Potential Difference Across Axon Membrane The axoplasm of an axon has a resistance Rax. The axon's membrane has both a resistanc
e (Rmem) and a capacitance (Cmem). A single segment of an axon can be modeled by a circuit with Rmem and Cmem in parallel with each other and in series with an open switch, a battery, and Rax. Imagine the voltage of the battery is ΔV. Part A We can model the firing of an action potential by the closing of the switch, which completes the circuit. Immediately after the switch closes, what is the potential difference across the membrane of this single segment?
1 answer:
Answer:
<em>check the diagram in the attachment below.</em>
After the switch closes, the voltage across the membrane will be zero. It is so due to the fact that the capacitor will be short circuited.
Explanation:
question
Potential Difference Across Axon Membrane The axoplasm of an axon has a resistance Rax. The axon's membrane has both a resistance (Rmem) and a capacitance (Cmem). A single segment of an axon can be modeled by a circuit with Rmem and Cmem in parallel with each other and in series with an open switch, a battery, and Rax. Imagine the voltage of the battery is ΔV. Part A We can model the firing of an action potential by the closing of the switch, which completes the circuit. Immediately after the switch closes, what is the potential difference across the membrane of this single segment?
ANSWER;
After the switch closes, the voltage across the membrane will be zero. It is so due to the fact that the capacitor will be short circuited.
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Answer:
Their speed in a vacuum is a constant value.
Explanation:
Electromagnetic waves consits of oscillations of electric field and magnetic field. The oscillations of these fields occur in a direction perpendicular to the direction of propagation of the waves, so they are transverse wave. Electromagnetic waves, contrary to mechanical waves, do not need a medium to propagate, so they can also travel through a vacuum. In a vacuum, their speed is constant and has always the same value, the speed of light:
Answer:
a) t = 0.74s
b) D = 4.76m
c) Vf = 5.35m/s
Explanation:
The ball starts rolling when Vf = ωf*R.
We know that:
Vf = Vo - a*t
ωf = ωo + α*t
With a sum of forces on the ball:
With a sum of torque on the ball:
Replacing both accelerations:
t=0.74s
The distance will be:
Final velocity:
Vf=5.35m/s
Answer:
The ball is in the air for 3.5 seconds
The initial horizontal component of velocity is 28.6 m/s
The vertical component of the final velocity is 34.3 m/s downward
The final velocity is 44.7 m/s in the direction 50.2° below the horizontal
Explanation:
A ball is thrown horizontally
That means the vertical component of the initial velocity
The initial velocity is the horizontal component
The ball is thrown from the top of a 60 m
That means the vertical displacement component y = 60 m
→ y = t +
gt²
where g is the acceleration of gravity and t is the time
y = -60 m , g = -9.8 m/s² ,
Substitute these values in the rule
→ -60 = 0 + (-9.8)t²
→ -60 = -4.9t²
Divide both sides by -4.9
→ 12.2449 = t²
Take √ for both sides
∴ t = 3.5 seconds
* <em>The ball is in the air for 3.5 seconds </em>
The initial velocity is the horizontal component
The ball lands 100 meter from the base of the building
That means the horizontal displacement x = 100 m
→ x = t
→ t = 3.5 s , x = 100 m
Substitute these values in the rule
→ 100 = (3.5)
Divide both sides by 3.5
→ = 28.57 m/s
<em>The initial horizontal component of velocity is 28.6 m/s</em>
The vertical component of the final velocity is
→ =
+ gt
→ = 0 , g = -9.8 m/s² , t = 3.5 s
Substitute these values in the rule
→ = 0 + (-9.8)(3.5)
→ = -34.3 m/s
<em>The vertical component of the final velocity is 34.3 m/s downward</em>
The final velocity v is the resultant vector of and
→ Its magnetude is
→ Its direction
→ = 28.6 ,
= -34.3
Substitute this values in the rules above
→
→ Its direction
The negative sign means the direction is below the horizontal
<em>The final velocity is 44.7 m/s in the direction 50.2° below the horizontal</em>