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Nata [24]
2 years ago
12

In a given neuron, the current membrane potential is 60 mv. what does this tell you about the current potential compared to the

resting membrane potential of 70 mv?
Physics
1 answer:
alekssr [168]2 years ago
3 0

The membrane is depolarized compared to the resting membrane potential.

Through conformational changes from closed, nonconducting states to an open, current-conducting state, membrane depolarization activates sodium channels. Na+ channels open slowly and change from an open state to a nonconducting, rapidly inactivated state as a result of delayed openings, which contribute to the declining fraction of INa induced by prolonged depolarization. Additionally, sodium channels can move swiftly from the closed state to the fast-inactivated state. When the membrane is depolarized, inactivated channels are prevented from opening.

The distribution of channels between the closed and slow-inactivated states, however, limits the number of excitable sodium channels as a function of the membrane potential since slow inactivation acts at greater negative potentials than fast inactivation.

To learn more about membrane potential please visit-
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Someone help me with this ASAP!!
alexandr402 [8]
It protects us from the sun so the answer would be A
3 0
3 years ago
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A mass free to vibrate on a level, frictionless surface at the end of a horizontal spring is pulled 35 cm from its equilibrium p
saul85 [17]

Answer:

0.67 s

Explanation:

This is a simple harmonic motion (SHM).

The displacement, x, of an SHM is given by

x = A\cos(\omega t)

A is the amplitude and \omega is the angular frequency.

We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or \frac{\pi}{4} radian.

From trigonometry, \sin A =\cos B if A and B are complementary.

At t = 0, x = 3.5

3.5 = A\cos(\omega \times0)

A =3.5

So

x = 3.5\cos(\omega t)

At t = 0.12, x = 1.5

1.5 = 3.5\cos(0.12\omega)

\cos(0.12\omega)=\dfrac{1.5}{3.5}=0.4286

0.12\omega =\cos^{-1}0.4286

0.12\omega = 1.13

\omega = 9.4

The period, T, is related to \omega by

T = \dfrac{2\pi}{\omega} = \dfrac{2\times3.14}{9.4}=0.67

5 0
3 years ago
Which planet do most known extrasolar planets most resemble?
tatuchka [14]
The awnser is A>Most known exoplanets resemble gas giants known as "hot Jupiters" as a result of being large objects orbiting close to they're host star.
Most known exoplanets that we have detected are likely or confirmed to be gas giants. This is because one of the most used detection techniques is the Transit technique.

This technique involves looking at a distance star's light curve (the changes in the brightness of star over a period of time). Most stars dim and brighten over time by a small bit. But if the light curve shows a periodic and large dip in brightness, this is a sign that something large is passing (or transiting) in front of it.

The reason most detected exoplanets are gas giants is simply because they're the easiest to detect. They cause a larger dip in the curve than a smaller planet would. A small planet's dip are masked by the star's normal dimming and brightening.

Imagine having a flashlight shining on a wall. If you pass a large object through the light beam, it has a large shadow. If you pass a smaller object through, it has a smaller shadow. The change in the light you see on the wall is similar to what you will see in a star's light curve. Now if you pass an object closer to flashlight, as opposed to closer to the wall, this changes the shadow as well.

This is why most exoplanets are "hot Jupiters". They're large gas giants that are hot because they pass very close to the star, resulting in a larger dip that is easier to see in the curve. They also have shorter orbits so they're more likely to be seen in a few weeks or month's time.


6 0
4 years ago
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TiliK225 [7]

is this the skater one? If so, it does support the law of conservation of energy because at the top of the ramp the skater has the same amount of potential energy their total energy and as the skater goes down the ramp their potential energy lowers as their kinetic energy rises.

6 0
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Vaselesa [24]

Answer:

21,000 N

Explanation:

You would use the formula F=ma

So you would do substitution...

F=700 kg(30 m/s/s)

F=21,000 N

6 0
3 years ago
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