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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Answer:
i guess it is D because,
y = A sin(wt)
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