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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 particular cylindrical bucket has a height of 36.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is em
Sergeeva-Olga [200]

Answer:

a. 0.000002 m

b. 0.00000182 m

Explanation:

36 cm = 0.36 m

15 cm = 0.15 m

a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:

P = \ro g h = 1000 * 10 * 20 = 200000 Pa

Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same

P_1V_1 = P_2V_2

Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:

V_1 = Ah

As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:

P_1Ah_1 = P_2Ah_2

h_2 = h_1\frac{P_1}{P_2} = 0.36\frac{1.105}{200000} = 0.000002 m

b) If the temperatures changes, we can still reuse the ideal gas equation above:

\frac{P_1Ah_1}{T_1} = \frac{P_2Ah_2}{T_2}

h_2 = h_1\frac{P_1T_2}{P_2T_1} = 0.36\frac{1.105 * 275}{200000*300} =0.00000182 m

3 0
2 years ago
A beam of electrons moves at right angles to a magnetic field of 4.5 × 10-2 tesla. If the electrons have a velocity of 6.5 × 106
defon

Answer:

4.7\cdot 10^{-14}N

Explanation:

For a charge moving perpendicularly to a magnetic field, the force experienced by the charge is given by:

F=qvB

where

q is the magnitude of the charge

v is the velocity

B is the magnetic field strength

In this problem,

q=1.6\cdot 10^{-19} C

v=6.5\cdot 10^6 m/s

B=4.5\cdot 10^{-2} T

So the force experienced by the electrons is

F=(1.6\cdot 10^{-19}C)(6.5\cdot 10^6 m/s)(4.5\cdot 10^{-2} T)=4.7\cdot 10^{-14}N

3 0
1 year ago
Which statements accurately describe Dmitri Mendeleev’s contributions to the development of the periodic table? Check all that a
Mama L [17]
Mendeleev produced the first orderly arrangement of known elements. Mendeleev used patterns to predict undiscovered elements.
6 0
1 year ago
Read 2 more answers
Two tiny particles having charges 20.0 μC and 8.00 μC are separated by a distance of 20.0 cm What are the magnitude and directio
Alecsey [184]

Answer:

The magnitude and direction of electric field midway between these two charges is 10.8\times10^{5}\ N/C along AB.

Explanation:

Given that,

First charge q_{1}= 20\mu C

second charge q_{2}= 8\mu C

Distance = 20 cm

We need to calculate the electric field

For first charge,

Using formula of electric field

E_{1}= \dfrac{kq_{1}}{r^2}

Put the valueinto the formula

E_{1}=\dfrac{9\times10^{9}\times20\times10^{-6}}{10\times10^{-2}}

E_{1}=18\times10^{5}\ N/C

Direction of electric field along AB

We need to calculate the electric field

For second charge,

Using formula of electric field

E_{2}= \dfrac{kq_{2}}{r^2}

Put the valueinto the formula

E_{2}=\dfrac{9\times10^{9}\times8\times10^{-6}}{10\times10^{-2}}

E_{2}=7.2\times10^{5}\ N/C

Direction of electric field along AO

We need to calculate the net electric field at midpoint

E_{net}=E_{1}-E_{2}

E_{net}=(18-7.2)\times10^{5}\ N/C

E_{net}=10.8\times10^{5}\ N/C

Direction of net electric field along AB

Hence, The magnitude and direction of electric field midway between these two charges is 10.8\times10^{5}\ N/C along AB.

8 0
1 year ago
Alex throws a 0.15-kg rubber ball down onto the floor. The ball’s speed just before impact is 6.5 m/s, and just after is 3.5 m/s
Jet001 [13]

Answer: Change in ball's momentum is 1.5 kg-m/s.

Explanation: It is given that,

Mass of the ball, m = 0.15 kg

Speed before the impact, u = 6.5 m/s

Speed after the impact, v = -3.5 m/s (as it will rebound)

We need to find the change in the magnitude of the ball's momentum. It is given by :

So, the change in the ball's momentum is 1.5 kg-m/s. Hence, this is the required solution.

Read more on Brainly.com - brainly.com/question/12946012#readmore

7 0
1 year ago
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