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1 year ago

After the depolarization phase of an action potential, the resting potential is restored by.

Physics

1 answer:

Anna007 [38]1 year ago

7 0

Depolarization causes sodium channels to close and voltage-gated potassium channels to open.

<h3>The resting potential is impacted by depolarization in what ways?</h3>

The cell must "reset" its membrane voltage to the resting potential when depolarization is finished. The Na+channels close in order to achieve this and cannot be opened. The neuron is now in its refractory phase, during which it is unable to generate another action potential, as a result of the sodium channels' refusal to open.

<h3>What renews potential while at rest?</h3>

ATPase makes an effort to restore the initial balance of ions by withdrawing Na+ from the cell and bringing K+ into the cell against its gradient of concentration.

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Answer: the maximum heigth of the stadium at ist back wall is 151.32 ft

Explanation:

1. use the position (x) equation in parobolic movement to find the time (t)

565 ft = [frac{176 ft}{1 s\\}[/tex] * cos (35°) * t

t= 3.92 s

2. use the position (y) equation in parabolic movement to find de maximun heigth the ball reaches at 565 ft from the home plate.

y= [[frac{176 ft}{1 s\\}[/tex] * sen (35°) * 3.92 s] - $\frac{32.2 ft/s^{2} *3.92 s^{2} }{2}$

y= 148.32 ft

3. finally add the 3 ft that exist between the home plate and the ball

148.32 ft + 3 ft = 151.32

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3 years ago

A person takes a trip, driving with a constant speed of 89.5km/h, except for a 22.0−min rest stop. If the person's average speed

Elanso [62]

Answer:

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4 years ago

A 1100 kg car is traveling around a flat 82.3 m radius curve. The coefficient of static friction between the car tires and the r

Novay_Z [31]

Answer:

The maximum speed of car will be 20.5m/sec

Explanation:

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3 years ago

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Answer:

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Explanation:

First of all, we can find the angle of the inclined plane.

We have:

L = 5 m the length of the incline

h = 1.2 m is the height

We also have the relationship

$h = L sin \theta$

where $\theta$ is the angle of the incline. Solving for the angle,

$\theta= sin^{-1} (\frac{h}{L})=sin^{-1} (\frac{1.2 m}{5 m})=13.9^{\circ}$

Now we can find the components of the weight of the box, which is the force that the box exerts on the plank. Calling W = 230 N the weight of the box, we have:

- Component parallel to the incline:

$W_{par} = W sin \theta = (230 N)(sin 13.9^{\circ}) =55.3 N$

- Component perpendicular to the incline:

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If the wavelength of a wave increases does its frequency also increase?

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