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3 years ago

HELP ME PLZ 30 POINT AND A MARK

Physics

1 answer:

patriot [66]3 years ago

8 0

Question 1:How do humans and animals obtain the nitrogen their bodies need to make proteins and nucleic acids?

Answer is :by consuming plants or other animals

Question 2:Which is a key element found in all carbohydrates, lipids, proteins, and nucleic acids

Answer is:Carbon

Question 3:How do decomposers affect the carbon cycle?

Answer is:They release the carbon that remains in the bodies of dead organisms.

Question 4:Which of the following processes does not produce carbon dioxide?

Answer is:Photosynthesis

Question 5:How does the carbon in grass become part of a lion

Answer is:The lion eats an herbivore that ate the grass.

And your Attempt Score=10 / 10 - A

Overall Grade (highest attempt)=10 / 10 - A

Hope I helped you all who is looking for the answers to the test:cycles in nature.

If you guys do get an A plz let me know in the comments,and mark me brainliest for helping thx bye!!!

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I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi

Masteriza [31]

Answer:

After it has traveled through 1 cm : $I(1 \ cm) = 0.5488 I_0$

After it has traveled through 2 cm : $I(2 \ cm) = 0.3012 I_0$

After it has traveled through 1 cm : $od( 1\ cm) = 0.2606$

After it has traveled through 2 cm : $od( 2\ cm) = 0.5211$

Explanation:

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

The optical density od is given by:

$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$ .

So, after travelling 1 cm:

$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$

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