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kakasveta [241]
2 years ago
14

The axoplasm of an axon has a resistance Rax. The axon's membrane has both a resistance (Rmem) and a capacitance (Cmem). A singl

e segment of an axon can be modeled by a circuit with Rmem and Cmem in parallel with each other and in series with an open switch, a battery, and Rax?.
Imagine the voltage of the battery is ?V?.

Imagine that after a very long time there is a current I going through the battery. The potential difference across the membrane can be written as I Rmem.

Write down another expression for the potential difference across the membrane using all or some of the given variables, except Rmem.

Physics
1 answer:
dybincka [34]2 years ago
8 0

Answer:

The expression is Vmem = ΔV - I*Rax

Explanation:

According to the picture, if switch S is closed, the Cmem will be short. If Vcap = 0, the current flows through the capacitor only (not through Rmem), thus Vmem = 0 after closing S.

When C is fully charged, then we have:

Vmem = ΔV - I*Rax

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A tank contains 100 gal of water and 50 oz of salt.water containing a salt concentration of 1 4 (1 1 2 sin t) oz/gal flows into
Alchen [17]

Answer:

Explanation:

Heres the possible full question and solution:

A tank contains 100 gal of water and 50 oz of salt. Water containing a salt concentration of ¼ (1 + ½ sin t) oz/gal flows ito the tank at a rate of 2 gal/min, and the mixture in the tank flows out at the same rate.

a. Find the amount of salt in the tank at any time.

b. Plot the solution for a time period long enough so that you see the ultimate behavior of the graph.

c. The long-time behavior of the solution is an oscillation about a certain constant level. What is this level? What is the amplitude of the oscillation?

solution

a)

Consider the tank contains 100gal of water and 50 oz of salt

Assume that the amount of salt in the tank at time t is Q(t).

Then, the rate of change of salt in the tank is given by \frac{dQ}{dt}.

Here, \frac{dQ}{dt}=rate of liquid flowing in the tank - rate of liquid flowing out.

Therefore,

Rate_{in} =2gal/min \times \frac{1}{4} (1+ \frac{1}{2}sin t)oz/gal\\\\\\ \frac{1}{2} (1+ \frac{1}{2}sin t)oz/min\\\\\\Rate_{out}=2gal/min \times\frac{Q}{100}oz/gal\\\\\frac{Q}{50}oz/min

Therefore,

\frac{dQ}{dt} can be evaluated as shown below:

\frac{dQ}{dt}=\frac{1}{2}(1+\frac{1}{2}\sin t)-\frac{Q}{50}\\\\\\\frac{dQ}{dt}+\frac{1}{50}Q=\frac{1}{2}+\frac{1}{4}\sin t

The above differential equation is in standard form:

\frac{dy}{dt}+Py=G

Here, P=\frac{1}{50},G=\frac{1}{2}+\frac{1}{4}\sin t

The integrating factor is as follows:

\mu(t)=e^{\int {P}dt}\\\mu(t)=e^{\int {\frac{1}{50}}dt}\\\mu(t)=e^{\frac{t}{50}}

Thus, the integrating factor is  \mu(t)=e^{\frac{t}{50}}

Therefore, the general solution is as follows:

y\mu(t)=\int {\mu (t)G}dt\\\\Qe^{\frac{t}{50}}=\int {e^{\frac{t}{50}}(\frac{1}{2}+\frac{1}{4}\sin t) dt}\\\\Qe^{\frac{t}{50}}=\frac{1}{2}\int {e^{\frac{t}{50}}dt + \frac{1}{4}\int {\sin t {e^{\frac{t}{50}}} dt}\\\\\Qe^{\frac{t}{50}}=25 {e^{\frac{t}{50}} + \frac{1}{4}\int {\sin t {e^{\frac{t}{50}}} dt}+C...(1)

Here, C is arbitrary constant of integration.

Solve \int {\sin te^{\frac{t}{50}}} dt}

Here u = e^{\frac{t}{50}} and v =\sin t.

