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2 years ago

During action potential, positively charged ________ ions move inside the cell.

Chemistry

1 answer:

Tems11 [23]2 years ago

8 0

During action potential, positively charged sodium ions move inside the cell.

So option D is correct one.

The sodium ion moves inside the cell during a action potential. The stage of action potential is called depolarization . This open voltage gated sodium channel.

Action potentials ( those electrical impulse that send signals around body ) is nothing but more than temporary shift ( from negative to positive ) in the neuron's membrane potential caused by ions suddenly flowing in and out of the neuron.

It consists of phases:

Depolarization
overshoot
repolarization

An active potential propagates along the cell membrane of an axon until it reaches the terminal button.

to known more about action potential

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Write and balance an equation for the following reaction. Identify the type of reaction.

Strike441 [17]

Answer: Synthesis/ Combination

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3 years ago

Which of the following is NOT a characteristic of a compound?

MissTica

Answer:

I think it's D

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Explanation:

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3 years ago

The structural level of a protein least affected by a disruption in hydrogen bonding is the

sergeinik [125]

I think the correct answer from the choices listed above is option A. The structural level of a protein least affected by a disruption in hydrogen bonding is the primary level. The other levels are very much affected by hydrogen bonding. Hope this answers the question.

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3 years ago

In a Dam, If we doubled the depth of the dam the hydrostatic force will be ?

AnnZ [28]

Answer: Option (A) is the correct answer.

Explanation:

Force acting on a dam is as follows.

F = $\frac{1}{2}\rho g\omega H^{2}$ .......... (1)

Now, when we double the depth then it means H is increasing 2 times and then the above relation will be as follows.

F' = $\frac{1}{2}\rho g\omega (2)^{2}$

F' = $\frac{1}{2}\rho g\omega 4$ ........... (2)

Now, dividing equation (1) by equation (2) as follows.

$\frac{F}{F'}$ = $\frac{\frac{1}{2}\rho g\omega H^{2}}{\frac{1}{2}\rho g\omega 4}$

Cancelling the common terms we get the following.

$\frac{F}{F'}$ = $\frac{1}{4}$

4F = F'

Thus, we can conclude that if doubled the depth of the dam the hydrostatic force will be 4F.

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3 years ago

How does the polarity of water make water more cohesive than other liquids?

MatroZZZ [7]

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3 years ago