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Rasek [7]
3 years ago
10

A nerve signal is transmitted through a neuron when an excess of Na+ ions suddenly enters the axon, a long cylindrical part of t

he neuron. Axons are approximately 10.0 μm in diameter, and measurements show that about 5.60×1011 Na+ions per meter (each of charge +e) enter during this process. Although the axon is a long cylinder, the charge does not all enter everywhere at the same time. A plausible model would be a series of nearly point charges moving along the axon. Let us look at a 0.100 mm length of the axon and model it as a point charge.1) What electric field (magnitude and direction) does the sudden influx of charge produce at the surface of the body if the axon is 5.00 cm below the skin? Note that distance to the skin is much greater than the length of the charged axon so that the axon charge is effectiely point-like.,E = (INSERT ANSWER) N/C2) If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a 0.100 mm length of the axon? Answer in C3) the field is directed away from the axon or the field is directed into the axon
Physics
1 answer:
ivolga24 [154]3 years ago
3 0

Answer:

1.32.225 N/C, direction is away from the point charge

2. 8.972*10^-12 C

3. the field is directed away from the axon

Explanation:

The electric field can be calculated as shown below:

E = k*|q|/r^2

Where:

E = electric field; k = 8.98755*10^9 N*m^2/C^2; r = distance between the measured field and point charge = 0.05 m; q = the point charge

For 0.100 m of the axon, the value of q is:

q = (5.6*10^11)*(+e)*(0.001)

+e = charge of an electron = 1.60217*10^-19 C

Thus:

q = (5.6*10^11)*(1.60217*10^-19)*(0.0001) = 8.972*10^-12 C

Therefore:

E = (8.98755*10^9)*(8.972*10^-12)/0.05^2 = 32.255 N/C

A positive point charge always produce an electric field that is directed away from the field while a negative point charge produces an electric field that is directed toward the field

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You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state
posledela

Answer:

a)  y₂ = 49.1 m ,    t = 1.02 s , b)   y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

            v_{y}² = v_{oy}² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

           y-yo = v_{oy}² /2 g
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The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
             y₂ = 5.1 + 44
             y₂ = 49.1 m
Let's use the other equation to find the time
              [tex]v_{y} = v_{oy} - g t

              t = v_{oy} / g

              t = 10 / 9.8

              t = 1.02 s

b) the maximum height

            y- 44.0 = v_{y}² / 2 g

            y - 44.0 = 5.1

            y = 5.1 +44.0

            y = 49.1 m

The time is the same because it does not depend on the initial height

              t = 1.02 s

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