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Sholpan [36]
3 years ago
6

The inner and outer surfaces of a cell membrane carry a negative and positive charge respectively. Because of these charges, a p

otential difference of about 70 mV exists across the membrane. The thickness of the membrane is 8 nm. If the membrane were empty (filled with air), what would the magnitude of the electric field inside the membrane
Physics
1 answer:
padilas [110]3 years ago
6 0

Answer:

The magnitude of the electric field inside the membrane is 8.8×10⁶V/m

Explanation:

The electric field due to electric potential at a distance Δs is given by

E=ΔV/Δs

We have to find the magnitude electric field in the membrane

Ecell= -ΔV/Δs

E_{cell}=-\frac{V_{in}-V_{out}}{s} \\E_{cell}=\frac{V_{out}-V_{in}}{s}\\E_{cell}=\frac{0.070V}{8*10^{-9}m } \\E_{cell}=8.8*10^{6}V/m

The magnitude of the electric field inside the membrane is 8.8×10⁶V/m

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Simlarity between longitudanal and transverse waves
babymother [125]

For transverse waves, the waves move in perpendicular direction to the source of vibration. For longitudinal waves, the waves move in parallel direction to the source of vibration . They are similar in the sense that energy is transferred in the form of waves.

8 0
3 years ago
A 81 kg man is riding on a 40 kg cart traveling at a speed of 2.3 m/s. He jumps off with zero horizontal speed relative to the g
Alexus [3.1K]

Answer:

\Delta v= 4.66\frac{m}{s}

Explanation:

In this case we have to use the Principle of conservation of Momentum:

<em>This principle says that in a system  the total momentum is constant if no external forces act in the system. The formula is:</em>

m_1v_1+m_2v_2=m_1u_1+m_2u_2

<em>Where:</em>

m_1: Mass of the first object.

m_2: Mass of the second object.

v_1: Initial velocity of the first object.

v_2: Initial velocity of the second object.

u_1: Final velocity of the first object.

u_2: Final velocity of the second object.

In <u>this problem</u> we have:

m_1=81kg\\m_2=40kg\\v_1_2=2.3\frac{m}{s}

u_1=0\frac{m}{s}

Observation: v_1_2: Is because the system has the same initial velocity.

First we have to find u_2,

m_1v_1+m_2v_2=m_1u_1+m_2u_2

We can rewrite it as:

(m_1+m_2)v_1_2=m_1u_1+m_2u_2

Replacing with the data:

(m_1+m_2)v_1_2=m_1u_1+m_2u_2\\\\(81kg+40kg)2.3\frac{m}{s}=81kg(0\frac{m}{s})+40kg(u_2)\\\\(121kg)2.3\frac{m}{s}=40kg(u_2)\\\\\frac{(121kg)2.3\frac{m}{s}}{40kg}=u_2\\\\\frac{278.3}{40}\frac{m}{s}=u_2\\\\6.96\frac{m}{s}=u_2

We found the final velocity of the cart, but the problem asks for the resulting change in the cart speed, this means:

\Delta v=u_2-v_2\\\Delta v=6.96\frac{m}{s}-2.3\frac{m}{s}\\\Delta v= 4.66\frac{m}{s}

Then, the resulting change in the cart speed is:

\Delta v= 4.66\frac{m}{s}

5 0
3 years ago
Find the magnitude of the net electric force exerted on this charge. Express your answer in terms of some or all of the variable
Yuri [45]

Answer:

Th steps is as shown in the attachment

Explanation:

from the diagram, its indicates that twelve identical charges are distributed evenly on the circumference of the circle. assuming one of gthe charge is shifted to the centre of he circle alomng the x axis, as such the charge is unbalanced and there is need ot balanced all the identical charges for the net force to be equal to zero.

The mathematical interpretation is as shown in the attachment.

3 0
3 years ago
A driver of a car traveling at -15m/s applies the brakes, causing a uniform acceleration of +2.0m/s2. If the brakes are applied
klemol [59]

Given :

Initial velocity, u = -15 m/s.

Acceleration , a = 2 m/s².

Time taken to applied brake, t = 2.5 s.

To Find :

The velocity of the car at the end of the braking period.

How far has the car moved during the braking period.

Solution :

By equation :

v = u+at\\\\v=-15 + 2\times 2.5\\\\v=-10 \ m/s

Now, distance covered by car is :

s=ut+\dfrac{at^2}{2}\\\\s=(-15)(2.5)+\dfrac{2(2.5)^2}{2}\\\\s=-31.25\ m

Hence, this is the required solution.

3 0
4 years ago
When did the Space Age begin?
Marta_Voda [28]
The Space Age began on October 4, 1957.
4 0
4 years ago
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