The hottest objects with temperatures in the millions of Kelvins, give off most of their radiation in which part of the electromagnetic spectrum? The hottest objects with temperatures in the millions of Kelvins, give off most of their radiation in which part of the electromagnetic spectrum? radio visible X-ray gamma ray ultraviolet

Answers

Answer 1

Answer:

Answer:

The hottest objects with temperatures in the millions of Kelvins, give off most of their radiation in the form of X-rays and gamma rays.

Explanation:

The electromagnetic (EM) spectrum contains radio waves, microwaves, infrared light, visible light, ultraviolet light, X-rays and gamma-rays. All these different types of radiation are made up of photons having specific wavelengths and different amounts of energy. In the EM spectrum, the photons of radio waves have the lowest energy and the energy of photons increases through microwaves, infrared, visible light, ultraviolet, X-rays, and the photons of gamma-rays have the highest energy (the energy of photons is measured in electron volts).

All warmer objects such as stars, planets, etc emit photons having a specific range of wavelengths and it depends on the surface temperature of those objects. The very hot objects with temperatures in the millions of Kelvins or more mainly emit photons with shorter wavelengths, such as gamma rays and X-rays while cooler objects emit radiation such as infrared or radio waves, having longer wavelengths.

The ultraviolet radiation has the energy in the range of a few electron volts to about 100 eV. The energy of X-ray photons is in the range of 100 eV to 100 keV and the energy of gamma-rays is greater than 100 keV. The nuclear explosions, radioactive decay, the hottest and most energetic objects in the universe such as neutron stars, supernova explosions, etc produce gamma rays.

Answer 2

Answer:

Objects with temperatures in the millions of Kelvins emit most of their radiation in the X-ray and gamma-ray parts of the electromagnetic spectrum.

Objects with temperatures in the millions of Kelvins primarily give off most of their radiation in the X-ray and gamma-ray parts of the electromagnetic spectrum. As an object's temperature increases, the wavelengths of radiation it emits become shorter. This phenomenon is described by Wien's displacement law.

At lower temperatures, such as those found on Earth or in stars like our Sun, objects emit most of their radiation in the visible and infrared parts of the spectrum. However, as temperatures rise to millions of Kelvins, the emitted radiation shifts to shorter wavelengths, eventually falling into the X-ray and gamma-ray regions.

In the X-ray and gamma-ray parts of the electromagnetic spectrum, radiation has extremely high energy and short wavelengths. These types of radiation are associated with the very high temperatures and intense energy found in extremely hot objects, such as the cores of massive stars, supernovae, and certain high-energy astrophysical phenomena. Scientists use X-ray and gamma-ray telescopes to study these extreme environments and the radiation they emit.

Learn more about Electromagnetic Spectrum here:

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a boy on a bicycle is resting at the top of a hill then he rides his bicycle down the hill describe how the boy potential and kineti energy differ at the top middle and bottom of the him

Answers

Answer:

Explanation:

As the boy is riding from top to bottom of the hill his Potential energy gradually decreasing and simultaneously converting into the Kinetic energy.

At the top, his Potential Energy is maximum and kinetic energy is minimum i.e. zero

In the middle of the hill, he possesses both Potential and kinetic energy with equal magnitude.

At the bottom whole Potential energy is converted into Kinetic energy.

Thus Potential Energy is maximum at the top while Kinetic energy is maximum at the bottom.

How much voltage would be necessary to generate 10 amps of current in a circuit that has 5 ohms ofresistance?

Answers

Voltage would be necessary to generate 10 amps of current in a circuit that has 5 ohms of resistance will be 50 V

What is ohm's law ?

Ohm's law is a law that states that the voltage across a resistor is directly proportional to the current flowing through the resistance

using ohm's law

V = IR

V = voltage in a circuit = ?

I = current = 10 A

R = Total resistance in the circuit = 5 ohm

V = IR

= 10 A * 5 ohm

= 50 V

voltage would be necessary to generate 10 amps of current in a circuit that has 5 ohms of resistance will be 50 V

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$U = R * I \rightarrow U = 10 * 5 = 50 V$

What makes a hypothesis testable

Answers

A hypothesis is testable when you can create an experiment to study the proposition contained within the hypothesis. For example, the hypothesis ‘Santa travels slower than a unicorn’ is testable in theory by measuring the speeds of both, but it is not truly testable because neither exists in reality.

Answer:

it must be able to be proved right or wrong - apex

Explanation:

Does anyone understand the Circuit Analysis paper. I'm completely lost. If anyone can help, PLEASE DO!!!

Answers

It's possible that I might have a good chance of understanding it,
and possibly even explaining it, if I could SEE it !

stone thrown vertically upward with speed of 15.5m/s from edge of cliff 75.0m high. how much later does it reach bottom of cliff?

Answers

Answer:

Approximately $5.66\; {\rm s}$ , assuming that $g = 9.81\; {\rm m\cdot s^(-2)}$ and that air resistance is negligible.

Explanation:

Let upward be the positive direction. Under the assumptions, acceleration of the stone would be $a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}$ (negative since the stone is accelerating downward.)

The duration of the flight can be found in the following steps:

Find velocity right before landing given displacement, initial velocity, and acceleration.
Find duration of the flight from acceleration and the change in velocity.

In SUVAT equation $v^(2) - u^(2) = 2\, a\, x$ :

$v$ is the final velocity right before landing (needs to be found,)
$u = 15.5\; {\rm m\cdot s^(-1)}$ is the initial velocity,
$a = (-9.81)\; {\rm m\cdot s^(-2)}$ is acceleration, and
$x = (-75.0)\; {\rm m}$ is displacement (downward because the stone landed below where it was launched.)

Rearrange this equation to find $v$ :

$\begin{aligned}v &= \sqrt{u^(2) + 2\, a\, x} \n &= \sqrt{(15.5)^(2) + 2\, (-9.81)\, (-75.0)} \; {\rm m\cdot s^(-1)} \n &\approx -40.05\; {\rm m\cdot s^(-1)}\end{aligned}$ .

In other words, the velocity of this stone has changed from the initial value of $u = (15.5)\; {\rm m\cdot s^(-1)}$ to $v \approx (-40.05)\; {\rm m\cdot s^(-1)}$ during the flight. Divide the change in velocity by acceleration $a = (-9.81)\; {\rm m\cdot s^(-2)}$ (the rate of change in velocity) to find the duration of the flight:

$\begin{aligned}t &= (v - u)/(a) \n &\approx ((-40.05) - (15.5))/((-9.81))\; {\rm s} \n &\approx 5.66\; {\rm s}\end{aligned}$ .

In other words, the stone would be in the air for approximately $5.66\; {\rm s}$ .

What is the biggest planet in our galaxy

Answers

We won't be able to answer that until we've seen all of them and
we know that there aren't any more that we haven't seen yet.

There are estimated to be between 200 billion and 400 billion stars
in our galaxy, and a large part of the stars in the galaxy aren't visible
to us. So you can see that it's going to take some time before we're
able to answer your question. Check back with us next Tuesday.

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