The movement of what ion is reflected in the rising phase of the action potential?
Answers
The movement of sodium ions (Na+) is reflected in the rising phase of the action potential.
During an action potential, the rising phase is the period of depolarization, where the membrane potential becomes more positive. This is mainly due to the influx of sodium ions into the cell.
At rest, the cell membrane is polarized, meaning the inside of the cell is negatively charged compared to the outside. When a stimulus is received, voltage-gated sodium channels open, allowing sodium ions to rush into the cell. This influx of positively charged sodium ions depolarizes the cell, making the inside less negative and more positive.
As more sodium ions enter the cell, the membrane potential continues to rise, leading to the peak of the action potential.
It is important to note that other ions, such as potassium (K+), also play a role in the action potential, but their movement primarily influences the repolarization and hyperpolarization phases, respectively.
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Related Questions
What is the acceleration of a 50N object traveling at terminal velocity0m/s/s50m/s/s10m/s/s-10m/s/s
Answers
The acceleration of an object traveling at terminal velocity is 0 m/s/s since it is not accelerating.
What is terminal velocity?Terminal velocity is the maximum speed achieved by an object as it falls through a fluid such as air or water. The object's weight, drag coefficient, and surface area all affect its terminal velocity. Terminal velocity increases as an object's weight increases and its drag coefficient and surface area decrease. Terminal velocity is greatest when an object reaches its equilibrium between the force of gravity and the fluid's drag force. For example, a human skydiver has a terminal velocity of about 120 mph.
Terminal velocity is the maximum velocity an object can reach and remain at a constant speed as it is subjected to a constant force such as gravity. If the object is subject to any additional force, it will accelerate.
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The coefficient of kinetic friction is 0.21, for a 8 kg block on the floor. It is being pulled by a 17 N force, what is the acceleration of the box?
Answers
The acceleration of the box is 0.025 m/s²
We'll begin by calculating the the frictional force
Coefficient of kinetic friction (μ) = 0.21
Mass (m) = 8 Kg
Acceleration due to gravity (g) = 10 m/s²
Normal reaction (N) = mg = 8 × 10 = 80 N
Frictional force (Fբ) =?Fբ = μN
Fբ = 0.21 × 80
Fբ = 16.8 NNext, we shall determine the net force acting on the boxFrictional force (Fբ) = 16.8 N
Force (F) = 17 N
Net force (Fₙ) =?Fₙ = F – Fբ
Fₙ = 17 – 16.8
Fₙ = 0.2 NFinally, we shall determine the acceleration of the boxMass (m) = 8 Kg
Net force (Fₙ) = 0.2 N
Acceleration (a) =?a = Fₙ / m
a = 0.2 / 8
a = 0.025 m/s²Therefore, the acceleration of the box is 0.025 m/s²
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In the parallel circuit of (Figure 1), the current amplitude is the same through the inductor branch, the capacitor branch, and the resistor branch. L
Answers
Thus angular frequency \(\omega =2\pi f =\frac{1}{\sqrt{LC}}=50 rad/s\) and the resistance is 2 Ohm.
What is inductance and resistance?
When a coil opposes the change in current flowing through it, self-inductance results. When a secondary coil resists a primary coil's change in current, mutual inductance results. Henrys are used to measure inductance. d) Resistance: A material's ability to obstruct the passage of electrical current through it.
the question is incomplete, the complete question is -
In the parallel circuit of, the current amplitude is the same through the inductor branch, the capacitor branch, and the resistor branch. L = 40.0 mH and C = 10.0 mF.
a)what is source angel frequency?
b)what is the resistance R?
Current in any any branch is given as
\(i=\frac{V}{Z}\)
In inductive circuit
\(i_{L}=\frac{V}{X_{L}}=\frac{V}{2\pi L}\)
in capacitive circuit
\(i_{ c}=V*2\pi fC\)
in resistive circuit
\(i_{ R}=\frac{V}{R}\)
According to question
both capacitive and inductive circuit has same current amplitude thus
\(V*2\pi fC =\frac{V}{2\pi fL} \Rightarrow f=\frac{1}{2\pi\sqrt{LC}}\)
Thus angular frequency
\(\omega =2\pi f =\frac{1}{\sqrt{LC}}=50 rad/s\)
Part B
again since the same amplitude is flowing through resistor branch
thus R= Xc=XL
R= 50*40 *10-3 Ohm = 2 Ohm
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A box having a mass of 1.5 kg is accelerated across a table at 1.5 m/s2. The coefficient of friction on the box is 0.3. What is the force being applied to the box? If this force were applied by a spring, what would the spring constant have to be in order for the spring to be stretched to only 0.08 m while pulling the box?
