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Rasek [7]
2 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]2 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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AleksandrR [38]

B is the answer because it takes millions of years to form these fossil fuels and everyday we use way more than we can find we may have a surplus for now but we may run out sooner than some think

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2 years ago
Lexy used the formula shown to calculate the force of gravity on a space shuttle. Fg = G What does 3 × 105 kg represent? the dif
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<h2>Answer:</h2>

<u>This term shows the </u><u>mass of the space shuttle</u>

<h2>Explanation:</h2>

We know that the mass of the Earth is 5.972 × 10^24 kg. Similarly the sum of  mass of earth and the mass of shuttle must be a greater number as compared to the number given. It simply means that the mass of earth is itself 5.972 × 10^24 kg and the value given is 3 × 105 kg so it is obvious that if was the sum then it must be greater than the mass of earth. Therefore we can say that this not the mass of earth, neither the sum of mass of earth and shuttle, but this is only the mass of space shuttle which is the last multiple choice.

5 0
1 year ago
A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.
Volgvan

Answer:

A=0.199

Explanation:

We are given that  

Mass of spring=m=450 g==\frac{450}{1000}=0.45 kg

Where 1 kg=1000 g

Frequency of oscillation=\nu=1.2Hz

Total energy of the oscillation=0.51 J

We have to find the amplitude of oscillations.

Energy of oscillator=E=\frac{1}{2}m\omega^2A^2

Where \omega=2\pi\nu=Angular frequency

A=Amplitude

\pi=\frac{22}{7}

Using the formula

0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2

A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398

A=\sqrt{0.0398}=0.199

Hence, the amplitude of oscillation=A=0.199

4 0
1 year ago
Now that we have a feel for the state of the circuit in its steady state, let us obtain the expression for the current in the ci
vesna_86 [32]

Answer:

i(t) = (E/R)[1 - exp(-Rt/L)]

Explanation:

E−vR−vL=0

E− iR− Ldi/dt = 0

E− iR = Ldi/dt

Separating te variables,

dt/L = di/(E - iR)

Let x = E - iR, so dx = -Rdi and di = -dx/R substituting for x and di we have

dt/L = -dx/Rx

-Rdt/L = dx/x

interating both sides, we have

∫-Rdt/L = ∫dx/x

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x = exp(-Rt/L + C)

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x = Aexp(-Rt/L) Substituting x = E - iR we have

E - iR = Aexp(-Rt/L) when t = 0, i(0) = 0. So

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i(t)R = E(1 - exp(-Rt/L))

i(t) = (E/R)(1 - exp(-Rt/L))

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Answer:

d. less than 20m/s

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3 0
1 year ago
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