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Fires in forests from lightning rid the forest of dead branches and leaves. if the trees survive and the soil remains intact, th

pogonyaev

It is called ecological succession, which can take up to a decade. It is a process of change in the species structure that occurs after a vast loss or extinction. Ecologists study how complexity is evident when a community is formed or occurrence takes place from an empty land or bare soil.

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3 years ago

Which can disrupt the cell cycle? A.mutation B.G0 phase C.replication D.cancer

Aneli [31]

Answer:

c

Explanation:

c

5 0

3 years ago

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It takes approximately _____ action· potentials from olfactory receptors for us to perceive· an odor· .

taurus [48]

Action potentials (i.e. nerve impulses) occur in several types of animal cells, called excitable cells, which include neurons, muscle cells, endocrine cells, and in some plant cells. It takes around 40 action potentials for a smell sensation to be reported.

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3 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 6890 years.

<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.

Available data:

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

λ = 0.0001209

Now, let us calculate the first term in the equation

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

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 ≅ 6890 years

The bones are approximately 6890 years.

You can learn more about dating organic matter with carbon14 at

brainly.com/question/4149380

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6 0

2 years ago

Questions 1 Compare the direction of the currents in the North Pacific Ocean with those of the South Pacific Ocean How are they

serg [7]

Answer:lol

Explanation:

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3 years ago

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