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malfutka [58]
3 years ago
10

Eral

Biology
1 answer:
EleoNora [17]3 years ago
3 0

Complete question: The chart shows the movement of a ball after several seconds.

Column A is on the x-axis, and Column B is on the y-axis. Which titles should replace A and B?

a. Column A should be “Time,” and Column B should be “Position.”

b. Column A should be “Position,” and Column B should be “Time.”

c. Column A should be “Velocity,” and Column B should be “Speed.”

d. Column A should be “Speed,” and Column B should be “Velocity.”

Answer:

Column A should be "Time," and Column B should be  "Position"

Explanation:

In general, the first column always refers to the x-axis, this is the independent variables. While the second column refers to the y-axis, the dependent variables.

Option A) names Time and Position to the columns. This seems to be correct as time is an independent variable, while the ball position depends on time. The x-axes is named "time", and the Y-axes is named "ball position".

Option B) is not possible because the Position is not the independent variable, and time is not the dependent variable.

Option C) and D) are not possible, as both of them are variables that depend on time.

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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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