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boyakko [2]
4 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]4 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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Therefore, by mass
Ca = 24.7 g, H = 1.2 g, C = 14.8 g, and O = 59.3 g

Converting these to moles

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Ca = 24.7*1/40.078 =0.6163 mol Ca
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Next, divide all the mols by the smallest value obtained.

Ca: 0.6163/0.6163 = 1 mol Ca
H: 1.1881/0.6163 = 2 mol H
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Therefore, empirical formula of the compound is

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