Answers
Force = mass x acceleration
Plugging in the given values, we get:
Force = 1.5 kg x 1.5 m/s²
Force = 2.25 N
Therefore, the force being applied to the box is 2.25 N.
To find the spring constant required to stretch the spring only 0.08 m while pulling the box, we can use Hooke's law:
Force = spring constant x displacement
Plugging in the given values, we get:
2.25 N = k * 0.08 m
Solving for the spring constant, we get:
k = 2.25 N / 0.08 m
k = 28.125 N/m
Therefore, the spring constant would have to be 28.125 N/m in order for the spring to be stretched to only 0.08 m while pulling the box with a force of 2.25 N.
Metric conversions
SHOW WORK!!!
1.) 45.23 L —> mL
2.) .035 hL—> cL
3.) 27.32 mm —> m
Answers
1:-
1L=1000mL\(\\ \sf\longmapsto 45.23L\)
\(\\ \sf\longmapsto 45.23\times 1000\)
\(\\ \sf\longmapsto 45230mL\)
2:-
1hL=10000cL\(\\ \sf\longmapsto .035hL\)
\(\\ \sf\longmapsto 0.035\times 10000\)
\(\\ \sf\longmapsto 350cL\)
3:-
1mm=0.0001m\(\\ \sf\longmapsto 27.32\)
\(\\ \sf\longmapsto 27.32\times 0.0001\)
\(\\ \sf\longmapsto 0.0027m\)
A baseball bat strikes a ball with an average force of 2.0 x 104 Newtons. if the bat stays in contact with the ball for a distance of 5.0 x 10-3 meter, what kinetic energy will the ball acquire from the bat?
1.0 x 102 J
2.0 x 102 J
2.5 x 101 J
4.0 x 102 J
Answers
A baseball bat strikes a ball with an average force of 2.0 x 104 Newtons. if the bat stays in contact with the ball for a distance of 5.0 x 10-3 meter , the kinetic energy is 2× 10² J
In classical mechanics, the kinetic energy of a non-rotating body of mass m moving with velocity v is 1/2 mv².
Favg= -1/2 MV²
Favg = -1 2 ( 5.0 x 10⁻³)²
Favg = - 1/2 ×25 × 10⁻⁶
Favg = 2× 10² J
Kinetic energy is the energy that an object has when it is in motion. It is defined as the work required to accelerate an object of specified mass from rest to a specified velocity. After the body acquires this energy during acceleration, it retains this kinetic energy as long as the velocity does not change.
The kinetic energy of an object is the energy it has due to its motion. It is defined as the work required to accelerate an object of specified mass from rest to a specified velocity.
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which energies change as the bobsled moves down the slope and hits the spring? select all that apply.
Answers
In our bobsleds, the bobsled's potential energy on the ramp was converted into kinetic energy as it descended the ramp, and then thermal energy as friction as it came to a stop.
What does science mean by a potential energy?The term "potential energy" describes the energy that is held within an object due to the object's position, state, and arrangement. It is one of the two essential types of energy; the other is energy. The object's position, status, or configuration changes, releasing the stored energy.
Describe a potential energy example.Due to its position, an object might accumulate energy. If a demolition machine's heavy ball is held in an elevated position, for instance, it can store energy. Potential energy is the name for this kinetic energy that is held in a position. As a result of its location, a drawn bow can also store energy.
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value of x
Guys plz answer
and get marked as brainliest :)
Answers
Answer:
80⁰
Explanation:
triangle AOB = 180
So, a = 50
triangle CAB = a+a+x=180
CAB = 50+50+X=180
CAB = X = 180-100
X=80
HOPE IT HELPS
Answer:
Explanation:
Since OA = OB (Radius of a circle)
angle OAB = angle OBA = 50
I'm not sure what to do beyond this...
