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saveliy_v [14]
3 years ago
9

Skin is the main barrier between internal organs and the outside environment. The outer layer of skin is composed mostly of epit

helial cells. Which is a characteristic of epithelial cells that makes them ideal for providing this type of protection?
The cells are packed tightly together.

The cells are surrounded by a lot of extracellular matrix.

The cells are separated by synapses.

The cells are organized into a mesh structure.
Physics
2 answers:
zhannawk [14.2K]3 years ago
4 0

Answer- A: The cells are packed tightly together.

vagabundo [1.1K]3 years ago
3 0
The characteristic of epithelial cells that makes them ideal for providing this type of protection is that the cells are packed tightly together. 
Skin, the body's largest organ,is our first and best defense against external aggressors. The many layers work hard to protect us, however when its condition is compromised, its ability to work as an effective barrier is impaired. 
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Answer:

Option C

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A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con
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Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5  x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \  \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm

The magnitude of the electric field is now given as;

E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

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