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3 years ago

How does the epidermis protect our skin

Physics

2 answers:

Darina [25.2K]3 years ago

8 0

From harmful radiation from our sun!!!!!

Ilya [14]3 years ago

6 0

The epidermis is the outermost epithelial layer over the skin. This is what you see and touch. It is made up mostly of keratinocytes.

Clearly this layer protects the layers beneath, i.e., the dermis and the subcutaneous layer which both have more delicate functions. It keeps out dirt, debris, bacteria, and basically everything else you don't want inside you. It also forms a protective barrier that protects from damage.

Within the epidermis, melanocytes create melanin to help protect from UV-rays. Lymphocytes and Langerhans cells help to neutralize pathogens.

You might be interested in

Why are there two fluoride ions in magnesium fluoride but only one fluoride ion in lithium fluoride

postnew [5]

This would be more of a chemistry question. Remember magnesium has a charge of 2+, and would need to hand off its two extra electrons. Fluorine can only take one electron at a time, so there needs to be two fluorines to take one magnesium's 2 electrons.

With lithium, it has a +1 charge, so it has one extra electron, which it can hand off to just 1 fluorine atom.

Another way of looking at this is: $Mg^{2+}$ + 2 $F^{-}$ = MgF2 (the charges must balance out to zero)

$Li^{+}$ + $F^{-}$ = LiF (the charges balance out to zero)

5 0

3 years ago

Read 2 more answers

A metal wire breaks when its tension reaches 100 newton. If the radius and length of the wire were both doubled then it would br

serg [7]

Answer:

200 N

Explanation:

Since Young's modulus for the metal, E = σ/ε where σ = stress = F/A where F = force on metal and A = cross-sectional area, and ε = strain = e/L where e = extension of metal = change in length and L = length of metal wire.

So, E = σ/ε = FL/eA

Now, since at break extension = e.

So making e subject of the formula, we have

e = FL/EA = FL/Eπr² where r = radius of metal wire

Now, when the radius and length are doubled, we have our extension as e' = F'L'/Eπr'² where F' = new force on metal wire, L' = new length = 2L and r' = new radius = 2r

So, e' = F'(2L)/Eπ(2r)²

e' = 2F'L/4Eπr²

e' = F'L/2Eπr²

Since at breakage, both extensions are the same, e = e'

So, FL/Eπr² = F'L/2Eπr²

F = F'/2

F' = 2F

Since F = 100 N,

F' = 2 × 100 N = 200 N

So, If the radius and length of the wire were both doubled then it would break when the tension reached 200 Newtons.

7 0

3 years ago

Please help on this one?

otez555 [7]

help me w mine and ill try to help with yours

3 0

3 years ago

A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th

photoshop1234 [79]

Answer:

The height is $h_c = 42.857$

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

From the question we are told that

The height is $h_s = 30 \ cm$