If the action potential starts with a potential difference of +40.0 mV, and the length constant of the axon is 2.00 mm, how far from the site of the stimulus will the potential difference across the membrane drop to 1.00 mV (assuming the action potential is not regenerated)?

= 7.38mm

Explanation:

The decrease in potential difference across membrane in the distance is given by the equation

where

= max voltage attained in action potential

l = length constant

x = the distance

taking natural log on both side

c. newton's third law of motion

d) physical means can be used to separate the components of air.