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boyakko [2]
3 years ago
7

The speed of propagation of the action potential (an electrical signal) in a nerve cell depends (inversely) on the diameter of t

he axon (nerve fiber). If the nerve cell connecting the spinal cord to your feet is 1.2 m long, and the nerve impulse speed is 27 m/s, how long (in s) does it take for the nerve signal to travel this distance?
Physics
1 answer:
sattari [20]3 years ago
3 0

Answer:

0.044 seconds

Explanation:

Parameters given:

Length of nerve cell = 1.2m

Nerve impulse speed = 27m/s

Speed is given as:

Speed = distance/time

Therefore, we can make time the subject of the formula. We have:

Time = distance/speed

Hence, time taken for the nerve cell to travel this distance is:

Time = 1.2 / 27

Time = 0.044 seconds

It takes 0.044 seconds for the nerve signal to travel the distance of the nerve connecting the spinal cord to the feet.

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Answer: n=4

Explanation:

We have the following expression for the volume flow rate Q of a hypodermic needle:

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Where the dimensions of each one is:

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Viscosity of the liquid \eta=\frac{M}{LT}

We need to find the value of n whicha has no dimensions, and in order to do this, we have to rewritte (1) with its dimensions:

\frac{L^{3}}{T}=\frac{\pi L^{n}(\frac{M}{LT^{2}})}{8(\frac{M}{LT}) L}  (2)

We need the right side of the equation to be equal to the left side of the equation (in dimensions):

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n}}{LT}  (3)

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n-1}}{T}  (4)

As we can see n must be 4 if we want the exponent to be 3:

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{4-1}}{T}  (5)

Finally:

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{3}}{T}  (6)

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Explanation:

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