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Anni [7]
3 years ago
12

If a positively charged ion is more concentrated outside the cell, the forces required to balance the chemical gradient would be

directed ________. Thus, the equilibrium potential for this ion would be ________ charged. If a positively charged ion is more concentrated outside the cell, the forces required to balance the chemical gradient would be directed ________. Thus, the equilibrium potential for this ion would be ________ charged. inward : negatively outward : positively outward : neutrally inward : positively outward : negatively
Physics
2 answers:
galben [10]3 years ago
8 0

Answer: If a positively charged ion is more concentrated outside the cell, the forces required to balance the chemical gradient would be directed OUTWARD. Thus, the equilibrium potential for this ion would be POSITIVELY charged. The correct answer is OUTWARD: POSITIVELY.

Explanation: Usually across a cell membrane there is a force that acts on it which is as a result of unequal distribution of charges. This force is known as electrochemical driving force. It is determined by the difference between the membrane potential ( that is, the electrical potential difference across the cell membrane) and the ion equilibrium potential. The membrane potential of a cell helps in signal transmission between different parts of the cell and results when there is unequal distribution across the cell.

Therefore If a positively charged ion is more concentrated outside the cell, the forces required to balance the chemical gradient would be directed outward.Thus, the equilibrium potential for this ion would be positively charged.

antoniya [11.8K]3 years ago
8 0

Answer:

Inward:negatively

Explanation:

This question is duplicated making it incorrectly asked. The correct question is:

If a positively charged ion is more concentrated outside the cell, the forces required to balance the chemical gradient would be directed ________. Thus, the equilibrium potential for this ion would be ________ charged.

inward : negatively, outward : positively, outward : neutrally, inward : positively, outward : negatively

The answer is Inward:negatively

Chemical gradient across a cell or biological membrane is the gradient of the concentration of the ions in the cell or the membrane. To balance the chemical gradient across a cell, an equal force in an opposite direction is required and equilibrium potential must be zero. For instance a positive charge ion more concentrated outside the cell will require a force directed inwardly with a negative charge ion to neutralise the positive charge ion

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The number of cans that would be considered lethal if 10g was lethal and there where 12oz in a can is 419 cans.

<h3>How to convert mass?</h3>

According to this question, caffeine concentration is 1.99 mg/oz.

1.99 milligrams can be converted to grams as follows:

1.99milligrams ÷ 1000 = 0.00199grams

This means that 0.00199grams per oz is the caffeine concentration.

If there were 12 oz in a can, then, 0.00199grams × 12 = 0.02388 grams in 1 can.

This means that if 10grams is considered lethal, 10grams ÷ 0.02388 grams = 419 cans would be lethal for consumption.

Therefore, the number of cans that would be considered lethal if 10g was lethal and there where 12oz in a can is 419 cans.

Learn more about conversion factor at: brainly.com/question/14479308

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To solve this problem we will use the linear motion description kinematic equations. We will proceed to analyze the general case by which the analysis is taken for the second car and the tenth. So we have to:

x = v_0 t \frac{1}{2} at^2

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a = Acceleration

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Since there is no initial velocity, the same equation can be transformed in terms of length and time as:

L = \frac{1}{2} a t_1 ^2

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2L \frac{1}{2} at_2^2

When the tenth car is aligned the length will be 9 times the initial therefore:

9L = \frac{1}{2} at_3^2

When the tenth car has passed the length will be 10 times the initial therefore:

10L = \frac{1}{2}at_4^2

The difference in time taken from the second car to pass it is 5 seconds, therefore:

t_2-t_1 = 5s

From the first equation replacing it in the second one we will have that the relationship of the two times is equivalent to:

\frac{1}{2} = (\frac{t_1}{t_2})^2

t_1 = \frac{t_2}{\sqrt{2}}

From the relationship when the car has passed and the time difference we will have to:

(t_2-\frac{t_2}{\sqrt{2}}) = 5

t_2 (\sqrt{2}-1) = 3\sqrt{2}

t_2= (\frac{5\sqrt{2}}{\sqrt{2}-1})^2

Replacing the value found in the equation given for the second car equation we have to:

\frac{L}{a} = \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2

Finally we will have the time when the cars are aligned is

18 \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_3^2

t_3 = 36.213s

The time when you have passed it would be:

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The difference between the two times would be:

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