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noname [10]
3 years ago
10

What can taphonomy tell scientists about fossils? (site 1)

Biology
2 answers:
Sergio039 [100]3 years ago
5 0
Taphonomy is the study of the process of fossilization, from the death of the organism to the moment of the fossil's discovery and everything in between. Fossilization is the process by which a plant or animal becomes a fossil, the process is extremely rare and only a small fraction of the plants and animals that have lived in the past 600 million years are preserved as fossils. Taphonomy entails the study of what happens in plants and animals after they die, including the processes of decay and fossilization.
Gnesinka [82]3 years ago
3 0

Answer:

The taphonomy tries to explain all the events that affected an organism from the past, from the time the individual died until their fossils are found in the stratigraphic record.

Therefore, the results are investigated, in this case the fossils and their state of conservation, to be able to postulate and verify the processes and the agents involved, over time, in the formation of a deposit.

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C

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Increased biodiversity helps diminish reliance on one species

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assume that life on Mars requires cell potential to be 100mV, and the extracellular concentrations of the three major species ar
svetlana [45]

Answer:

Explanation:

From the information given:

The cell potential on mars E = + 100 mV

By using Goldman's equation:

E_m = \dfrac{RT}{zF}In \Big (\dfrac{P_K[K^+]_{out}+P_{Na}[Na^+]_{out}+P_{Cl}[Cl^-]_{out} }{P_K[K^+]_{in}+P_{Na}[Na^+]_{in}+ P_{Cl}[Cl^-]_{in}}      \Big )

Let's take a look at the impermeable cell with respect to two species;

and the two species be Na⁺ and Cl⁻

E_m = \dfrac{RT}{zF} In \dfrac{[K^+]_{out}}{[K^+]_{in}}

where;

z = ionic charge on the species = + 1

F = faraday constant

∴

100 \times 10^{-3} = \Big (\dfrac{8.314 \times 298}{1\times 96485} \Big) \mathtt{In}  \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

100 \times 10^{-3} = 0.0257 \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

3.981= \mathtt{In} \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

exp ( 3.981) = \dfrac{4}{[K^+]_{in}} \\ \\  53.57 = \dfrac{4}{[K^+]_{in}}

[K^+]_{in} = \dfrac{4}{53.57}

[K^+]_{in}  = 0.0476

For [Cl⁻]:

100 \times 10^{-3} = -0.0257 \  \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}}   \Big)

-3.981 =  \  \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}}   \Big)

0.01867 =  \dfrac{120}{[Cl^-]_{in}}

[Cl^-]_{in} = \dfrac{120}{0.01867}

[Cl^-]_{in} =6427.4

For [Na⁺]:

100 \times 10^{-3} = 0.0257 \Big ( \dfrac{145}{[Na^+]_{in}}   \Big)

53.57= \Big ( \dfrac{145}{[Na^+]_{in}}   \Big)

[Na^+]_{in}= 2.70

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What life-threatening infection can result from allowing a urinary tract infection to go untreated in an older adult?
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Hi!

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