same area

T = 3000K

       P₁ = σ A T₁⁴

T = 12000K

       P₂ = σ A T₂⁴

       P₂ / P₁ = T₂⁴ / T₁⁴

       P₂ / P₁ = (12000/3000)⁴

       P₂ / P₁ = 256

This indicates that the hottest body emission is 256 times the coldest body emission.

Let's calculate the maximum emission wavelength

Body 1

T = 3000K

       λ T = 2,898 10-3

       λ₁ = 2.89810-3 / T

       λ₁ = 2,898 10-3 / 3000

       λ₁ = 0.966 10-6 m

      λ₁ = 966 nm

T = 12000K

      λ₂ = 2,898 10-3 / 12000

      λ₂ = 0.2415 10-6 m

      λ₂ = 214 nm

a) The colder body (3000k) emits more light in the infrared, since the emission of the hot body is at a minimum (emission tail)

b) The two bodies have emission in this region, the body of 3000K in the part of rise of the emission and the body to 12000K in the descent of the emission even when this body emits 256 times more than the other, so this body should have the highest broadcast in this area

c) The emission of the hottest 12000K body is mainly in UV

d) The hottest body emits more energy in UV and visible

e) No body has greater emission in all zones

5 0
3 years ago
Starting from rest, a person runs with a constant acceleration, traveling 40 meters in 10 seconds. What is their final velocity?
Assoli18 [71]

Answer:

Final velocity v = 8.944 m/sec

Explanation:

We have given distance S = 40 meters

Time t = 10 sec

As it starts from rest so initial velocity u = 0

From second equation of motion s=ut+\frac{1}{2}at^2

40=0\times 10+\frac{1}{2}a10^2

a=0.8944m/sec^2

Now from first equation of motion v=u+at, here v is final velocity, u is initial velocity, a is acceleration and t is time

So v=u+at=0+0.8944\times 10=8.944m/sec

6 0
3 years ago
A child drops a ball from a window. The ball strikes the ground in 3.0 seconds. What is the velocity of the ball the instant bef
inessss [21]

Answer:

29.396988 m/s

Explanation:

Really, it depends on where the child is when he drops the ball - e.g., which planet he is on, and his distance from the center of that planet.

I'll assume that the child is on Earth at sea level at the equator, so that his distance from the geocenter is 6378000 meters.

The acceleration, g, is found from

g = GM/r²

G = 6.6743e-11 m³ kg⁻¹ sec⁻²

M = 5.9724e+24 kg

r = 6.378e+6 m

g = 9.799086 m sec⁻²

An approximate answer is found from an equation from constant acceleration kinematics:

v = gt

t = 3.0 sec

v = 29.397259 m/s

Now, the above method is an approximation that makes the technically incorrect assumption that the acceleration of gravity is a constant throughout the entire fall. You get away with it because the drop is very short. In another situation, it might not be. So it would be nice to develop a more accurate method that does not assume constant gravitational acceleration. For that, we begin with the Vis Viva equation:

v = √[GM(2/r − 1/a)]

Here,

a = the semimajor axis of a plunge orbit, which is equal to half of the apoapsis distance of 6378000+h, where

h = the altitude from which the ball is dropped

We can (using some math) develop the following equation:

t − t₀ = √[d/(2GM)] { √(rd−r²) + d arctan √(d/r−1) }

t − t₀ = 3 sec

r = 6378000 meters

d = r + h

Using an iterative method (e.g. Newton's or Danby's), we can determine that the altitude,

h = 44.0954 meters

So,

d = 6378044.09538 meters

a = d/2 = 3189022.04769 meters

Now we can calculate that

v = 29.396988 m/s

This is the more nearly correct answer because it takes into account the variability of the gravitational acceleration during the fall.

5 0
3 years ago
Sketch the solid whose volume is given by the integral and evaluate the integral. $ \int_{0}^{{\color{red}6}}\int_{0}^{2\pi }\in
Serggg [28]

Answer:

Explanation:

Attached is the evaluation

8 0
3 years ago
Two balls are thrown from a cliff. One is thrown directly up, the other directly down. Both balls have the same initial speed, a
Otrada [13]

Answer:

both the same

Explanation:

When a ball is thrown vertically upwards, it experiences that same acceleration due to gravity as an object thrown directly downwards.

This means that if we ignore the effects of air resistance, and the two balls have the same initial speed, they are expected both expected to hit the ground at the same speed as a result of the principle of conservation of energy.

8 0
3 years ago
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