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3 years ago

The fatty layer contains many nerves and blood vessels. True or False

Physics

1 answer:

mylen [45]3 years ago

5 0

Answer:

False , The fatty layer do not contains many nerves and blood vessels. The fatty layer is related to skin layers.

Explanation:

The fatty layer is the most under a layer of skin. It made up of a system of collagen and fat cells. It helps maintain the body's temperature and shields the body from harm by serving as a shock absorber.

The dermis is the central layer of the skin. The dermis is maintained collectively by a protein termed collagen. This layer gives skin elasticity and power. The dermis also holds shock and feel receptors.

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Block A, with a mass of 6.0 kg, is sliding across a frictionless track at 65 m/s when it collides with Block B ( 9.0 kg) which i

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The correct answer is a I hope that helped enjoy the rest of your weekend

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3 years ago

What do scientists use to divide the geologic time scale?

dmitriy555 [2]

The use Major changes in mineral deposits in the rock layers.

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3 years ago

A 1-kilogram mass is attached to a spring whose constant is 21 N/m, and the entire system is then submerged in a liquid that imp

amm1812

Answer:

the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Explanation:

Given that,

mass, m = 1kg

spring constant k = 21N/M

damping force = $-\beta\frac{dx}{dt} = \frac{-10dx}{dt}$

$\beta = 10$

By Newtons second law ,

The diffrential equation of motion with damping is given by

$m\frac{d^2x}{dt^2} = -kx-\beta\frac{dx}{dt}$

substitute the value of m =1kg, k = 21N/M, and $\beta = 10$

$1\frac{d^2x}{dt^2} = -21x=10\frac{dx}{dt}$

$\frac{d^2x}{dt^2} + 10\frac{dx}{dt} + 21x = 0$

suppose the equation of the form $x =e^m^t$ ,

and the auxilliary equation is given by

$m^2 + 10m + 21 = 0\\\\m^2 + 7m+3m+21=0\\\\m(m+7)+3(m+7)=0\\\\(m+7)(m+3)=0\\\\m=-7\\m=-3$

The general solution for the above differential equation is

$x(t) =C_1e^{-3t}+C_2e^{-7t}$

Derivate with respect to t

$x'(t)=-3C_1e^{-3t}-7C_2e^{-7t}$

(a)

since time is 0 then mass is one meter below

so x(0) = 1

Also it start from rest , that implies , velocity is 0 and time is 0

$x'(0) = 0$

substitute the initial condition

$C_1 +C_2 = 1$

$-3C_1-7C_2=0$

Solve the above equation to get C₁ and C₂

$C_1 =\frac{7}{4} and C_2 = -\frac{3}{4}$

substitute for C₁ and C₂ in general solution

$x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Thus the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

3 0

3 years ago

Read 2 more answers

A field measuring 12 meters by 16 meters is to have a brick paver walkway installed all around it, increasing the total area to

kolezko [41]

Answer:

1.5 m

Explanation:

Length. L = 12 m

Width, W = 16 m

Area, A = 12 x 16 = 192 m^2

Let the width of pavement be d.

The new length, L' = 12 + 2d

the new width, W' = 16 + 2d

New Area, A' = L' x W' = (12 + 2d)(16 + 2d) = 192 + 56 d + 4d^2

Difference in area = A' - A

285 = 192 + 56 d + 4d^2 - 192

93 = 56 d + 4d^2

4d^2 + 56 d - 93 = 0

$d = \frac{-56\pm \sqrt{56^{2}+4\times 4\times 93}}{8}$

$d=\frac{-56\pm 87.72}{8}$ \

d = 1.5 m

Thus, the width of the pavement is 1.5 m.

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3 years ago

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Softa [21]

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2 years ago