I hope this piece of information was helpful.
The acceleration of free fall is about 10 m/s^2. Why does the unit second appear twice?
Answers
In free fall, the fact that in the acceleration value (10 m/s^2) the unit of the second appears twice indicates that the object is moving at a rate of 10 meters per second every second.
The acceleration is the physical quantity that indicates the variation of velocity as a function of time, and it is expressed in units of distance per time squared e.g.: m/sec2 ; km/h2
What is free fall?It is when the object or mobile falls from a height (h) with a positive acceleration equal to the gravity, describing a vertical rectilinear travel.
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A periodic wave transfers...
A: energy, only
B: mass, only
C: both energy and mass
D: neither energy nor mass
Answers
The answer is that a periodic wave transfers energy, only.
A wave is a disturbance that travels through a medium, transferring energy but not mass. As a wave travels through the medium, the particles of the medium oscillate back and forth, but they do not travel with the wave. The energy of the wave is transferred from particle to particle, but the particles themselves do not move with the wave. Therefore, the correct answer to your question is A.
In summary, a periodic wave transfers energy but not mass through the medium. This is an important concept to understand in many fields, including physics, engineering, and biology.
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1
Which device can only be used for heating a space?
an air conditioner
a heat pump
a furnace
a heat engine
Answers
Explanation:
heat engine is correct answer
Answer:
The answer is:
C) A Furnace
When the temperature of water raises from ___ degrees Celsius, it becomes a liquid.
Answers
Answer:
At temperatures above 32°F (0°C), pure water ice melts and changes state from a solid to a liquid (water); 32°F (0°C) is the melting point. For most substances, the melting and freezing points are about the same temperature.
An isolated charged point particle produces an electric field with magnitude e at the distance 2m from it. At the distance 1m from the particle the magnitude of the field is:.
Answers
While the electric field caused by a dipole changes inversely with the cube of the distance, the electric field caused by an isolated point charge changes inversely with the square of the distance.
Where is an isolated evenly charged object's electric field?The electric field strength is zero throughout the whole cavity of a spherically charged, isolated shell. The length of the electric field is endlessly long and uniformly charged inside a cylindrical shell that is separated from the outside world.
The size of the electric field produced by a point charge Q is determined by this equation. The distance r in the denominator is the separation between the point of interest and the point charge, Q, or the centre of a spherical charge.
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Let’s begin by calculating the total momentum of two moving objects. Suppose a small compact car with a mass of 1000 kg is traveling north on Morewood Avenue at a speed of 15 m/s. At the intersection of Morewood and Fifth Avenues, it collides with a truck with a mass of 2000 kg that is traveling east on Fifth Avenue at 10 m/s. Treating each vehicle as a particle, find the total momentum (magnitude and direction) just before the collision.
If the car is moving at 20 m/s , how fast must the truck move for the total momentum vector to make a 45 ∘ angle with the +x axis (and 45 ∘ with the +y axis)?
Answers
The truck must be moving eastward at 10 m/s for the total momentum vector to make a 45° angle with both the +x and +y axes.
To find the total momentum just before the collision, we can use the conservation of momentum principle, which states that the total momentum of an isolated system remains constant unless acted upon by an external force.
In this case, we can assume that there are no external forces acting on the system of the car and truck before the collision.
The momentum of an object is given by its mass times its velocity, so the momentum of the car before the collision is:
p_car = m_car * v_car = 1000 kg * 15 m/s = 15000 kg·m/s (north)
The momentum of the truck before the collision is:
p_truck = m_truck * v_truck = 2000 kg * 10 m/s = 20000 kg·m/s (east)
To find the total momentum just before the collision, we can use the Pythagorean theorem to combine the magnitudes of these momenta, and the tangent function to find the direction:
\(p_total = sqrt(p_car^2 + p_truck^2) ≈ 24594 kg·m/s\)
theta = arctan(p_car/p_truck) ≈ 36.9° north of east
Therefore, the total momentum just before the collision is about 24594 kg·m/s, with a direction of 36.9° north of east.
