What is the strength of the electric field inside the membrane just before the action potential?
1 answer:
Answer:
Incomplete question, check attachment for the graph needed to solve problem.
A 8.1nm........
Explanation:
Electric Field is given as
E=V/d
Where V is voltage
And d is the distance apart
E is the electric field
The voltage V just before action of potential is -70mV,
The value d=8.1nm
d=8.1×10^-9m
E=V/d
E=-70×10^-3/8.1×10^-9
E=-8.6×10^6 N/C
Then the magnitude of the electric field is 8.6×10^6N/C
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Depth of Death valley is 85 m below sea level,
The summit of nearby Mt. Whitney has an elevation of 4420 m,
Mass of the hiker, m = 65 kg
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So, the change in potential energy of the hiker is . Hence, this is the required solution.
Given
Three 7 ohm resistor are in series.
The battery is V=10V
To find
The equivalent resistance
Explanation
When the resistance are in series then the resistance are added to find its equivalent.
Thus the equivalent resistance is:
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The equivalent resistance is 21 ohm
Explanation:
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