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3 years ago

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A biophysics experiment uses a very sensitive magnetic field probe to determine the current associated with a nerve impulse trav

eling along an axon.
If the peak field strength 1.3mm from an axon is 9.0pT , what is the peak current carried by the axon?

1 answer:

fenix001 [56]3 years ago

7 0

Answer:

The peak current carried by the axon is 5.85 x 10⁻⁸ A

Explanation:

Given;

distance of the field from the axon, r = 1.3 mm

peak magnetic field strength, B = 9 x 10⁻¹² T

To determine the peak current carried by the axon, apply the following equation;

$B = \frac{\mu I}{2\pi r}$

where;

B is the peak magnetic field

r is the distance of the magnetic field from axon

μ is permeability of free space = 4π x 10⁻⁷

I is the peak current

Re-arrange the equation and solve for "I"

$B = \frac{\mu I}{2\pi r} \\\\I = \frac{B*2\pi r}{\mu} \\\\I = \frac{9*10^{-12}*2*\pi *1.3*10^{-3}}{4\pi *10^{-7}} \\\\I = 5.85 *10^{-8} \ A$

Therefore, the peak current carried by the axon is 5.85 x 10⁻⁸ A

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Answer:

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Explanation:

Given the data in the question;

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(69.5 kg + 0.0475 kg) × 2.35 m/s = 163.4366 kg.m/s

Now, When he throws it at 31.5 m/s, these constitutes a momentum of;

(0.0475 kg )(31.5 m/s) = 1.49625 kg.m/s

hence his momentum now is: 163.4366 - 1.49625 = 161.94035 kg.m/s

To get his velocity, we say;

161.94035 = mv

{ he lost weight of the snow ball so, m = 69.5 kg )

161.94035 = 69.5 × v

v = 161.94035 / 69.5

v = 2.33 m/s

Therefore, After throwing the snow, velocity of the thrower is 2.33 m/s

Next is the Receiver;

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he has no momentum initially and after he catches the snow ball;

1.49625 kg.m/s = mv

1.49625 kg.m/s = ( 57.5 kg + 0.0475 kg ) × v

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To increase the centripetal acceleration to $2.00 m/s^2$ , you can double the speed or decrease the radius by 1/4

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The acceleration of an object in uniform circular motion is called centripetal acceleration, and it is given by

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r is the radius of the circular path

In the problem, the original centripetal acceleration is

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We want to increase it by a factor of 4, i.e. to

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We notice that the centripetal acceleration is proportional to the square of the speed and inversely proportional to the radius, so we can do as follows:

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This way, the new acceleration is

$a'=\frac{(2v)^2}{r}=4(\frac{v^2}{r})=4a$

so, 4 times the original acceleration

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So the new acceleration is

$a'=\frac{v^2}{(r/4)}=4(\frac{v^2}{r})=4a$

so, the acceleration has increased by a factor 4 again.

Learn more about centripetal acceleration:

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Which two of the following involve the same energy transfer. Assume that the same substance and the same mass is involved in all

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B. evaporation
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A 2-lb slider is propelled upward at A along the fixed curved bar which lies in a vertical plane. If the slider is observed to h

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To develop this problem it is necessary to apply the concepts given in the balance of forces for the tangential force and the centripetal force. An easy way to detail this problem is through a free body diagram that describes the behavior of the body and the forces to which it is subject.

PART A) Normal Force.

$F_n = \frac{mv^2}{r}$

$N+mgcos\theta = \frac{mv^2}{r}$

Here,

Normal reaction of the ring is N and velocity of the ring is v

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$-W sin\theta = \frac{W}{g} a_t$

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Negative symbol indicates deceleration.

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