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likoan [24]
3 years ago
6

The membrane that surrounds a certain type of living cell has a surface area of 5.1 x 10-9 m2 and a thickness of 1.4 x 10-8 m.

Physics
1 answer:
ziro4ka [17]3 years ago
6 0

a) The charge on the outer surface is 1.2\cdot 10^{-12} C

b) The number of ions is 7.5\cdot 10^6

Explanation:

a)

The membrane behaves as a parallel plate capacitor, whose capacity is given by the equation

C=\frac{k\epsilon_0 A}{d}

where

k = 4.3 is the dielectric constant

\epsilon_0 =8.85\cdot 10^{-12} F/m is the vacuum permittivity

A=5.1\cdot 10^{-9} m^2 is the surface area

d=1.4\cdot 10^{-8} m is the distance between the two plates

Substituting,

C=\frac{(4.3)(8.85\cdot 10^{-12})(5.1\cdot 10^{-9})}{1.4\cdot 10^{-8}}=1.4\cdot 10^{-11} F

The capacity of the membrane is related to the potential difference between the two surfaces by

C=\frac{Q}{\Delta V}

where here we have

Q = excess charge on one surface

\Delta V = 85.5 mV = 0.0855 V is the potential difference between the two surfaces

Solving for Q, we find

Q=C\Delta V=(1.4\cdot 10^{-11})(0.0855)=1.2\cdot 10^{-12} C

b)

We said that the net charge on the outer surface is

Q=1.2\cdot 10^{-12} C

The charge of one K+ ions is equal to the electron charge

+e=1.6\cdot 10^{-19} C

Therefore, the number of ions on the outer surface can be found by dividing the total charge by the charge of a single ion:

N=\frac{Q}{e}=\frac{1.2\cdot 10^{-12}}{1.6\cdot 10^{-19}}=7.5\cdot 10^6

Learn more about capacity:

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#LearnwithBrainly

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The International Space Station (ISS) orbits Earth at an altitude of 400 km. Using this information, plus the mass and radius of
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Answer:

v = 7671.57 m/s

T = 1.55 hours

Explanation:

mass of Earth, M = 6 x 10^24 kg

Radius of earth, R = 6400 km = 6.4 x 10^6 m

height, h = 400 km

Velocity is given by

v=\sqrt{\frac{GM}{R+h}}

where, G be the universal gravitational constant.

G = 6.657 x 10^-11 Nm^2/kg^2

v=\sqrt{\frac{6.67\times 10^{-11}\times 6\times 10^{24}}{6800\times 10^{3}}}

v = 7671.57 m/s

Let T b the period

T=\frac{2\pi (R+h)}{v}

T=\frac{2\times 3.14(6800\times 1000)}{7671.57}

T = 5566.53 second

T = 1.55 hours

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A) an electron has an initial speed of 226000 m/s. if it undergoes an acceleration of 4.0 x 1014 m/s2, how long will it take to
KIM [24]

initial speed of 226000 m/s

acceleration of 4.0 x 1014 m/s2,

speed of 781000 m/s

What is Acceleration?

  • Acceleration is a rate of change of velocity with respect to time with respect to direction and speed.
  • A point or an object moving in a straight line is accelerated if it speeds up or slows down.
  • Acceleration formula can be written as,

                    a = (v - u ) / t m/s²

As we have to find the time taken, the formula can be altered as,

t = \frac{v-u}{a}

where, t - time taken to reach a final speed

v - final velocity

u - initial velocity

a - acceleration.

Substituting all the given values,

t =\frac{781000 - 226000} {4* 1014}

= 1.3875 × 10⁻⁹ seconds.

So, taken to reach the final speed is found to be 1.3 × 10⁻⁹ 8iH..

7 0
2 years ago
If you're walking on the ice cream at 5 ounces per toaster, and your bicycle loses a sock, how much gravy will you need to repai
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You observe that a mass suspended by a spring takes 0.25 s to make a full oscillation. What is the frequency of this oscillation
Katarina [22]

Answer:

Frequency of oscillation, f = 4 Hz

time period, T = 0.25 s

Angular frequency, \omega = 25.13 rad/s

Given:

Time taken to make one oscillation, T = 0.25 s

Solution:

Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:

f = \frac{1}{T}

f = \frac{1}{0.25}

Frequency of oscillation, f = 4 Hz

The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.

Therefore, time period, T = 0.25 s

Angular frequency of oscillation is given by:

\omega = 2\pi \times f

\omega = 2\pi \times 4

\omega = 25.13 rad/s

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