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qwelly [4]
1 year ago
5

How do you compare the energies of the different parts of the electromagnetic spectrum?

Physics
1 answer:
faltersainse [42]1 year ago
5 0

The different types of radiation in electromagnetic spectrum are compared by the amount of energy found in the photons.

Radio waves have photons with low energies, microwave photons have a little more energy than radio-waves. Infrared photons still have more energy, then comes visible, ultraviolet, x-rays and the most energetic of all, gamma rays.

The energy associated with electromagnetic radiation is proportional to frequency and inversely proportional to wavelength. So, electromagnetic waves with shorter wavelengths have more energy.

On one end of the electromagnetic spectrum are radio waves, which have wavelengths billions of times longer than those of visible light. On the other end of the spectrum are gamma rays with wavelengths billions of times smaller than those of visible light.

To know more about electromagnetic spectrum:

brainly.com/question/27839167

#SPJ4

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You apply a net force on a soccer ball of 15 N. If the acceleration it has is 5 m/s2 what is the mass of the ball?​
Ipatiy [6.2K]

Answer:

<h2>3 kg </h2>

Explanation:

The mass of the ball can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

We have

m =  \frac{15}{5}  = 3 \\

We have the final answer as

<h3>3 kg</h3>

Hope this helps you

4 0
3 years ago
Antifreeze is a "must have" in an Albertan winter because it helps keep your vehicle engine from freezing in cold temperatures.
Natasha_Volkova [10]

Products such as antifreeze are composed of organic compounds that are classified as <em>alcohols</em>. (a)

Maybe those other classes of chemicals also lower the freezing temperature of water, just like alcohol does.  I don't know.  But alcohol is what's used to make anti-freeze.  I'm guessing alcohol must be cheaper, less toxic, and less corrosive inside the engines' cooling systems than any of that other stuff is.

4 0
3 years ago
a stone attached to 1m long string is moving with the speed of 5ms in a circle find the centripetal acceleration of the stone​
Dafna1 [17]

Answer:

The centripetal acceleration of the stone is 5 m/s²

Explanation:

The length of the string to which the stone is attached, r = 1 m

The speed with which the string is rotated, v = 5 m/s

The centripetal acceleration, a_c, is given as follows;

a_c = \dfrac{v^2}{r}

Therefore, the centripetal acceleration of the stone found as follows;

a_c = \dfrac{(5 \ m/s)^2}{1 \ m} = 5 \ m/s^2

The centripetal acceleration of the stone, a_c = 5 m/s².

5 0
3 years ago
How do Newton's laws of motion explain why it is important to keep the ice smooth on a hockey rink so that players can
lord [1]

Answer:

I'm not sure..but please refer to your teacher later.

Answer: Based on Newton's First law of motion (where inertia is involved), smooth ice increases the forceused to accelerate the hockey puck.

Explanation;

  • smooth ice reduces the resistances between the surface of the figure skates and the ice itself.
  • based on inertia theory ; the heavier the weight, the larger the inertia.. which explains it takes alot of force to move a heavier object than the lighter ones.. it also hard to *stop* the motion of heavier objects than the lighter ones.
  • now let's look at the design of the player shoe itself, they have a sharp blade at the bottom of the figure stakes.. which takes us to the law of the force.. the smaller the surface area, the more forces acting on it. So, players force (weight, F= mg) acts on the tip of the blade and on the ice
  • high inertia (run fast) and high force (attack opponent and pass puck) enables them to perform well in playing hockey
  • Thus if there's no resistance and the inertia of the player is high then they could run and pass the puck quickly
6 0
3 years ago
Read 2 more answers
Light of wavelength 623.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 76.5 cm from the slit.
Dvinal [7]

Answer:

Explanation:

wave length of light λ = 623 x 10⁻⁹ m .

Distance of screen D = 76.5 x 10⁻² m

width of slit        =      d

Distance on the screen between the second order minimum and the central maximum       =  2  λ D / d

1.11 x 10⁻²  = (2 x 623 x 10⁻⁹ x 76.5 x 10⁻² )/ d

d =  ( 2 x 623 x 10⁻⁹ x 76.5 x 10⁻²) / 1.11 x 10⁻²

= 85872.97 x  10⁻⁹

=  85.87297 x  10⁻⁶

= 85.87 μm

width a of the slit is = 85.87 μm

6 0
3 years ago
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