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zzz [600]
3 years ago
7

At what time of day would you be most likely to find that the air over water is significantly warmer than the air over land near

by? A. 5:00 AM B. 11:00 AM C. 4:00 PM D. 9:00 PM
Biology
2 answers:
Vaselesa [24]3 years ago
7 0

Answer:

5:00 A.M

Explanation:

on E-d-g-e-n-u-i-t-y there is a lab and I took the test

MissTica3 years ago
5 0
I could have to say a but I’m not to sure
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In this Landsat image of a rain forest, healthy vegetation appears red and deforested areas appear light blue. What explains the
lutik1710 [3]

Answer:

Colors are assigned to different reflected bands electromegnetic waves.

Explanation:

7 0
2 years ago
The bones from an animal found at an archaeological dig have a C614 activity of 0.10 Bq per gram of carbon. The half-life of C61
erastova [34]

C14 is an isotope used in radiocarbon dating techniques to date organic matter remains. The age of these bones is approximately<u> 6890 years.</u>

<h3>What is Carbon 14?</h3>

Carbon 14, also known as radiocarbon, is a radioactive carbon isotope.

Isotopes are the atoms of the same element -carbon- that vary in neutrons and, hence, in their massic number. They are alternative forms of the same element.

The radioactive C14 nucleus contains 6 protons and 8 neutrons and has a half-life of 5730 years.

The term half-life is a reference. It means that an organism that has been dead for 5730 years has half the C14 amount or concentration than the same organism had when it was alive.

Knowing the half-life of an element is useful to determine the age of the dead matter.

C14 is used in radiocarbon dating techniques or methods to estimate the age of fossils. This is a reliable technique used for dating organic samples that are less than 50,000 years old.

<u>Available data</u>:

  • The half-life of C14 is 5730 years
  • Bones activity of 0.10 Bq per gram of carbon

To answer this question we can make use of the following equation

Ln (C14T₁/C14 T₀) = - λ T₁

Where,

  • C14 T₀ ⇒ Amount of carbon in a living body. We know, by bibliography, that living organism activity is 0.23 Bq per gram of carbon. So, C14 T₀ = 0.23 Bq/g
  • C14T₁ ⇒ Amount of carbon in the dead body. C14T₁ = 0.1 Bq/g
  • λ ⇒ radioactive decay constant = (Ln2)/T₀,₅
  • T₀,₅ ⇒ The half-life of carbon 14 = 5730 years
  • T₀ = Time when the organism was alive
  • T₁ = Age of bones

Let us first calculate the radioactive decay constant.

λ = (Ln2)/T₀,₅

λ = 0.693/5730

<u>λ = 0.0001209</u>

Now, let us calculate the first term in the equation

Ln (C14T₁/C14 T₀) = Ln (0.1/0.23) = Ln 0.4347 =<u> - 0.833</u>

Finally, let us replace the terms, clear the equation, and calculate the value of T₁.

Ln (C14T₁/C14 T₀) = - λ T₁

- 0.833 = - 0.0001209 x T₁

T₁ = - 0.833 / - 0.0001209

T₁ =  6889.99 ≅ <u>6890 years</u>

The bones are approximately<u> 6890 years.</u>

You can learn more about dating organic matter with carbon14 at

brainly.com/question/4149380

#SPJ1

6 0
2 years ago
Which characteristic do gnetophytes and cycads have in common? are all deciduous may grow in tropical climates grow as vines are
jonny [76]

Answer AND Explanation:

Gnetophytes are gymnosperms , they are woody  and the embryo has two cotyledons . Cycads are woody plants that produce seeds found in Subtropical and warm temperate regions and also they produce motile sperm cells.

4 0
3 years ago
Which level of organization is best represented by the liver?
siniylev [52]

Answer:

Organ level– an organ is a structure composed of at least two different tissue types that perform a specific function within the body. Examples include the brain, stomach, and liver. Complex functions begin to emerge at this level.

Explanation:

3 0
3 years ago
The leaves of a sensitive plant close whenever any part of the plant is touched because of a???
nadya68 [22]
It is thigmotropism because it is not affected by gravity or light.
6 0
3 years ago
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