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eimsori [14]
3 years ago
10

Which three statements describe electromagnetic waves?

Physics
2 answers:
meriva3 years ago
7 0

Answer:

I think option A ,B,D are correct

o-na [289]3 years ago
3 0

Answer:

It is most likely option A B and C

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There are___<br> periods included in the periodic table.
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A sample of chloroform is found to contain 12.0 g of carbon, 106.4 g of chlorine, and 1.01 g of hydrogen. If a second sample of
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Given:

Sample 1:

Chloroform is CHCl_{3}

12 g Carbon

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Sample 2:

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3 years ago
A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that
Marat540 [252]

Answer:

Part(a): The value of the spring constant is 3.11 \times 10^{2}~Kg~s^{-2}.

Part(b): The work done by the variable force that stretches the collagen is 1.5 \times 10^{-6}~J.

Explanation:

Part(a):

If 'k' be the force constant and if due the application of a force 'F' on the collagen '\Delta l' be it's increase in length, then from Hook's law

F = k~\Delta l....................................................................(I)

Also, Young's modulus of a material is given by

Y = \dfrac{F/A}{\Delta l/l}...............................................................(II)

where 'A' is the area of the material and 'l' is the length.

Comparing equation (I) and (II) we can write

&& Y = \dfrac{l~k}{A}\\&or,& k = \dfrac{Y~A}{l}\\&or,& k = \dfrac{Y~(\pi~r^{2})}{l}

Here, we have to consider only the circular surface of the collagen as force is applied only perpendicular to this surface.

Substituting the given values in equation (III), we have

k = \dfrac{3.10 \times 10^{6}~N~m^{-2} \times \pi \times (0.00093)^{2}~m^{2}}{.027~m} = 3.11 \times 10^{2}~Kg~s^{-2}

Part(b):

We know the amount of work done (W) on the collagen is stored as a potential energy (U) within it. Now, the amount of work done by the variable force that stretches the collagen can be written as

W = \dfrac{1}{2}k~x^{2} = \dfrac{(\dfrac{F}{k})^{2}k}{2} = \dfrac{F^{2}}{2~k}...................................(IV)

Substituting all the values, we can write

W = \dfrac{(3.06 \times 10^{-2})^{2}~N^{2}}{2 \times 3.11 \times 10^{2}~Kg~s^{-2}} = 1.5 \times 10^{-6}~J

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