A neuron is stimulated with an electric pulse at t=0 s. The resulting action pot

Question

A neuron is stimulated with an electric pulse at t=0 s. The resulting action potential is detected at t= 0.0052 s later at a point 2.5 cm down the axon. The action potential is detected again at a point 5.7 cm down the axon at t= 0.0065 s.
a) Determine the speed of the electric pulse as it travels down the axon.
b)Determine the delay time between the application of the electric pulse and the onset of the activation potential.
c)Describe in two to four bullet points the physical mechanisms causing an axon to go from resting potential to activation potential when a stimulus is applied.

Explanation / Answer

The initial distance, x1 = 2.5 cm = 0.025 m The final distance, x2 = 5.7 cm = 0.057 m x = 0.057 - 0.025 = 0.032 m Initial time, t1 = 0.0052s Final time, t2 = 0.0065 s t = 0.065-0.052 = 0.013 s a) The speed of the electric pulse, v = x/t = 2.46 m/s b) We have a formula, x = vt + 0.5gt^2 The above can be written as, 4.9t^2 + 2.46t - 0.032 = 0 By solving the above equation we get the value of delay time So, t = 0.013 sec c) I dont have an Idea

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