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2 years ago

10

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 0.076 V exists across the membrane. The thickness of the membrane is 8.9 x 10-9 m. What is the magnitude of the electric field in the membrane?

1 answer:

BlackZzzverrR [31]2 years ago

3 0

Answer:

E = 8.5 * 10^6 V/m

Explanation:

In general we have the following relation between the Electric Field and the Elecric Potential:

$\int\limits^2_1 {\vec{E} \cdot\vec{dl}} = V_{2} -V_{1}$

Due to the vector nature of the electric filed, we can only know the mean Electric field E across the membrane, and take it out from the integral, that is:

E = (ΔV)/L

Where L is the thickness of the membrane and ΔV is the potential difference.

Therefore:

E = 8.53933*10^6 V/m

rounding to the first tenth:

E = 8.5 * 10^6 V/m

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A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c

Artist 52 [7]

Answer:

Approximately $25\; {\rm N}$ (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

When the $6.0\; {\rm cm}$ -spring in this question is stretched to $10\; {\rm cm}$ , the displacement is $x = (10\; {\rm cm} - 6.0\; {\rm cm})$ .
Likewise, if this spring is stretched to $20\; {\rm cm}$ , the displacement would be $(20\; {\rm cm} - 6\; {\rm cm})$ .

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of $F_{\text{a}}$ displaces this spring by $x_{\text{a}}$ , while a force of $F_{\text{b}}$ displaces this spring by $x_{\text{b}}$ , then:

$\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}$ .

In this question, it is given that a force of $F_{\text{a}} = 7.0 \; {\rm N}$ would stretch this spring by $x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm})$ . Thus, the force $F_{\text{b}}$ required to stretch this spring by $x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm})$ would satisfy:

$\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}$ .

Rearrange and solve for $F_{\text{b}}$ :

$\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}$ .

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1 year ago

An object with a mass of 2.0 kg is accelerated at 5.0 m/s/s. the net force acting on the mass is

Vitek1552 [10]

First we need to know the equation:
F= Mass times acceleration
F = 2.0 kg times 5.0 m/s^2
multiply them to get the net force!
F= 10 N
N is newton
Hope this heps

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3 years ago

Read 2 more answers

Think about the way optical fibers are designed. why are the ends of the fibers glowing very brightly while the sides of the fib

Effectus [21]

Answer:

D. because the light is reflected back into the fiber along its sides

Explanation:

The fiber is constructed in a way that the light is bent/reflected/refracted toward the center core of glass. So, from the center core, there is a layer above it that has a different propagation than the core, and above that the same thing. To give you a real world visual example, if you look down in a pool of water, then stick a straight stick into it, you see that the straight stick appears to bend. That is what is happening to the light as it travels through a different medium (air to water). This same effect is incorporated in the fiber optic cable construction.

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2 years ago

A 100 mH inductor whose windings have a resistance of 6.0 Ω is connected across a 12 Vbattery having an internal resistance of 3

Alexus [3.1K]

Answer:

The store energy in the inductor is 0.088 J

Explanation:

Given that,

Inductor = 100 mH

Resistance = 6.0 Ω

Voltage = 12 V

Internal resistance = 3.0 Ω

We need to calculate the current

Using ohm's law

$V = IR$

$I=\dfrac{V}{R+r}$

Put the value into the formula

$I=\dfrac{12}{6.0+3.0}$

$I=1.33\ A$

We need to calculate the store energy in the inductor

$U=\dfrac{1}{2}LI^2$

$U=\dfrac{1}{2}\times100\times10^{-3}\times(1.33)^2$

$U=0.088\ J$

Hence, The store energy in the inductor is 0.088 J

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2 years ago

A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.

Tcecarenko [31]

Answer:

a) 1.89 m/s b) 172.1 N

Explanation:

a)

Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
This work, is just 4,475 J.
So we can write the following equation:

$\Delta K = \frac{1}{2} * m*v^{2} = 4,475 J$

where m= mass of the truck = 2.5*10³ kg.
So, we can find the speed v, as follows:

$v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} } = 1.89 m/s$

b)

The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

$W = F*d$

We can solve for F, as follows:

$F = \frac{W}{d} = \frac{4,475 J}{26.0m} = 172.1 N$

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2 years ago