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mars1129 [50]
3 years ago
7

The speed of propagation of electrical signals in a nerve cell depends on the diameter of the axon (nerve fiber). If the nerve c

ell connecting the spinal cord to your feet is 1.1 m long, and the nerve impulse speed is 18 m/s, how long (in milliseconds) does it take for the nerve signal to travel this distance
Physics
1 answer:
gayaneshka [121]3 years ago
6 0

Answer:

61 ms

Explanation:

Given that

Length of the nerve call connecting the spinal cord to your feet, s is 1.1 m

Speed of the impulse of the nerve, v = 18 m/s

Time taken by the nerve to travel the distance, t = ?

In solving this, we'd apply the very most basic formula in physics. Formula of speed, with respect to time and distance.

Speed, v = distance, s / time, t

time, t = distance, s / speed, v

time, t = 1.1 / 18

time, t = 0.061 s, converting to milliseconds, we have, time, t = 61 ms

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Makovka662 [10]

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Explanation:

Because the IV (Independent Variable) is always on the x-axis.

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An Olympic discus thrower (~100 kg) launches the 2.0 kg discus by spinning rapidly (~4 times per second) with arm outstretched (
vladimir1956 [14]

Answer:

F = 1263.03 N

Explanation:s

given,                      

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angular speed of the disk  = 4 rev/s

arm outstretched = 1 m                  

centripetal force of the disk in the circular path

F = m ω² r                        

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F = m ω² r                      

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F = 1263.03 N                                              

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6 0
3 years ago
What is the difference between a graph of linear motion and a graph of harmonic motion?
fredd [130]
Hope this helps :)

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The Amazon River is in Europe. True or False
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Welcome to this IE. You may navigate to any page you've seen already using the IE Outline tab on the right. A particle beam is m
Genrish500 [490]

Answer:

the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m

Explanation:

Given the data in the question;

Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J

Mass of proton = 1.673 × 10⁻²⁷ kg

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we know that

Kinetic Energy = Charge of proton × Potential difference ΔV

so

Potential difference ΔV = Kinetic Energy / Charge of proton

we substitute

Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )

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Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;

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Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m

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3 years ago
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