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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

You might be interested in
A system of two paint buckets connected by a lightweight rope is released from rest with the 12.0-kg bucket 2.00 m above the flo
NISA [10]

Explanation:

The given data is as follows.

    Mass of small bucket (m) = 4 kg

    Mass of big bucket (M) = 12 kg

    Initial velocity (v_{o}) = 0 m/s

    Final velocity (v_{f}) = ?

  Height H_{o} = h_{f} = 2 m

and,    H_{f} = h_{o} = 0 m

Now, according to the law of conservation of energy

         starting conditions = final conditions

  \frac{1}{2}MV^{2}_{o} + Mgh_{o} + \frac{1}{2}mv^{2}_{o} + mgh_{o} = \frac{1}{2}MV^{2}_{f} + Mgh_{f} + \frac{1}{2}mv^{2}_{f} + mgh_{f}

     \frac{1}{2}(12)(0)^{2} + (12)(9.81)(2) + \frac{1}{2}(4)(0)^{2} + (4)(9.81)(0) = \frac{1}{2}(12)V^{2}_{f} + (12)(9.81)(0) + \frac{1}{2}(4)V^{2}_{f} + (4)(9.81)(2)

                 235.44 = 8V^{2}_{f} + 78.48

                V_{f} = 4.43 m/s

Thus, we can conclude that the speed with which this bucket strikes the floor is 4.43 m/s.

3 0
1 year ago
An object begins at position x = 0 and moves one-dimensionally along the x-axis witļi a velocity v
Liula [17]

Answer:

The answer is "between 20 s and 30 s".

Explanation:

Calculating the value of positive displacement:

\ (x_{+ve}) = \frac{1}{2} \times 15 \times  20 \\\\

          = \frac{1}{2} \times 300 \\\\=  150 \\\\

Calculating the value of negative displacement upon the time t:

(x_{-ve}) = \frac{1}{2} \times 5 \times 20- 20(t-20) \\\\

          = \frac{1}{2} \times 100- 20t+ 400 \\\\= 50- 20t+ 400 \\\\

\to X= X_{+ve} + X_{-ve} \\\\

\to  150 - 50 -20t+400 =0\\\\\to 100 -20t+400 =0 \\\\\to 500 -20t =0\\\\\to 20t =500 \\\\\to t=\frac{500}{20}\\\\\to t=\frac{50}{2}\\\\\to t= 25

That's why its value lie in "between 20 s and 30 s".

6 0
2 years ago
In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec
Bezzdna [24]

Answer with Explanation:

We are given that

r=0.053 nm=0.053\times 10^{-9} m

1 nm=10^{-9} m

Charge on proton,q=1.6\times 10^{-19} C

a.We have to find the electric  potential of the proton at the position of the electron.

We know that the electric potential

V=\frac{kq}{r}

Where k=9\times 10^9

V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

V=27.17 V

B.Potential energy of electron,U=\frac{kq_e q_p}{r}

Where

q_e=-1.6\times 10^{-19} c=Charge on electron

q_p=q=1.6\times 10^{-19} C=Charge on proton

Using the formula

U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

U=-4.35\times 10^{-18} J

8 0
2 years ago
Un tren emplea cierto tiempo en recorrer 240 km. Si la velocidad hubiera sido 20 km por hora mas que la que llevaba hubiera tard
podryga [215]

Answer:

A train takes some time to travel 240 km. If the speed had been 20 km per hour more than the one it was carrying, it would have taken 2 hours less to travel this distance. In what time did he cover the 240 km

Explanation:

Given that,

A train travelled a distance of 240km

Let the initial speed be

S_1 = x km/hr

Let assume the time spent on the first journey is

t_1 = a

Now if he increase the speed to

S_2 = (x + 20) km/hr

Then, he would have take 2hrs less time

Then, time t_2 = a - 2

The common data fore the two journey is the distance

Speed = distance / time

For the first stage

S_1 = d / t_1

d = S_1 × a

d = x × a

240 = x•a

x = 240 / a Equation 1

For stage two

d = S_2 × t_2

d = (x+20) × (a - 2)

240 = (x+20) × (a - 2). Equation 2

Substitute equation 1 into 2

240 = (240/a + 20) × (a -2)

240 = 240 - 480/a + 20a - 40

240 - 240 + 40 = - 480/a + 20a

240 - 240 + 40 = (-480 + 20a²) / a

40 = (-480 + 20a²) / a

40a = -480 + 20a²

20a² - 40a -480 = 0

Divided through by 20

a² - 2a - 24 = 0

a² + 4a - 6a - 24 = 0

a(a+4) -6(a+4) = 0

(a-6)(a+4) = 0

(a-6) = 0 or (a+4) = 0

So, a = 6 or a = -4

The time cannot be negative, then, the time is a = 6hours

So, t_1 = a = 6hours,

So, the time used in the first journey is 6hours

So, in the second journey the time use is 2hours less than the first journey

Then, t_2 = 6 - 2 = 4 hours

t_1 = 6 hours

t_2 = 4 hours

Spanish

Un tren recorrió una distancia de 240 km.

Deje que la velocidad inicial sea

S_1 = x km / h

Supongamos que el tiempo dedicado al primer viaje es

t_1 = a

Ahora si aumenta la velocidad a

S_2 = (x + 20) km / h

Entonces, habría tomado 2 horas menos de tiempo

Entonces, el tiempo t_2 = a - 2

Los datos comunes para los dos viajes son la distancia.

Velocidad = distancia / tiempo

Para la primera etapa

S_1 = d / t_1

d = S_1 × a

d = x × a

240 = x • a

x = 240 / a Ecuación 1

Para la etapa dos

d = S_2 × t_2

d = (x + 20) × (a - 2)

240 = (x + 20) × (a - 2). Ecuación 2

Sustituye la ecuación 1 en 2

240 = (240 / a + 20) × (a -2)

240 = 240 - 480 / a + 20a - 40

240 - 240 + 40 = - 480 / a + 20a

240 - 240 + 40 = (-480 + 20a²) / a

40 = (-480 + 20a²) / a

40a = -480 + 20a²

20a² - 40a -480 = 0

Dividido entre 20

a² - 2a - 24 = 0

a² + 4a - 6a - 24 = 0

a (a + 4) -6 (a + 4) = 0

(a-6) (a + 4) = 0

(a-6) = 0 o (a + 4) = 0

Entonces, a = 6 o a = -4

El tiempo no puede ser negativo, entonces, el tiempo es a = 6 horas

Entonces, t_1 = a = 6 horas,

Entonces, el tiempo utilizado en el primer viaje es de 6 horas

Entonces, en el segundo viaje, el uso del tiempo es 2 horas menos que el primer viaje

Entonces, t_2 = 6 - 2 = 4 horas

t_1 = 6 horas

t_2 = 4 horas

5 0
2 years ago
2.0 kg of solid gold (Au) at an initial temperature of 1000K is allowed to exchange heat with 1.5 kg of liquid gold at an initia
Elanso [62]

Answer:

Explanation:

The specific heat of gold is 129 J/kgC

It's melting point is 1336 K

It's Heat of fusion is 63000 J/kg

Assuming that the mixture will be solid, the thermal energy to solidify the gold has to be less than that needed to raise the solid gold to the melting point. So,

The first is E1 = 63000 J/kg x 1.5 = 94500 J

the second is E2 = 129 J/kgC x 2 kg x (1336–1000)K = 86688 J

Therefore, all solid is not correct. You will have a mixture of solid and liquid.

For more detail, the difference between E1 and E2 is 7812 J, and that will melt

7812/63000 = 0.124 kg of the solid gold

8 0
2 years ago
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