Answer:

Explanation:
We are given that
Surface area of membrane=
Thickness of membrane=
Assume that membrane behave like a parallel plate capacitor.
Dielectric constant=5.9
Potential difference between surfaces=85.9 mV
We have to find the charge resides on the outer surface of membrane.
Capacitance between parallel plate capacitor is given by

Substitute the values then we get
Capacitance between parallel plate capacitor=

V=


Hence, the charge resides on the outer surface=
Answer:
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Explanation:
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Answer:
Explanation:
D = 8.27 m ⇒ R = D / 2 = 8.27 m / 2 = 4.135 m
ω = 0.66 rev/sec = (0.66 rev/sec)*(2π rad/1 rev) = 4.1469 rad/s
We can apply the equation
Ff = W ⇒ μ*N = m*g <em>(I)</em>
then we have
N = Fc = m*ac = m*(ω²*R)
Returning to the equation <em>I</em>
<em />
μ*N = m*g ⇒ μ*m*ω²*R = m*g ⇒ μ = g / (ω²*R)
Finally
μ = (9.81 m/s²) / ((4.1469 rad/s)²*4.135 m) = 0.1379
Answer: the first answer is correct
Explanation:
she goes back to a stable feeling