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Gwar [14]
3 years ago
5

Nerve cells transmit electric signals through their long tubular axons. These signals propagate due to a sudden rush of Na+ ions

, each with charge +e, into the axon. Measurements have revealed that typically about 5.7×10^11 Na+ ions enter each meter of the axon during a time of 7.0 ms. What is the current during this inflow of charge in a meter of axon?
Physics
1 answer:
iren [92.7K]3 years ago
3 0

Answer:

I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A

Explanation:

For this case we have the following info given:

Number of Na+ ions 5.7 x10^{11} ions

Each ion have a charge of +e and the crage of the electron is 1.6 x10^{-19}C

The time is given t = 7 ms if we convert this into seconds we got:

t = 7ms * \frac{1s}{1000 ms}= 0.007s

Now we can use the following formula given from the current passing thourhg a meter of nerve axon given by:

Q = Ne

Where N represent the number of ions, e the charge of the electron and Q the total charge

If we replace on this case we have this:

Q= 5.7x10^{11} * (1.6 x10^{-19}C) = 9.12x10^{-8} C

And from the general definition of current we know that:

I =\frac{Q}{t}

And since we know the total charge Q and the time we can replace:

I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A

The current during the inflow charge in the meter axon for this case is 13.02 \mu A

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So, Voltage = Power/Current

Put the values, 

V = 240/2

V = 120 V

In short, Your Final Answer would be: 120 Volts

Hope this helps!
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3 years ago
A(n) 2602 kg van runs into the back of a(n)
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8.5 m/s

Explanation:

please see paper for the work!

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2 years ago
During a trial run, race car A starts from rest and accelerates uniformly along a straight level track for a particular interval
kondor19780726 [428]
Answer: car B has travelled 4times as far as Car A

d=vi*t+1/2at^2

No initial velocity so equation becomes;

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Car A d=1/2a(1)^2

Car B d=1/2a(2)^2

Car A d= 1^2=1

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5 0
3 years ago
51. Suppose you measure the terminal voltage of a 3.200-V lithium cell having an internal resistance of 5.00Ω by placing a 1.00-
Tanya [424]

Explanation:

Given that,

Terminal voltage = 3.200 V

Internal resistance r= 5.00\ \Omega

(a). We need to calculate the current

Using rule of loop

E-IR-Ir=0

I=\dfrac{E}{R+r}

Where, E = emf

R = resistance

r = internal resistance

Put the value into the formula

I=\dfrac{3.200}{1.00\times10^{3}+5.00}

I=3.184\times10^{-3}\ A

(b). We need to calculate the terminal voltage

Using formula of terminal voltage

V=E-Ir

Where, V = terminal voltage

I = current

r = internal resistance

Put the value into the formula

V=3.200-3.184\times10^{-3}\times5.00

V=3.18\ V

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\dfrac{Terminal\ voltage}{emf}=\dfrac{3.18}{3.200 }

\dfrac{Terminal\ voltage}{emf}= \dfrac{159}{160}

Hence, This is the required solution.

5 0
3 years ago
a cart's initial velocity is +3.0 meters per second. What is its final velocity after accelerating at a rate of 1.5m/s2 for 8.0
Andreas93 [3]

The final velocity is +15.0 m/s

Explanation:

The motion of the cart is a uniformly accelerated motion (=at constant acceleration), therefore we can use the following suvat equation:

v=u+at

where

v is the velocity at time t

u is the initial velocity

a is the acceleration

t is the time

For the cart in this problem, we have:

u = +3.0 m/s (initial velocity)

a=1.5 m/s^2 (acceleration)

t = 8.0 s (time)

Substituting, we find the final velocity:

v=3.0+(1.5)(8.0)=+15.0 m/s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

4 0
3 years ago
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