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Ugo [173]
3 years ago
11

Can we use a copper wire instead of eureka wire

Physics
1 answer:
natka813 [3]3 years ago
8 0
Yes you can amd it will actually work better than eureka wire copper wire holds more charge
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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.
ziro4ka [17]

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:

brainly.com/question/10427437

brainly.com/question/8892837

brainly.com/question/9617400

#LearnwithBrainly

6 0
3 years ago
Write one general properties of non contact force​
Gekata [30.6K]

Answer:    Gravitational force between the two masses does not depend on the medium separating two masses.

3 0
2 years ago
To see why an MRI utilizes iron to increase the magnetic field created by a coil, calculate the current needed in a 400-loop-per
gtnhenbr [62]

Answer:

B = 4.059 x 10¹⁵ T

Explanation:

Given,

Number of loop, N = 400

radius of loop, r = 0.65 x 10⁻¹⁵ m

Current, I = 1.05 x 10⁴ A

Magnetic field at the center of the loop

B = \dfrac{\mu_0NI}{2R}

B = \dfrac{4\pi\times 10^{-7}\times 400 \times 1.05 \times 10^4}{2\times 0.65\times 10^{-15}}

B = 4.059 x 10¹⁵ T

7 0
3 years ago
Convert (a) 50 oF, (b) 80 oF, (c) 95 oF to Celsius
storchak [24]
I really need these points thx a lot
5 0
2 years ago
A heat source could be considered a source of pollution Please select the best answer from the choices provided
jeyben [28]

The answer is true, is called Thermal pollotion:).

5 0
2 years ago
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