To find the speed of the truck needed for the total momentum vector to make a 45° angle with both the +x and +y axes, we can use trigonometry. Let's call the required speed of the truck "v_truck".
We know that the momentum vector must have equal x and y components, so:
p_total,x = p_total,y
Using the momentum values from before and the fact that the car is moving at 20 m/s:
m_car * 20 m/s = m_truck * v_truck
Solving for v_truck, we get:
v_truck = (m_car/m_truck) * 20 m/s
Substituting the given masses, we get:
v_truck = (1000 kg/2000 kg) * 20 m/s = 10 m/s
Therefore, the truck must be moving eastward at 10 m/s for the total momentum vector to make a 45° angle with both the +x and +y axes.
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Sound travels faster in air if the air temperature is :_____
a) warm.
b) average.
c) cold.
Answers
Answer: A) warm
Explanation:
The higher the temperature the faster it travels.
When a drop of ink mixes with water, what happens to the entropy of the system? A) The system's entropy increases, and the total entropy of the universe increases. B) The system's entropy decreases, and the total entropy of the universe increases. C) The system's entropy increases, and the total entropy of the universe decreases. D) The system's entropy decreases, and the total entropy of the universe decreases.
Answers
option A) The system's entropy increases, and the total entropy of the universe increases is correct.
When a drop of ink mixes with water, the system's entropy increases, and the total entropy of the universe increases. This statement is true Entropy is a measure of disorder or randomness. A substance with a higher entropy is more disorderly, while a substance with a lower entropy is more orderly.
The second law of thermodynamics is the law that governs entropy. It states that the overall entropy of a system always rises over time. According to this law, energy naturally flows from areas of high concentration to areas of low concentration until a state of equilibrium is achieved. This process is irreversible.A drop of ink will diffuse in water when it is added. The ink will be evenly distributed throughout the water, resulting in an increase in entropy.
When the entropy of a system rises, its capacity to do work decreases. Because the entropy of the universe has increased, the total entropy has also increased, according to the second law of thermodynamics.
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Explain the advantages and disadvantages of having a racecar with a large mass in reference to Newton’s 1st law of motion. Use the term inertia in your answer.
Answers
Answer:
read the explanation
Explanation:
My answer to research question one is a car with a higher mass has more inertia than a car with a lower mass. A car with a higher mass needs more force to move than a car with a lower mass. The advantage of a car with greater mass is that it takes less impact when crash happens and has more protection.
The advantage of a race car with large mass is that it will continue moving without stopping once it is already in motion.
The disadvantage of a race car with large mass is that it will be difficult to get it start moving.
The Newton's first law of motion states that, an object in a state of rest or uniform motion in a straight line will continue in that form unless it is acted upon by an external force.
This first law is also known as law of inertia because it depends on mass of the object.
Inertia is reluctance of an object to move or stop moving once it is in motion.
Thus, the advantage of a race car with large mass is that it will continue moving without stopping once it is already in motion.
The disadvantage of a race car with large mass is that it will be difficult to get it start moving.
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5). A body of mass 75kg lying on a surface of
μ = 0.4 move in the direction of a horizontal force
of 300N applied to it.
a). Calculate the
i). friction force
ii). acceleration of the body
b). Explain the result in aii).
Answers
Explanation:
a) i) Calculation of the friction force:
The friction force can be determined using the equation:
friction force = coefficient of friction * normal force
The normal force is equal to the weight of the object, which can be calculated as:
normal force = mass * gravitational acceleration
where the gravitational acceleration is approximately 9.8 m/s².
normal force = 75 kg * 9.8 m/s² = 735 N
friction force = 0.4 * 735 N = 294 N
ii) Calculation of the acceleration of the body:
Now, we can calculate the acceleration using Newton's second law:
net force = mass * acceleration
Since the applied force and the friction force act in opposite directions, the net force can be calculated as:
net force = applied force - friction force
net force = 300 N - 294 N = 6 N
mass = 75 kg
6 N = 75 kg * acceleration
acceleration = 6 N / 75 kg = 0.08 m/s²
b) Explanation:
In part (a), we calculated the friction force to be 294 N and the acceleration of the body to be 0.08 m/s². The positive acceleration indicates that the body is moving in the direction of the applied force.
