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Reil
1 month ago
9

A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik

e a parallel plate capacitor, each with a charge density of 10?5C/m2, and the outer wall is positively charged. Although unrealistic, assume that the space between cell walls is filled with air.
PART A. What is the magnitude of the electric field between the membranes?

1×106N/C

1×10?15N/C

5×10?5N/C

9×10?2N/C

PART B. What is the magnitude of the force on a K+ ion between the cell walls?

2×10?13N

9×10?13N

2×10?11N

3×10?12N

PART C. What is the potential difference between the cell walls?

6×10?3V

1×107V

10V

1×10?2V

PART D. What is the direction of the electric field between the walls?

There is no electric field.

Toward the inner wall.

Parallel to the walls.

Toward the outer wall.

PART E. If released from the inner wall, what would be the kinetic energy of a 3fC charge at the outer wall? 1fC=10?15C.

3×10?2J

3×10?8J

3×10?17J

3×10?14J
Physics
1 answer:
serg [3.5K]1 month ago
5 0

Response:

The provided options are estimations of the precise answers:

A) 1\times10^6N/C

B) 2\times10^{-13}N

C) 1\times10^{-2}V

D) Towards the inner wall

E) 3\times10^{-17}J

Clarification:

A) The electric field within a parallel plate capacitor is described by the formula E=\frac{\sigma}{k\epsilon_0}, where \epsilon_0=8.85\times10^{-12}C/Vm and in our scenario \sigma=10^5C/m^2 and for air,k=1.00059, thus resulting in:

E=\frac{10^5C/m^2}{(1.00059)(8.85\times10^{-12}C/Vm)}=1.13\times10^6N/C

B) The K+ ion features one excess elemental charge, hence its charge is q=1.6\times10^{-19}C, and the force that a charge experiences in an electric field E is represented by F=qE, thus we have:

F=(1.6\times10^{-19}C)(1.13\times10^6N/C)=1.81\times10^{-13}N

C) The potential difference across two points that are separated by a distance d in a uniform electric potential E is expressed as \Delta V=dE, leading to:

\Delta V=(10\times10^{-9}m)(1.13\times10^6N/C)=1.13\times10^{-2}V

D) The electric field moves from positive to negative charges, directing it toward the inner wall.

E) The work accomplished by an electric field across a potential difference \Delta V affecting a charge Q is represented by W=Q\Delta V, corresponding to the kinetic energy it receives, yielding:

K=(3\times10^{-15}C)(1.13\times10^{-2}V)=3.39\times10^{-17}J

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The answer is letter d, 8.3.

 

Here’s a solution for the given problem:

 

We have:

B = 6i - 8j 


Let A be unknown; we'll denote A as = mi + nj 



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and we also know the magnitude of A+B is equivalent to the magnitude of A,

 

therefore, mag(A+B)=K=sqrt(m^2+n^2), or K^2 = m^2+n^2. 



Using vector addition, A+B becomes (m+6)i + (n-8)j.

 

Since we know A+B = Ki + 0j, we can establish that: 

m + 6 = K 


n - 8 = 0, which gives n=8. 

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Answer:

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Explanation:

a) learners can model this system as a simple pendulum, where the angular velocity is given by

         w = √ g / l

Here, angular velocity, frequency, and period are interconnected:

         w = 2π f = 2π / T

Substituting yields:

         T = 2π√ l / g

Using this formula, students can calculate the gravitational acceleration by measuring the period for several pendulum lengths and plotting:

        T² = 4π²  l / g

We plot T² against l.

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The slope is determined from the line's values rather than experimental data.

2) To perform the experiment, the string is secured to the sphere, then the pendulum length from the pivot to the sphere's center is measured using a tape measure. A slight angle (less than 10 degrees) is released, allowing the first swing to occur. Generally, the time for several oscillations, usually 10 or 20, is tracked to find the period:

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3) The independent variable, which is the length of the pendulums, is plotted on the x-axis, while the dependent variable, the squared period, is on the y-axis.

4) Referring to the line equation:

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5) Once the spring is cut, the sphere continues to be influenced by gravitational acceleration. The harmonic motion ceases, and the sphere moves vertically.

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