PHYSICS COLLEGE

The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a potential difference of about 0.078 V exists across the membrane. The thickness of the membrane is 7.1 x 10-9 m. What is the magnitude of the electric field in the membrane?

Answers

Answer 1

Answer:

Answer:

$10.99* 10^6\ V/m$

Explanation:

Given:

Potential difference across the membrane (ΔV) = 0.078 V

Thickness of the membrane (Δx) = 7.1 × 10⁻⁹ m

Magnitude of electric field (|E|) = ?

We know that, the electric field due to a potential difference (ΔV) across a distance of Δx is given as:

$E=-(\Delta V)/(\Delta x)$

So, the magnitude of the electric field is calculated by ignoring the negative sign and thus is given as:

$|E|=(\Delta V)/(\Delta x)$

Plug in the given values and solve for '|E|'. This gives,

$|E|=(0.078\ V)/(7.1* 10^(-9)\ m)\n\n|E|=10.99* 10^6\ V/m$

Therefore, the magnitude of the electric field in the membrane is $10.99* 10^6\ V/m$ .

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A positively charged rod is held near a neutral conducting sphere as illustrated below. A positively charged particle is moved from point A to point B. Thee electrostatic work done on the positively charged particle during the motio

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The movement of a positively charged particle from point A to point B. the motion-induced electrostatic work done on the positively charged particle.

Whether positively or negatively charged, an object that is neutral will interact with it in a pleasing way. Both positively charged and neutral items attract one another, as do negatively charged and neutral objects. These electrons gather on the further surface of sphere B, depleting the electron supply in sphere A. Therefore, sphere A (which is closer to the rod) obtains a positive charge and sphere B acquires a negative charge when the two spheres separate in the presence of the rod. The change in the particle's electrostatic potential energy in the external field equals the work done by the external force. When a charge is pushed from point A to point B, its potential energy changes, representing the labor of an outside force.

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The Mistuned Piano Strings Two identical piano strings of length 0.800 m are each tuned exactly to 480 Hz. The tension in one of the strings is then increased by 1.0%. If they are now struck, what is the beat frequency between the fundamentals of the two strings? SOLUTION Conceptualize As the tension in one of the strings is changed, its fundamental frequency changes. Therefore, when both strings are played, they will have different frequencies and beats be heard. Categorize We must combine our understanding of the waves model for strings with our new knowledge of beats.

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

The answer is

$T_2 = 1.008$ % higher than $T_1$

$T_2 = 0.99$ % lower than $T_1$

Explanation:

From the question we are told that

The first string has a frequency of $f_1 = 230 Hz$

The period of the beat is $t_(beat) = 0.99s$

Generally the frequency of the beat is

$f_(beat) = (1)/(t_(beat))$

Substituting values

$f_(beat) = (1)/(0.99)$

$= 1.01 Hz$

From the question

$f_2 - f_1 = f_(beat)$ for $f_2$ having a higher tension

$f_2 - 230 = 1.01$

$f_2 = 231.01Hz$

From the question

$(f_2)/(f_1) = \sqrt{(T_2)/(T_1) }$

$(T_2)/(T_1) = (f_2^2)/(f_1^2)$

Substituting values

$(T_2)/(T_1) = ((231.01)^2)/((230)^2)$

$T_2 = 1.008$ % higher than $T_1$

For $f_2$ having a lower tension

$f_1 - f_2 = f_(beat)$

$230 - f_2 = 1.01$

$f_2 = 230 -1.01$

$= 228.99$

From the question

$(f_2)/(f_1) = \sqrt{(T_2)/(T_1) }$

$(T_2)/(T_1) = (f_2^2)/(f_1^2)$

Substituting values

$(T_2)/(T_1) = ((228.99)^2)/((230)^2)$

$T_2 = 0.99$ % lower than $T_1$

What displacement do I have if I travel at 10 m/s E for 10 s? A. 1 m E B. 1 m C. 100 m D. 100 m E Scalar quantities include what 2 things? A. Number and direction B. Numbers and units C. Units and directions D. Size and direction What measures distance in a car? A. Odometer B. Pressure gauge C. Speedometer D. Steering wheel What displacement do I have if I travel 10 m E, then 6 m W, then 12 m E? A. 28 m E B. 16 m E C. 16 m D. 28 m

Answers

Hope this will help you

Final answer:

The displacement is 100 m to the east.

Explanation:

The displacement can be calculated using the formula:

Displacement = Velocity × Time

In this case, the velocity is 10 m/s to the east and the time is 10 seconds.

So, Displacement = 10 m/s × 10 s = 100 m to the east.

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A power P is required to do work W in a time interval T. What power is required to do work 3W in a time interval 5T? (a) 3P (b) 5P (c) 3P/5 (a) P (e) 5P/3

Answers

Answer:

Explanation:

The formula to calculate the power is:

$P=(W)/(T)$

where

W is the work done

T is the time required for the work to be done

In the second part of the problem, we have

Work done: 3W

Time interval: 5T

So the power required is

$P=(3W)/(5T)=(3)/(5)(W)/(T)=(3)/(5)P$

Students have four identical, hollow, uncharged conducting spheres, W, X, Y, and Z.Sphere Z is given a positive charge of +40 C. Sphere Z is touched first to sphere W, then sphere X, and finally to sphere Y. What is the resulting charge on sphere Y?

a. +5 με

b. +10 μC

c. +20 μC

d. +40 με

Answers

Explanation:

because they made contact that means their new force will be the same

Final answer:

Sphere Z is initially charged with +40 C. When it is touched to three other spheres, the charge is evenly distributed among them. The resulting charge on sphere Y is +10 μC.

Explanation:

The initial charge on sphere Z is +40 C. When sphere Z is touched to sphere W, the charge is evenly distributed between the two spheres, resulting in each sphere having a charge of +20 C. Then, when sphere Z is touched to sphere X, the total charge is evenly distributed between all three spheres, resulting in each sphere having a charge of +13.33 C. Finally, when sphere Z is touched to sphere Y, the total charge is evenly distributed between all four spheres, resulting in each sphere having a charge of +10 C. Therefore, the resulting charge on sphere Y is +10 μC (option b).

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Match each planet to an accurate characteristic

Answers

Answer:

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mercury - 1

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