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kiruha [24]
2 years ago
9

BRAINLIEST+10PTS: does ozone absorb just harmful UV rays? Or just a random portion of UV?

Physics
2 answers:
emmasim [6.3K]2 years ago
4 0

Answer:As sunlight passes through the atmosphere, all UVC and most UVB is absorbed by ozone, water vapour, oxygen and carbon dioxide. UVA is not filtered as significantly by the atmosphere. Is there a connection between ozone depletion and UV radiation? Ozone is a particularly effective absorber of UV radiation.

Explanation:

I had old notes

Umnica [9.8K]2 years ago
3 0

Answer:

The ozone layer absorbs a random portion of UV rays.

Explanation:

The ozone layer absorbs a random portion of UV rays because the only way UV rays could be harmful to your skin is if there is too much of it. All UV rays are bad, but the only reason you get harmed from them is if you absorb an excessive amount of them.

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If a car's mass is 439 kg, what is its weight on earth?
trasher [3.6K]
The correct answer is B.
8 0
3 years ago
What does the angular momentum quantum number determine? Check all that apply.
amm1812
I think the correct answers from the choices listed above are options 1, 5 and 7. Angular momentum quantum number determine the energy of an orbital, the shape of the orbital and <span>the overall size of an orbital. Hope this answers the question.</span>
8 0
3 years ago
Read 2 more answers
When the distance increases, the electrostatic force____; we call this relationship proportional
Alik [6]

Answer:

I think decreases inversely

Explanation:

the third

5 0
2 years ago
Read 2 more answers
When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a
sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = \frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}

  = 4.59\times 10^{-19} \ J

or,

  = \frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}

  = 2.87 \ eV

(b)

As we know,

⇒ Vq=\frac{hc}{\lambda}-\Phi_0

By substituting the values, we get

⇒ 1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0

⇒                       \Phi_0=2.3\times 10^{-19} \ J

or,

⇒                            =\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}

⇒                            =1.4375 \ eV

5 0
3 years ago
When you and a friend move a couch to another room you exert a force of 75 N over 5m how much work will you do
sesenic [268]

The work done is 375 J

Explanation:

The work done by a force in moving an object is given by

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and the displacement

In this problem,

F = 75 N is the force applied to the couch

d = 5 m is the displacement

Assuming the force applied to the couch is parallel to the motion, \theta=0

And so, the work done is

W=(75)(5)(cos 0)=375 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

6 0
3 years ago
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