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RoseWind [281]
3 years ago
14

Assume an axon has an internal diameter of 1μm and a myelin sheath 1μm thick. The internal specific resistance is 100 Ω cm. For

the myelin sheath use a dielectric constant of 8 and a transmembrane resistance of 2 x 10^5 Ω cm^2 .
A) Calculate the internal resistance per unit length.
Please show equation and each variable meaning.
Physics
1 answer:
SpyIntel [72]3 years ago
8 0

Answer:

1.27\times 10^{12}\Omega/m

Explanation:

We are given that

Diameter=d=\mu m

Thickness=1\mu m

Radius=r=\frac{d}{2}=\frac{1}{2}\mu m=0.5\times 10^{-6} m

Using 1\mu m=10^{-6} m

Dielectric constant=8

Resistance =R=2\times 10^5\Omega cm^2

Internal specific resistance=r=100 ohm cm=100\times \frac{1}{100}\Omega-m=1\Omega m

Using 1 m=100 cm

Internal resistance per unit length=\frac{r}{A}=\frac{1}{\pi r^2}=\frac{1}{3.14\times (0.5\times 10^{-6})^2}=1.27\times 10^{12}\Omega/m

Using \pi=3.14

Internal resistance per unit length=1.27\times 10^{12}\Omega/m

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BartSMP [9]

Answer:

A. Final pressure P2

P2/P1 = (T2/T1)^n/n-1

P1 = 4bar

T1 = 438K

T2 = 300K

Polytropic index, n, = 1.3

P2 = 4 (300/438)^1.3/1.3-1

P2 = 4 (300/438)^4.333

P2 = 4 * 0.19400

P2 = 0.776bar.

B. The work done is;

W = mR/ n-1 (T1 -T2)

Where, R = 0.1889kJ/kg.K, m = 1

W = 1 * 0.1889/ 1.3-1 * (438-300)

W = 86.89kJ/kg.

C. The heat transfer, Q

Q = W + ΔU

Q = W + mCv(T2-T1), where Cv of nitrogen is 0.743kj/kgk

Q = 86.89 + 1 * 0.743 (300-438)

Q = 86.89 + (-102.534)

Q = -15.644kJ/K

Q = 15.64kJ/K

3 0
3 years ago
A thin rod of length L and total charge Q has the nonuniform linear charge distribution λ(x)=λ0x/L, where x is measured from the
marissa [1.9K]

Answer:

Explanation:

λ(x) = λo x/ L

(a) The total charge is Q.

Q=\int_{0}^{L}dq

Q=\int_{0}^{L}\frac{\lambda _{0}x}{L}dx

Q=\frac{\lambda _{0}}{2L}\left ( x^{2} \right )_{0}^{L}

Q=\frac{\lambda _{0}}{2L}

λo = 2Q/L

(b)

Let at a distance x from the origin the charge is dq.

so, dq = (2Q/L) x/ L dx

dq=\frac{2Qx}{L^{2}}dx

The potential due to this small charge at a distance d to the left of origin

dV = \frac{KdQ}{d+x}

\int_{0}^{V}dV = \frac{2KQ}{L^{2}}\int_{0}^{L}\frac{xdx}{d+x}

V = \frac{2KQ}{L^{2}}\int_{0}^{L}\left ( 1- \frac{d}{d+x}\right )dx

V = \frac{2KQ}{L^{2}}\times \left ( x-dln(d+x) \right )\int_{0}^{L}

V = \frac{2KQ}{L^{2}}\times \left ( L-dln(d+L)-0+dlnd \right )

V = \frac{Q}{4\pi \epsilon _{0}L^{2}}\times \left ( L+d\times ln\left (\frac{d}{d+L}  \right )\right )

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3 years ago
Two people singing off key in a choir
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Linda, a biker, is moving along a circular path at a constant speed of 10 km/h. In a neighboring arena Kevin, another biker, is
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Answer:

the answer is  C

Explanation:

i took the test just now so i am glad i can help someone in the future❣❣❣❣

<u><em>Also here is the explanation</em></u>

Note that the magnitude of acceleration (a) is given by a =  

v2

R

, where v is the speed and R is the radius. Because acceleration is the same for both bikers, the ratio 

v2

R

, also has to be the same for both. Given that Kevin’s speed is higher, the radius of his path must also be higher to maintain the same ratio. Therefore, Kevin’s circular path has a bigger radius than Linda’s path.

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The universe will collapse
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