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Vinil7 [7]
3 years ago
15

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?

Physics
1 answer:
Korolek [52]3 years ago
6 0

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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aivan3 [116]

Answer:

d) 1000 times

Explanation:

As we know that difference of sound level is given as

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so here we need to find the ratio of two intensity

it is given as

Log\frac{I_2}{I_1} = \frac{(L_2- L_1)}{10}

Log\frac{I_2}{I_1} = \frac{60 - 30}{10}

Log\frac{I_2}{I_1} = 3

now we have

\frac{I_2}{I_1} = 10^3

so it is

d) 1000 times

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3 years ago
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natka813 [3]

Answer:

Speed of the satellite V = 6.991 × 10³ m/s

Explanation:

Given:

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Radius r = √[GMm / F]

Radius r = √[(6.67 × 10⁻¹¹ )(5.97 × 10²⁴)(500) / (3,000)

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Speed of the satellite V = 6.991 × 10³ m/s

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