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3 years ago

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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?

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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At the lowest point, we can take the potential energy as zero, so the mechanical energy is just kinetic energy:

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At the highest point in the trajectory, the rod is stationary, so the kinetic energy will be zero, and the mechanical energy will simply be equal to the gravitational potential energy:

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Answer:

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$a = - 0.8798 \ g$

$Tb = 6.64 \ s$

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Explanation:

<h3>a and b</h3>

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<h3>c</h3>

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<h3>d</h3>

$\frac{Tb}{Tr} = \frac{ 6.64 \ s }{ 0.4 s }$

$\frac{Tb}{Tr} = 16.6$

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