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Mrrafil [7]
3 years ago
10

Tendons. Tendons are strong elastic fibers that attach muscles to bones. To a reasonable approximation, they obey Hooke’s law. I

n laboratory tests on a particular tendon, it was found that, when a 250-g object was hung from it, the tendon stretched 1.23 cm. (a) Find the force constant of this tendon in N>m. (b) Because of its thickness, the maximum tension this tendon can support without rupturing is 138 N. By how much can the tendon stretch without rupturing, and how much energy is stored in it at that point?
Physics
1 answer:
Zolol [24]3 years ago
7 0

Answer:

Explanation:

Given

mass of object(m)=250 gm

Stretching(x)=1.23 cm

Suppose force constant to be k

then kx=mg

k\times 1.23\times 10^{-2}=250\times 10^{-3}\times 9.8

k=\frac{9.8\times 100}{4\times 1.23}

k=199.18 N/m

(b)Maximum Tension is 138 N

138=kx

x=\frac{138}{199.18}=0.692\approx 69.2 cm

Energy stored=\frac{kx^2}{2}=\frac{199.18\times (0.692)^2}{2}=47.80 J

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2 years ago
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
alexandr1967 [171]

Answer:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

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k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

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e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

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m_e is the mass of the electron

Solving for v, we find

v=\sqrt{k\frac{e^2}{m_e r}}

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

r=5.3\cdot 10^{-11}m

while the electron mass is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

Substituting into the formula, we find

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

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