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aliina [53]
3 years ago
10

The most distal part of a nail is called the ____________ . The most proximal part of the nail that is embedded in the skin is t

he ____________ . The layer of edpidermis that is under the nail body is called the ____________ . The actively growing part of the nail is the ____________ . The whitish semilunar area of the proximal end of the nail body is the ____________ .
Physics
1 answer:
ankoles [38]3 years ago
7 0

Answer:

1.Free edge

2.Nail Root

3.Nail matrix

4.lunula

Note: The numbering coincides respectively with the dashes in the questions.

Explanation:

Free edge is the distal white nail ending.

Nail Root is the proximal part of the nail embedded in the skin.

Nail matrix is the actively growing part of the nail were the nail root thickens

Lunula the whitish semilunar area of the proximal end of the nail body it appears whitish because stratum basala obscures. the underlying blood vessels.

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(a) Consider the initial-value problem dA/dt = kA, A(0) = A0 as the model for the decay of a radioactive substance. Show that, i
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Answer:

a) t = -\frac{ln(2)}{k}

b) See the proof below

A(t) = A_o 2^{-\frac{t}{T}}

c) t = 3T \frac{ln(2)}{ln(2)}= 3T

Explanation:

Part a

For this case we have the following differential equation:

\frac{dA}{dt}= kA

With the initial condition A(0) = A_o

We can rewrite the differential equation like this:

\frac{dA}{A} =k dt

And if we integrate both sides we got:

ln |A|= kt + c_1

Where c_1 is a constant. If we apply exponential for both sides we got:

A = e^{kt} e^c = C e^{kt}

Using the initial condition A(0) = A_o we got:

A_o = C

So then our solution for the differential equation is given by:

A(t) = A_o e^{kt}

For the half life we know that we need to find the value of t for where we have A(t) = \frac{1}{2} A_o if we use this condition we have:

\frac{1}{2} A_o = A_o e^{kt}

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Applying natural log we have this:

ln (\frac{1}{2}) = kt

And then the value of t would be:

t = \frac{ln (1/2)}{k}

And using the fact that ln(1/2) = -ln(2) we have this:

t = -\frac{ln(2)}{k}

Part b

For this case we need to show that the solution on part a can be written as:

A(t) = A_o 2^{-t/T}

For this case we have the following model:

A(t) = A_o e^{kt}

If we replace the value of k obtained from part a we got:

k = -\frac{ln(2)}{T}

A(t) = A_o e^{-\frac{ln(2)}{T} t}

And we can rewrite this expression like this:

A(t) = A_o e^{ln(2) (-\frac{t}{T})}

And we can cancel the exponential with the natural log and we have this:

A(t) = A_o 2^{-\frac{t}{T}}

Part c

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So we can use the following equation:

\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}

Simplifying we got:

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t = T \frac{ln(8)}{ln(2)}

We can rewrite this expression like this:

t = T \frac{ln(2^3)}{ln(2)}

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t = 3T \frac{ln(2)}{ln(2)}= 3T

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