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3 years ago

differentiate among the types of electromagnetic waves by comparing frequency, wavelength, and energy

Physics

1 answer:

Gemiola [76]3 years ago

3 0

Answer:

The electromagnetic spectrum encompasses a continuous range of frequencies or wavelengths of electromagnetic radiation, ranging from long wavelength, low energy radio waves to short wavelength, high frequency, high-energy gamma rays. The electromagnetic spectrum is traditionally divided into regions of radio waves, microwaves, infrared radiation, visible light, ultraviolet rays, x rays, and gamma rays.

Explanation:Exploration of the electromagnetic spectrum quickly resulted practical advances. German physicist Henrich Rudolph Hertz regarded Maxwell's equations as a path to a "kingdom" or "great domain" of electromagnetic waves. Based on this insight, in 1888, Hertz demonstrated the existence of radio waves. A decade later, Wilhelm Röent gen's discovery of high-energy electromagnetic radiation in the form of x rays quickly found practical medical use.This was on my notes

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Define fundamental and derived quantities with two example of each.

stich3 [128]

Fundamental quantity : quantities which are independent on other physical quantity .

ex: length,mass,time, current, amount of substance, luminous intensity, thermodynamic temperature,

Derived quantity : quantities which are depend on fundamental quantities.

ex: Area, volume, density, speed, acceleration, force, velocity etc.

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3 years ago

Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible fricti

Bas_tet [7]

Answer:

v = 0.059 m/s

Explanation:

To find the final speed of Olaf and the ball you use the conservation momentum law. The momentum of Olaf and the ball before catches the ball is the same of the momentum of Olaf and the ball after. Then, you have:

$mv_{1i}+Mv_{2i}=(m+M)v$ (1)

m: mass of the ball = 0.400kg

M: mass of Olaf = 75.0 kg

v1i: initial velocity of the ball = 11.3m/s

v2i: initial velocity of Olaf = 0m/s

v: final velocity of Olaf and the ball

You solve the equation (1) for v and replace the values of all variables:

$v=\frac{mv_{1i}}{m+M}=\frac{(0.400kg)(11.3m/s)}{0.400kg+75.0kg}=0.059\frac{m}{s}$

Hence, after Olaf catches the ball, the velocity of Olaf and the ball is 0.059m/s

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3 years ago

What are the parts of the scientific method?

wlad13 [49]

1. Ask a question
2. Form a hypothesis
3. Experiment
4.Record data
5.Draw Conclusion
6. Share Results

7 0

3 years ago

Read 2 more answers

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

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3 years ago

Cooling causes a material to

e-lub [12.9K]

Answer:

whats the question?

Explanation:

4 0

3 years ago

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