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
You might be interested in
Answer:
the magnitude of the velocity of one particle relative to the other is 0.9988c
Explanation:
Given the data in the question;
Velocities of the two particles = 0.9520c
Using Lorentz transformation
Let relative velocity be W, so
v = ( u + v ) / ( 1 + ( uv / c²) )
since each particle travels with the same speed,
u = v
so
v = ( u + u ) / ( 1 + ( u×u / c²) )
v = 2(0.9520c) / ( 1 + ( 0.9520c )² / c²) )
we substitute
v = 1.904c / ( 1 + ( (0.906304 × c² ) / c²) )
v = 1.904c / ( 1 + 0.906304 )
v = 1.904c / 1.906304
v = 0.9988c
Therefore, the magnitude of the velocity of one particle relative to the other is 0.9988c
Answer: 0.333 h
Explanation:
This problem can be solved using the <u>Radioactive Half Life Formula</u>:
(1)
Where:
is the final amount of the material
is the initial amount of the material
is the time elapsed
is the half life of the material (the quantity we are asked to find)
Knowing this, let's substitute the values and find from (1):
(2)
(3)
Applying natural logarithm in both sides:
(4)
(5)
Clearing :
(6)
Finally:
This is the half-life of the Bismuth-218 isotope
Answer:
Te ayudo con una de prueba $
Explanation:
