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Temka [501]
4 years ago
9

What does electrons and the nucleus have in common?

Physics
1 answer:
timama [110]4 years ago
8 0
<span>All atoms of the same element have the same number of protons in the nucleus and consequently have the same atomic number. All atoms of the same neutral element have the same number of electrons as well.</span>
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Fill up the blank :- Weight is measured as the downward force of __________.
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A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that
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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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