The friction force opposes the motion of the body and acts in the opposite direction to the applied force. In this case, the applied force of 300 N is greater than the friction force of 294 N. As a result, the net force acting on the body is 6 N in the direction of the applied force.
The small net force of 6 N, compared to the body's mass of 75 kg, results in a relatively low acceleration of 0.08 m/s². This indicates that the body will accelerate slowly in the direction of the applied force due to the presence of friction.
Overall, the friction force and the resulting acceleration of the body are determined by the coefficient of friction (μ) and the mass of the object. In this case, the body experiences a relatively high friction force, leading to a small acceleration.
the λmax of s157c at ph 4.5 is greater than at ph 10. how does the proposed bonding of deprotonated c157 with y63 at ph 10 account for this observation?
Answers
The λmax of S157C at pH 4.5 is greater than at pH 10 due to the proposed bonding of deprotonated C157 with Y63 at pH 10.
At pH 4.5, the side chain of C157 is protonated, which means that it carries a positive charge. This protonated form of C157 is less likely to form a bond with Y63, resulting in a higher λmax value.
On the other hand, at pH 10, the side chain of C157 is deprotonated, meaning it loses a proton and becomes uncharged. In this state, the deprotonated C157 is more likely to form a bond with Y63. This bonding affects the electronic environment and results in a lower λmax value compared to the value at pH 4.5.
In summary, the proposed bonding of deprotonated C157 with Y63 at pH 10 accounts for the observation that the λmax of S157C at pH 4.5 is greater than at pH 10 because the bonding alters the electronic environment, leading to a lower λmax value at the higher p H.
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part 1 of 2 An ideal gas is compressed to half its original volume while its temperature is held con-stant. If 758 J of energy is removed from the gas during the compression, how much work is done on the gas? Answer in units of J.
Answers
The work done on the ideal gas is 758 joule.
What is internal energy?The amount of energy required to move a thermodynamic system from its initial internal state to the current internal state of interest, accounting for energy gains and losses resulting from changes in the internal state, including factors like magnetization, is the amount of internal energy in the system.
As the decrease in internal energy is 758 Joule but the temperature of the ideal gas does not change, total work done is = total loss in internal energy = 758 joule.
Hence, the work done on the ideal gas is 758 joule.
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3. Light travels from the Sun to Earth in 8.3 min. Given that the speed of light is 3.00108 m/s, what is the distance in meters between the Sun and Earth?
Answers
Answer:
13
Explanation:
13.0120481928 it is the distance
What is the radial distance between the 500 v equipotential surface and the 1000 v surface?.
Answers
Answer: the radial distance between the 500-v equipotential surface and the 1000 v surface will be 8.91*106 times the charge Q.
Explanation: To find the answer, we have to know more about the equipotential surfaces.
What are equipotential surfaces?An equipotential surface is the locus of all points which have the same potential due to the charge distribution.Any surface in an electric field, at every point of which, the direction of electric field is normal to the surface can be regarded as equipotential.We have the equation for electric potential as,\(V=\frac{Qk}{r}\) , where k is equal to 1/(4π∈₀) =\(8.99*10^9\) .
equation for radial distance will be,\(r_d=r_1-r_2\)
How to solve the problem?For the first surface, we can write the equation of potential as,\(500V=\frac{Qk}{r_1} \\thus, r_1=\frac{Qk}{500V}= (1.79*10^7 *Q) m\)
For the second surface, we can write the equation of potential as,\(1000V=\frac{Qk}{r_2}\\ r_2= \frac{Qk}{1000V} =(8.99*10^6*Q) m\)
Thus, the radial distance will be,\(r_d=1.79*10^7Q-8.99*10^6Q=(8.91*10^6*Q)m\)
Thus, we can conclude that, the radial distance between the equipotential surface of 500V and 1000V will be,8.91*106 times the charge Q.
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which part of body is exercised by drill
Answers
Answer:
Muscles in the lower body
...
have a good rest of your day
Answer:
legs
Explanation:
legs because drill is related with leg so it helps to exercise by drill
A ball at rest rolls across a frictionless floor at 12.0 m/s/s. How far will it travel in
8.00 s?
