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snow_lady [41]
4 years ago
9

A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik

e a parallel plate capacitor, each with a charge density of 10?5C/m2, and the outer wall is positively charged. Although unrealistic, assume that the space between cell walls is filled with air.
PART A. What is the magnitude of the electric field between the membranes?

1×106N/C

1×10?15N/C

5×10?5N/C

9×10?2N/C
Physics
1 answer:
Nina [5.8K]4 years ago
4 0

Answer:

1 × 10⁶ N/C

Explanation:

The magnitude of the electric field between the membrane = surface density / permittivity of free space = 10 ⁻⁵C/ m² / (8.85 × 10⁻¹²N⁻¹m⁻²C²) = 1.13 × 10⁶ N/C approx 1 × 10⁶ N/C

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Answer:

F = 4856.32 N

Explanation:

Given,

A satellite is orbiting earth at a distance from Earth surface, h = 35000 m

The mass of the satellite, m = 500 Kg

The radius of the Earth, R = 6.371 x 10⁶ m

The mass of the Earth, M = 5.972 x 10²⁴ Kg

The gravitational constant, G = 6.67408 x 10 ⁻¹¹ m³ kg⁻¹ s⁻²

The force between the Earth and the satellite is given by the formula

                                    F = GMm/(R+h)²  N

Substituting the values in the above equation

        F = (6.67408 x 10 ⁻¹¹ X 5.972 x 10²⁴ X 500) / (6.371 x 10⁶ + 35000)²

                                        = 4856.32 N

Hence,  the force between the planet and the satellite is, F = 4856.32 N

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3 years ago
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igomit [66]

Answer:

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We know that brightness is given by

B=\frac{P}{4\pi d^2}

So \frac{3.8\times 10^{26}}{4\pi (6\times 10^{16})^2}=\frac{0.20}{4\pi d^2}

3.8\times 10^{26}d^2=7.2\times 10^{32}

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3 0
3 years ago
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Answer:

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