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Contact [7]
2 years ago
5

Nerve impulses are carried along axons, the elongated fibers that transmit neural signals. We can model an axon as a tube with a

n inner diameter of 10 μm. The tube wall is insulated, but the fluid inside it has a resistivity of 0.50 Ω⋅m.What is the resistance of a 2.0-mm-long axon?
Physics
1 answer:
jeka942 years ago
8 0

Answer:

The resistance of the axon is 1.27\times 10^7\ \Omega.

Explanation:

Given that,

Inner diameter of the model of an axon, d=10\ \mu m

Radius of the model, r=5\ \mu m=5\times 10^{-6}\ m

Resistivity of the fluid inside the tube wall, \rho=0.5\ \Omega -m

Length of the axon, l = 2 mm = 0.002 m

We know that the resistance in terms of resistivity of an object is given by :

R=\rho\dfrac{l}{A}\\\\R=0.5\times \dfrac{0.002}{\pi (5\times 10^{-6})^2}\\\\R=1.27\times 10^7\ \Omega

So, the resistance of the axon is 1.27\times 10^7\ \Omega. Hence, this is the required solution.

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When a vertical beam of light passes through a transparent medium, the rate at which its intensity I decreases is proportional t
antoniya [11.8K]

Answer:

Intensity of beam 18 feet below the surface is about 0.02%

Explanation:

Using Lambert's law

Let dI / dt = kI, where k is a proportionality constant, I is intensity of incident light and t is thickness of the medium

then dI / I = kdt

taking log,

ln(I) = kt + ln C

I = Ce^kt

t=0=>I=I(0)=>C=I(0)

I = I(0)e^kt

t=3 & I=0.25I(0)=>0.25=e^3k

k = ln(0.25)/3

k = -1.386/3

k = -0.4621

I = I(0)e^(-0.4621t)

I(18) = I(0)e^(-0.4621*18)

I(18) = 0.00024413I(0)

Intensity of beam 18 feet below the surface is about 0.2%

3 0
2 years ago
A book rests on the shelf of a bookcase. The reaction force to the force of gravity acting on the book is 1. The force of the sh
Hoochie [10]

Answer:

1. The force of the shelf holding the book up.

Explanation:

The free body diagram of the book is as follows:

1 - The weight of the book towards downwards

2 - The normal force that the shelf exerts on the book towards upwards.

Since the book is at rest, these two forces are equal to each other and according to Newton's Third Law the reaction force to the force of gravity is equal but opposite to the weight of the book. This reaction force is the one that holds the book up on the shelf.

6 0
2 years ago
Read 2 more answers
A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by
aleksklad [387]

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The position of the car as a function of time t is given by

x(t)=\alpha t^2 - \beta t^3

where

\alpha = 1.50 m/s^2

\beta = 0.05 m/s^3

The average velocity is given by the ratio between the displacement and the time taken:

v=\frac{\Delta x}{\Delta t}

The position at t = 0 is:

x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0

The position at t = 2.00 s is:

x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m

So the displacement is

\Delta x = x(2)-x(0)=5.6-0=5.6 m

The time interval is

\Delta t = 2.0 s - 0 s = 2.0 s

And so, the average velocity in this interval is

v=\frac{5.6 m}{2.0 s}=2.8 m/s

b)

The position at t = 0 is:

x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0

While the position at t = 4.00 s is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

So the displacement is

\Delta x = x(4)-x(0)=20.8-0=20.8 m

The time interval is

\Delta t = 4.0 - 0 = 4.0 s

So the average velocity here is

v=\frac{20.8}{4.0}=5.2 m/s

c)

The position at t = 2 s is:

x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m

While the position at t = 4 s is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

So the displacement is

\Delta x = 20.8 - 5.6 = 15.2 m

While the time interval is

\Delta t = 4.0 - 2.0 = 2.0 s

So the average velocity is

v=\frac{15.2}{2.0}=7.6 m/s

Learn more about average velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

6 0
2 years ago
Tiana jogs 1.5 km along a straight path and then turns and jogs 2.4 km in the opposite direction. She then turns back and jogs 0
vichka [17]

Answer:

Distance: 4.6km Displacement= -0.2km

Explanation:

Total distance: 1.5+2.4+0.7= 4.6 km

Displacement: 1.5-2.4+0.7= -0.2km

The displacement may also be 0.2km, it just depends on if it wants it negative or not.

7 0
2 years ago
Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i
SCORPION-xisa [38]

Answer:

The value is  r =  5.077 \  m

Explanation:

From the question we are told that

   The  Coulomb constant is  k =  9.0 *10^{9} \  N\cdot  m^2  /C^2

   The  charge on the electron/proton  is  e =  1.6*10^{-19} \  C

    The  mass of proton m_{proton} =  1.67*10^{-27} \  kg

    The  mass of  electron is  m_{electron } =  9.11 *10^{-31} \ kg

Generally for the electron to be held up by the force gravity

   Then    

       Electric force on the electron  =  The  gravitational Force

i.e  

            m_{electron} *  g  = \frac{ k *  e^2  }{r^2 }

         \frac{9*10^9 *  (1.60 *10^{-19})^2  }{r^2 }  =     9.11 *10^{-31 }  *  9.81

         r =  \sqrt{25.78}

         r =  5.077  \  m

7 0
2 years ago
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