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3 years ago

The epidermis is the top layer of skin compared to the underlying dermis of the

Physics

2 answers:

madam [21]3 years ago

8 0

Answer:

Although you may not typically think of the skin as an organ, it is in fact made of tissues that work together as a single structure to perform unique and critical functions. The skin and its accessory structures make up the integumentary system, which provides the body with overall protection. The skin is made of multiple layers of cells and tissues, which are held to underlying structures by connective tissue (Figure 5.1.1). The most superficial layer of the skin is the epidermis which is attached to the deeper dermis. Accessory structures, hair, glands, and nails, are found associated with the skin. The deeper layer of skin is well vascularized (has numerous blood vessels) and is superficial to the hypodermics. It also has numerous sensory, and autonomic and sympathetic nerve fibers ensuring communication to and from the brain.

Explanation:

Describe the layers of the skin and the functions of each layer

-Identify the components of the integumentary system

-Describe the layers of the skin and the functions of each layer

-Describe the layers of the epidermis and dermis

-Identify and describe the hypodermis and fascia

-Describe the role of keratinocytes and their life cycle

-Describe the role of melanocytes in skin pigmentation

elena-14-01-66 [18.8K]3 years ago

4 0

Answer:

just as rea was situated at a distance in picture and a half dozen years old in his first comment in picture the actor was the first thing he had ever heard about it he reati was the only person who has

Explanation:

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NeTakaya

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4 0

3 years ago

Which parameter best defines the primary difference between weather and climate?

LenKa [72]

Answer:

time

Explanation:

weather is the atmospheric condition of a place over a short period of time, while climate is the weather condition prevailing in an area over a long period of time. From the two definitions above we can see that weather is the condition over a short period of time while climate is over longer periods, therefore the primary difference between them is time.

3 0

3 years ago

Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1

hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

$\sum F=ma$

We know that the force between two bodies due to gravity is given by the following equation:

$F_{g} = G\frac{m_{1}m_{2}}{r^{2}}$

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

$F_{g} =G\frac{Mm}{r^{2}}$

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

$F_{c}=ma_{c}$

where the centripetal acceleration is given by:

$a_{c}=\omega ^{2}r$

where

$\omega = \frac{2\pi}{T}$

Where T is the period, and $\omega$ is the angular speed of the planet, so:

$a_{c} = ( \frac{2\pi}{T})^{2}r$

or:

$a_{c}=\frac{4\pi^{2}r}{T^{2}}$

so:

$F_{c}=m(\frac{4\pi^{2}r}{T^{2}})$

so now we can do the sum of forces:

$\sum F=ma$

$F_{g}=ma_{c}$

$G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})$

in this case we can get rid of the mass of the planet, so we get:

$G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})$

we can now solve this for $T^{2}$ so we get:

$T^{2} = \frac{4\pi ^{2}r^{3}}{GM}$

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

$T_{1}^{2}>T_{2}^{2}$

So let's see what's going on there, we'll call:

$M_{1}$ = mass of Star 1

$M_{2}$ = mass of Star 2

So:

$\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}$

we can get rid of all the constants so we end up with:

$\frac{1}{M_{1}}>\frac{1}{M_{2}}$

and let's flip the inequality, so we get:

$M_{2}>M_{1}$

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.