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frozen [14]
3 years ago
6

Water cycle

Physics
1 answer:
tatuchka [14]3 years ago
8 0
Type into google: " water cycle" and this comes up.... put it in to your own words. This is very easy.


the cycle of processes by which water circulates between the earth's oceans, atmosphere, and land, involving precipitation as rain and snow, drainage in streams and rivers, and return to the atmosphere by evaporation and transpiration.
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Answer:

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

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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

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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}

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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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