Answers
Answer:
The distance, d travelled by the ball is 768 metres.
Explanation:
In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.
This simply means that, acceleration is given by the subtraction of final speed from the initial speed all over time.
Hence, if we subtract the final speed from the initial speed and divide that by the time, we can calculate an object’s acceleration.
Mathematically, acceleration is given by the equation;
\(Acceleration (a) = \frac{initial \; speed - final \; speed}{time}\)
\(a = \frac{v - u}{t}\)
Where,
a is acceleration measured in \(ms^{-2}\)
v and u is initial and final speed respectively, measured in \(ms^{-1}\)
t is time measured in seconds.
Given the following data;
Acceleration = 12.0m/s²
Time, t = 8secs
Velocity =?
First, we would calculate its velocity;
\(a = \frac{v - u}{t}\)
Since the ball rolls at rest, initial velocity is zero (0).
\(V = a * t\)
\(V = 12 * 8\)
\(Velocity = 96ms^{-1}\)
We can now solve for the distance;
\( Velocity = \frac{distance}{time}\)
Therefore,
\(Distance = velocity * time\)
\(Distance = 96 * 8\)
Distance = 768m.
if a 2kg object is falling at 3 m/s at what rate is gravity working on a object
Answers
Answer:9.81 m/s2.
Explanation:
If a 2 kg object is falling at 3 m/s, at what rate is gravity working on the object ? The object is falling at 3 m/s. Gravity is working on the object at a rate of 9.81 m/s2.
15. A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s. How much mass must be added to the object to change the period to 2.00 s?
The answer is 0.389 kg but please show your work.
Answers
Let's look at relationship
\(\\ \rm\rightarrowtail T=2\pi\sqrt{\dfrac{m}{k}}\)
\(\\ \rm\rightarrowtail T\propto \sqrt{m}\)
Hence
\(\\ \rm\rightarrowtail \dfrac{T_1}{T_2}=\sqrt{\dfrac{m1}{m2}}\)
\(\\ \rm\rightarrowtail \dfrac{1.5}{2}=\sqrt{\dfrac{0.5}{m2}}\)
\(\\ \rm\rightarrowtail 0.75^2=\dfrac{0.5}{m2}\)
\(\\ \rm\rightarrowtail m_2=\dfrac{0.5}{0.75^2}\)
\(\\ \rm\rightarrowtail m_2=0.889\)
Hence
Mass needs to added =0.889-0.500=0.389kgAnswer
Mass needs to added =0.889-0.500=0.389kg
Explanation:
g a 1.75 kg falcon catches a 0.415 kg dove from behind in midair. what is their velocity after impact if the falcon's velocity is initially 27.5 m/s and the dove's velocity is 5.35 m/s in the same direction?
Answers
Their velocity after impact will be 24.89 m/s.
Conservation of momentum will take place when two object moving with different velocity and they impact.
It means total momentum before impact would be same as the total momentum after impact.
before impact (M1V1+M2V2) = after impact (M1V1+M2V2)
putting the appropriate values, we get
1.75x27.5 + 0.415x5.35 = 1.75xV1 + 0.415xV2
since after impact the velocity would be same for both, so V1=V2=V
so, V = 53.89/2.165
V = 24.89 m/s
Hence the velocity after impact is 24.89 m/s
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A rocket explodes as it begins to launch. As a result, the rocket is destroyed
and debris is scattered around the launch site. Which question would a
scientist seek to answer?
A. How much will it cost to clean up the launch site?
B. Why did the rocket explode?
C. How will the explosion affect public opinion of the space
program?
D. For what purpose was the rocket being launched?
Answers
Answer:
probably B
Explanation:
it's not their job to sort out the cost of cleaning it up or worry about the public opinion. they should only make sure it doesn't happen again so they should find out why it happened
What i the velocity of the ice kater after
throwing the nowball? Diregard the friction
between the kate and the ice
Answers
Kinetic Friction, or Sliding Friction, is the friction between skates and ice. When two items slide past each other, this form of friction develops.
