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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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A 0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa
gregori [183]

The magnitude of the change in momentum of the stone is about 18.4 kg.m/s

\texttt{ }

<h3>Further explanation</h3>

Let's recall Impulse formula as follows:

\boxed {I = \Sigma F \times t}

<em>where:</em>

<em>I = impulse on the object ( kg m/s )</em>

<em>∑F = net force acting on object ( kg m /s² = Newton )</em>

<em>t = elapsed time ( s )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of ball = m = 0.500 kg

initial speed of ball = vo = 20.0 m/s

final kinetic energy = Ek = 70% Eko

<u>Asked:</u>

magnitude of the change of momentum of the stone = Δp = ?

<u>Solution:</u>

<em>Firstly, we will calculate the final speed of the ball as follows:</em>

Ek = 70\% \ Ek_o

\frac{1}{2} m v^2 = 70\% \ ( \frac{1}{2} m (v_o)^2 )

v^2 = 70 \% \ (v_o)^2

v = - v_o \sqrt{70 \%} → <em>negative sign due to ball rebounds</em>

v = - v_o \sqrt{0.7} \texttt{ m/s}

\texttt{ }

<em>Next, we could find the magnitude of the change of momentum of the stone as follows:</em>

\Delta p_{stone} = - \Delta p_{ball}

\Delta p_{stone} = - [ mv - mv_o ]

\Delta p_{stone} = m[ v_o - v ]

\Delta p_{stone} = m[ v_o + v_o\sqrt{0.7} ]

\Delta p_{stone} = mv_o [ 1 + \sqrt{0.7} ]

\Delta p_{stone} = 0.500 ( 20.0 ) [ 1 + \sqrt{0.7} ]

\Delta p_{stone} \approx 18.4 \texttt{ kg.m/s}

\texttt{ }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Average Speed of Plane : brainly.com/question/12826372
  • Impulse : brainly.com/question/12855855
  • Gravity : brainly.com/question/1724648

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

8 0
2 years ago
A ball is thrown with a velocity of 35 meters per second at an angle of 30° above the horizontal. which quantity has a magnitude
enot [183]
The quantity that has a magnitude of zero when the ball is at the highest point in its trajectory is the vertical velocity.

In fact, the motion of the ball consists of two separate motions:
- the horizontal motion, on the x-axis, which is a uniform motion with constant velocity v_x=v_0 cos 30^{\circ}, where v_0=35 m/s
- the vertical motion, on the y-axis, which is a uniformly accelerated motion with constant acceleration g=9.81 m/s^2 directed downwards, and with initial velocity v_y=v_= sin 30^{\circ}. Due to the presence of the acceleration g on the vertical direction (pointing in the opposite direction of the initial vertical velocity), the vertical velocity of the ball decreases as it goes higher, up to a point where it becomes zero and it reverses its direction: when the vertical velocity becomes zero, the ball has reached its maximum height. 
5 0
2 years ago
The force shown in the attached figure is the net eastward force acting on a ball. The force starts rising at t = 0.012 s, falls
kifflom [539]
Impulse = Integral of F(t) dt from 0.012s to 0.062 s

Given that you do not know the function F(t) you have to make an approximation.

The integral is the area under the curve.

The problem suggest you to approximate the area to a triangle.

In this triangle the base is the time: 0.062 s - 0.012 s = 0.050 s

The height is the peak force: 35 N.

Then, the area is [1/2] (0.05s) (35N) = 0.875 N*s

Answer> 0.875 N*s
6 0
2 years ago
Starting at point 0, you travel 500 m on a straight road that slopes upward at a constant angle of 5 degrees. What is your heigh
ira [324]

Answer:

43.58 m

Explanation:

If you travel 500 m on a straight road that slopes upward at a constant angle of 5 degrees

Using trigonometry ratio

Sin 5 = opposite/hypothenus

Where the hypothenus = 500m

Opposite = height h

Sin 5 = h/500

Cross multiply

500 × sin 5 = h

h = 500 × 0.08715

h = 43.58m

Therefore, the height above the starting point is equal to 43.58m

5 0
2 years ago
The mass m1 enters from the left with velocity v0 and strikes a mass m2 &gt; m1 which is initially at rest. The collision betwee
enot [183]

Answer:

1. False 2) greater than. 3) less than 4) less than

Explanation:

1)

  • As the collision is perfectly elastic, kinetic energy must be conserved.
  • The expression for the final velocity of the mass m₁, for a perfectly elastic collision, is as follows:

        v_{1f} = v_{10} *\frac{m_{1} -m_{2} }{m_{1} +m_{2}}

  • As it can be seen, as m₁ ≠ m₂, v₁f ≠ 0.

2)

  • As total momentum must be conserved, we can see that as m₂ > m₁, from the equation above the final momentum of m₁ has an opposite sign to the initial one, so the momentum of m₂ must be greater than the initial momentum of m₁, to keep both sides of the equation balanced.

3)    

  • The maximum energy stored in the in the spring is given by the following expression:

       U =\frac{1}{2} *k * A^{2}

  • where A = maximum compression of the spring.
  • This energy is always the sum of the elastic potential energy and the kinetic energy of the mass (in absence of friction).
  • When the spring is in a relaxed state, the speed of the mass is maximum, so, its kinetic energy is maximum too.
  • Just prior to compress the spring, this kinetic energy is the kinetic energy of m₂, immediately after the collision.
  • As total kinetic energy must be conserved, the following condition must be met:

       KE_{10} = KE_{1f}  + KE_{2f}

  • So, it is clear that KE₂f  < KE₁₀
  • Therefore, the maximum energy stored in the spring is less than the initial energy in m₁.

4)

  • As explained above, if total kinetic energy must be conserved:

        KE_{10} = KE_{1f}  + KE_{2f}

  • So as kinetic energy is always positive, KEf₂ < KE₁₀.
4 0
2 years ago
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