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rewona [7]
3 years ago
12

Which condition is a neuron in when the outside of the neuron has a net positive charge and the inside has a net negative charge

?. Threshold. Action potential . Resting potential
Physics
2 answers:
White raven [17]3 years ago
8 0
The condition is a neuron in when the outside of the neuron has a net positive charge and the inside has a net negative charge (due to accumulation of more sodium ions) is C. resting potential. T<span>he </span>resting membrane<span> </span>potential<span> of a </span>neuron<span> is approximately -70 mV (mV=</span><span>millivolt)</span>
kotegsom [21]3 years ago
5 0

Answer:

Resting potential

Explanation:

As we know that all the charges are stored inside and outside the neuron in form of positive and negative charge on inner and outer surface.

These charges are stored and remains at rest on the inner and outer surface due to there mutual electric field.

So here the energy must be in the form of electrostatic potential energy where charge is in stationary condition.

So Neuron in this case will hold this charge and hence it will store this electrostatic energy in it due to this stationary charge distribution.

So here the correct answer will be

RESTING POTENTIAL

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Assume a change at the source of sound reduces the wavelength of a sound wave in air by a factor of 3.
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3 years ago
Como estan formados los musculos?
VikaD [51]

Answer:

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3 0
3 years ago
Read 2 more answers
A force is applied to a block sliding along a surface (Figure 2). The magnitude of the force is 15 N, and the horizontal compone
Lemur [1.5K]

Answer:

Fy = 14.3 [N]

Explanation:

To be able to solve this problem we must know that the force is a vector and has magnitude and direction, therefore it can be decomposed into the force in the X & y components:

When we have the components on the horizontal and vertical axes we must use the Pythagorean theorem.

F = \sqrt{F_{x}^{2} +F_{y}^{2}  }

where:

F = 15 [N]

Fx = horizontal component = 4.5 [N]

Fy = vertical component [N]

15=\sqrt{4.5^{2}+F_{y}^{2}}\\ 15^{2}= (\sqrt{4.5^{2}+F_{y}^{2}})^{2} \\225 = 4.5^{2}+F_{y} ^{2}\\  F_{y}^{2} =225 -4.5^{2}\\ F_{y}^{2}=204.75\\F_{y}=\sqrt{204.75}\\  F_{y}=14.3 [N]

7 0
3 years ago
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