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2 years ago

PPLLLZZZZ HELP ME!!! I AM TIMED!! WILL REWARD!!!

Biology

2 answers:

AlekseyPX2 years ago

8 0

Answer:

Options (1), (3), (4) and (5)

Explanation:

The factors that are needed to study in order to understand better about the long-term climate changes are the position of the earth with respect to the sun, type, and intensity of the tectonic and volcanic activities and also the amount of solar energy entering into the earth's atmosphere.

If the position of the earth changes with respect to the sun and also if changes occur in its plane of orbit, then there will be significant changes in the climate of the earth.

Rapid and high intensity tectonic and volcanic activities can also lead to the increase in the earth's global temperature as it will produce a large amount of gases and dust particles from the volcanoes that are trapped in the atmosphere and does not allow the sunlight to reach the ground surface. Thus, affecting the earth's climate.

The increasing amount of solar radiation can also enhance the earth's climate as it will significantly warm the surface of the earth.

Thus, the correct answers are options (1), (3), (4) and (5).

kari74 [83]2 years ago

6 0

the answer is :

1)Earth’s position

3)tectonic activity

4)volcanic activity

5)solar energy

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Explanation:

From the information given:

The cell potential on mars E = + 100 mV

By using Goldman's equation:

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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⁻]:

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

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

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

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

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For [Na⁺]:

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

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

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