Substitute u , v in the below formula:

\int{u,v}dt=u\int{v}dt-\int\frac{du}{dt}\int{v}dt\dot dt\\\\\int {e^{\frac{t}{50}}\sin t}dt=-e^{\frac{t}{50}}\cos t + \frac{1}{50}\int{e^{\frac{t}{50}}\cos t}dt...(2)

Now, take u = e^{\frac{t}{50}}, v =\sin t

Therefore, \int{e^{\frac{t}{50}}\cos t} dt=\int {e^{\frac{t}{50}}\sin t}dt - \frac{1}{50}\int{e^{\frac{t}{50}}\sin t}dt...(3)

Use (3) in equation(2)

\int {e^{\frac{t}{50}}\sin t}dt=-e^{\frac{t}{50}}\cos t + \frac{e^{\frac{t}{50}}}{50}\sin t - \frac{1}{2500}\int{e^{\frac{t}{50}}\sin t}dt\\\\\frac{2501}{2500}\int{e^{\frac{t}{50}}\sin t}dt={e^{\frac{t}{50}}\cos t}+\frac{e^{\frac{t}{50}}}{50}\sin t\\\\\int{e^{\frac{t}{50}}\sin t}dt=\frac{2500}{2501}{e^{\frac{t}{50}}\cos t}+\frac{50}{2501}e^{\frac{t}{50}}\sin t...(4)

Use (4) in equation(l) .

Qe^{\frac{t}{50}}=25 e^{\frac{t}{50}} - \frac{625}{2501}e^{\frac{t}{50}}\cos t +\frac{25}{5002}e^{\frac{t}{50}}\sin t+C

Apply the initial conditions t =0, Q = 50.

50=25-\frac{625}{2501}+c\\\\c=\frac{63150}{2501}

So, Qe^{\frac{t}{50}}=25 e^{\frac{t}{50}} - \frac{625}{2501}e^{\frac{t}{50}}\cos t +\frac{25}{5002}e^{\frac{t}{50}}\sin t+\frac{63150}{2501}

Therefore, the amount of salt in the tank at any time is as follows:

Qe^{\frac{t}{50}}=25 e^{\frac{t}{50}} - \frac{625}{2501}e^{\frac{t}{50}}\cos t +\frac{25}{5002}e^{\frac{t}{50}}\sin t+\frac{63150}{2501}e^{\frac{-t}{50}}

b)

sketch the solution curve as shown in attachment as graph 1:

CHECK COMMENT FOR C

3 0
2 years ago
Two particles carrying charges q1 and q2 are separated by a distance r and exert an electric force F⃗ E on each other. If q1 is
zepelin [54]

Answer:

q2 must also be doubled

r may also be halved

Explanation:

According to Coulumbs law

F= K q1 q2/r^2

If q1 is doubled, we must necessarily double q2 and r may also be halved in order to maintain F at the same value. Once the value of F is thus kept constant and E is also constant, the product FE must remain constant.

5 0
2 years ago
A star orbiting a black hole in a clockwise direction at a radial distance of 1.0 × 106 km is acted upon by a counterclockwise f
snow_tiger [21]

Answer:

2.5 \times 10^{11} N-m upwards        

Explanation:

Torque is the vector cross product of the force and radial distance.

\tau = rF sin \theta

\tau = (1.0\times 10^6 \times 10^3) m \times 250 N\times sin 90^o \\\Rightarrow \tau= 2.5 \times 10^{11} N-m

The direction of the torque would be perpendicular to the direction of the force and radial distance. The direction of the force is counter-clockwise. The direction of the torque would be upwards.

4 0
2 years ago
If low CVP precipitates a suction alarm, rapid infusion of volume can remedy the situation after dropping the P-level.
motikmotik

Answer:

d

Explanation:

8 0
2 years ago
Which occurrence would lead you to conclude that lights are connected in a
skelet666 [1.2K]

Answer:B When one bulb burns out, all the others lights stay lit.

Explanation:

3 0
2 years ago
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