How to do 4th question?

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How to do 4th question?

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Answer 1 · 2018-03-15T09:44:50

See below.

Explanation:

Period is given by

$T \approx 2 π \sqrt{\frac{L}{g}}$ $T approx 2pi sqrt(L/g)$

now taking natural log to both sides

$ln T = ln (2 π) + \frac{1}{2} (ln L - ln g)$ $lnT = ln(2pi)+1/2 (ln L - ln g)$

assuming $g$ $g$ constant and deriving

$\frac{δ T}{T} = \frac{1}{2} \frac{δ L}{L}$ $(deltaT)/T = 1/2 (deltaL)/L$

now assuming an isotropic material the volumetric thermal expansion coefficient is three times the linear coefficient: or

$α_{V} = 3 α_{L}$ $alpha_V = 3 alpha_L$ and then

$δ L = α_{L} (40^{\circ} - 20^{\circ}) = \frac{1}{3} α_{V} (40^{\circ} - 20^{\circ}) = \frac{1}{2} (\frac{36}{3}) \times 20 \times 10^{- 6}$ $delta L = alpha_L (40^@-20^@) = 1/3 alpha_V(40^@-20^@) =1/2(36/3) xx 20 xx10^-6$ [s]

and finally

$\frac{δ T}{T} \approx 120 \times 10^{- 6}$ $(delta T)/T approx 120 xx 10^-6$ [s/s]

per day it gives

$δ T = 24 \times 60 \times 60 \times 120 \times 10^{- 6} = 10.38$ $delta T = 24 xx 60 xx 60 xx 120 xx 10^-6 = 10.38$ [s] delay/day.

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How to do 4th question?

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