Sliding Friction is created when the skate blade passes through the ice. The beautiful curves are governed by the second law. Acceleration is required for a change in velocity, which includes both speed and direction. Because of the constantly changing direction, traveling in a circle at a constant speed includes acceleration. Because everyone is initially at rest, the system's linear momentum is zero. Because there is no external force acting on the system when the guy throws the ball, linear momentum must be conserved.
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Action Potential Worksheet. Ohms Law I =gV Describe parameters A-D for each each phase of the action potential listed below (1-4): (A) Relative conductance (g) for Na+ and for K+ (B) Relative voltage driving force for both Na+ and K+ [VM-Ex] and [VM-ENJ (C) Relative amount of K+ and Na+ current (lk and Iya) (D) What types of channels are open? 1. Neuron at rest (A) (D) 2. Rising Phase (A) (B) (C) (D) 3. Falling Phase (A) (B) (C) 4. Undershoot (A) (B) (C) 5. What is responsible for the Absolute Refractory Period? 6. What is responsible for the Relative Refractory Period?
Answers
The Absolute Refractory Period is primarily caused by the inactivation of voltage-gated Na+ channels. During this period, the membrane is unable to generate a new action potential regardless of the strength of the stimulus. The Relative Refractory Period is primarily caused by the increased activation and opening of voltage-gated K+ channels, as well as the slow closing of voltage-gated Na+ channels. During this period, a stronger-than-usual stimulus is required to generate a new action potential.
To provide the requested information for each phase of the action potential, we will go through each phase individually:
Neuron at rest:
(A) Relative conductance (g) for Na+ and for K+: The relative conductance for Na+ is low, while the relative conductance for K+ is high.
(D) What types of channels are open?: The voltage-gated Na+ channels are closed, and the voltage-gated K+ channels are closed.
Rising Phase:
(A) Relative conductance (g) for Na+ and for K+: The relative conductance for Na+ is high, while the relative conductance for K+ is low.
(B) Relative voltage driving force for both Na+ and K+ [VM-Ex] and [VM-ENJ]: The voltage driving force for Na+ (VM-Ex) is positive, while the voltage driving force for K+ (VM-ENJ) is negative.
(C) Relative amount of K+ and Na+ current (lk and Iya): The Na+ current (Iya) is large, and the K+ current (lk) is small.
(D) What types of channels are open?: The voltage-gated Na+ channels are open, and the voltage-gated K+ channels are closed.
Falling Phase:
(A) Relative conductance (g) for Na+ and for K+: The relative conductance for Na+ is decreasing, while the relative conductance for K+ is increasing.
(B) Relative voltage driving force for both Na+ and K+ [VM-Ex] and [VM-ENJ]: The voltage driving force for Na+ (VM-Ex) is decreasing, while the voltage driving force for K+ (VM-ENJ) is increasing.
(C) Relative amount of K+ and Na+ current (lk and Iya): The Na+ current (Iya) is decreasing, and the K+ current (lk) is increasing.
(D) What types of channels are open?: The voltage-gated Na+ channels are closed or inactivated, and the voltage-gated K+ channels are open.
Undershoot:
(A) Relative conductance (g) for Na+ and for K+: The relative conductance for Na+ is low, while the relative conductance for K+ is high.
(B) Relative voltage driving force for both Na+ and K+ [VM-Ex] and [VM-ENJ]: The voltage driving force for Na+ (VM-Ex) is negative, while the voltage driving force for K+ (VM-ENJ) is negative.
(C) Relative amount of K+ and Na+ current (lk and Iya): The Na+ current (Iya) is very low or almost absent, and the K+ current (lk) is high.
(D) What types of channels are open?: The voltage-gated Na+ channels are closed or inactivated, and the voltage-gated K+ channels are open.
The Absolute Refractory Period is primarily caused by the inactivation of voltage-gated Na+ channels. During this period, the membrane is unable to generate a new action potential regardless of the strength of the stimulus.
The Relative Refractory Period is primarily caused by the increased activation and opening of voltage-gated K+ channels, as well as the slow closing of voltage-gated Na+ channels. During this period, a stronger-than-usual stimulus is required to generate a new action potential.
Note: It is important to consult relevant sources or materials for specific and accurate information when studying action potentials, as the details and terminology may vary depending on the source and level of study.
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