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nalin [4]
3 years ago
7

Many biological systems are well-described by the laws of statistical physics. A simple yet often powerful approach is to think

of a system as having only two states. For example, an ion channel may be open or closed. In this problem, consider a simple model of membrane channels for ions: The system is described by a Boltzmann distribution with only two states, with energies ε1 (open) and ε2 (closed). Assume the "open" state is the state of higher energy, so that ε1 > ε2.
If the probability of finding an ion channel open is popen and the probability of finding the ion channel closed is pclosed, which of the expressions below best represents the relative probability of open to closed, R = popen/pclosed? Use the notation z1 = e-ε1/kBT and z2 = e-ε2/kBT

a. z1-z2
b. z2-z1
c. z1/z2
d. z2/z1
e. Something else
Physics
1 answer:
GuDViN [60]3 years ago
8 0

Answer:

z1/z2

Explanation:

we have no quantum effects therefore we can make use of Maxwell Boltzmann distribution in the description of this system.

using the boltzman distribution the probability of finding a particle in energy state

P_{ei}  = \frac{gie^{-ei/kol} }{z}

we have

gi to be degeneration of the ith state

ei to be energy of ith state

z=e^{-ei/kbt} summation

P_{ope} = \frac{e^{-ei/kBt} }{z} = \frac{Z_{1} }{Z}

We have R to be equal to

\frac{P_{ope} }{P_{Close} } = \frac{Z1}{Z2}

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A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for
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Answer:

(a) a_s=0.62\frac{m}{s^2}

(b) a_s=0.13\frac{m}{s^2}

(c) x_f=2.6m

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}

(c) Using the kinematics equation:

x_f=x_0+v_0t \pm  \frac{at^2}{2}

For the girl, we have x_0=0 and v_0=0. So:

x_f_g=\frac{a_gt^2}{2}(1)

For the sled, we have v_0=0. So:

x_f_s=x_0_s-\frac{a_st^2}{2}(2)

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s

Now, we solve (1) for x_f_g

x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m

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3 years ago
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