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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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Hence;

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Answer:

F_a=1470\ N

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<u>Friction Force</u>

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.  

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

\displaystyle F_a-F_{r1}-F_{r2}=m.a=0

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

\displaystyle F_a=F_{r1}+F_{r2}.....[1]

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\displaystyle F_{r2}-T=0

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\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g

\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g

Simplifying

\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)

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\displaystyle F_{a}=0.25(9.8)[400+2(100)]

\boxed{F_a=1470\ N}

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