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sergejj [24]
3 years ago
10

Which best describes the ovary?

Biology
2 answers:
alukav5142 [94]3 years ago
7 0
It produces progesterone!!
klio [65]3 years ago
6 0
It is a to my knowledge, but I will give an update when i find out
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The external iliac vein ________. the external iliac vein ________. empties directly into the inferior vena cava drains the pelv
svetlana [45]
The correct answer is:  [D]:  "receives venous blood from the lower extremity" .
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8 0
3 years ago
Plants, monera, and animals are all A.Have cell walls B. In the same kingdom C. In different kingdoms D. Have chloroplast
kupik [55]
A i think would be the best choice
6 0
2 years ago
What is the function of the organs of the digestive system
ohaa [14]

Answer:

The digestive system is a collection of organs that work together to digest and absorb food. Digestion is the process your body uses to break the foods you eat down into molecules your body can use for energy and nutrients. The following organs work together to help your body process the foods you eat.

3 0
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<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
Community ecologists study the factors that determine the abundance of different species. The Abundance of three algae species (
Tema [17]

Answer:

Explanation:

Given that:

The abundance of three algal species in Lake A is now represented by the vectors:

[97, 84, 43] and [100, 80, 50]

Now if we look at Lake A, the change occurring in the vector of algae species can be determined as:

a = [100 -97, 80 - 84. 50 - 43]

a = [3, -4, 7]

Thus, the magnitude of that change is:

|a| = \sqrt{(3)^2 +(-4)^2+(7)^2}

|a| = \sqrt{9+16+49}

|a| = \sqrt{74}

|a| =8.6 \ mg/mL

The abundance of three algal species in Lake B is now represented by the vectors:

[25, 59, 22] and [20, 63, 15]

At Lake B, the change occurring in the vector of algae species can be determined as:

b = [20 -23, 63 - 59. 15 - 22]

b = [-3, -4, -7]

Thus, the magnitude of that change is:

|b| = \sqrt{(-3)^2 +(4)^2+(-7)^2}

|b| = \sqrt{9+16+49}

|b| = \sqrt{74}

|b| =8.6 \ mg/mL

Hence, for both Lake A and B, the magnitude of change is the same.

7 0
2 years ago
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