Question

\(\theta = s(\frac{3}{4} - \frac{1}{8} - \frac{3}{4} - \frac{1}{28} + \frac{3}{10}) = 0.139 \delta\\ \tau_w = \varphi \upsilon_o^2 \frac{d\theta}{dx}\\ \delta = 4.64 \sqrt{\frac{2\gamma x}{\upsilon_o}}\\ \mu \frac{3}{2} \frac{\upsilon_o}{s} = \rho \upsilon_o^2 \cdot 0.139 \frac{d\delta}{dx}\\ \delta d\delta = \frac{3}{2(0.139)} \upsilon_o^2 \frac{d\gamma}{dx}\\ \frac{\delta^2}{2} = \frac{3}{2(0.139)} \frac{\gamma x}{\upsilon_o} + c\)

          \(\theta = s(\frac{3}{4} - \frac{1}{8} - \frac{3}{4} - \frac{1}{28} + \frac{3}{10}) = 0.139 \delta\\ \tau_w = \varphi \upsilon_o^2 \frac{d\theta}{dx}\\ \delta = 4.64 \sqrt{\frac{2\gamma x}{\upsilon_o}}\\ \mu \frac{3}{2} \frac{\upsilon_o}{s} = \rho \upsilon_o^2 \cdot 0.139 \frac{d\delta}{dx}\\ \delta d\delta = \frac{3}{2(0.139)} \upsilon_o^2 \frac{d\gamma}{dx}\\ \frac{\delta^2}{2} = \frac{3}{2(0.139)} \frac{\gamma x}{\upsilon_o} + c\)

θ = s((3)/(4) - (1)/(8) - (3)/(4) - (1)/(28) + (3)/(10)) = 0.139 δ
= φ^2 (dθ)/(dx)
δ = 4.64 √((2γ x)/())
μ(3)/(2)()/(s) = ρ^2 · 0.139 (dδ)/(dx)
δ dδ = (3)/(2(0.139))^2 (dγ)/(dx)
(δ^2)/(2) = (3)/(2(0.139))(γ x)/() + c

Added by Irene C.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

06/10/2023

Step 1

In this case, it seems like the equation involving the delta variable is on the right side of the notes. Show more…

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I need help with these notes. I need to understand how they isolated the delta variable and evaluated to get that delta solution. Everything I do does not make sense mathematically, and I get a different delta solution than what they get. I'm talking abt the right side equations. I don't even think I understand how they separated the equation like this, are they doing whatever they can to get the delta and d(delta) terms together? idk this is very confusing. 3 4 + 0.134.8 z8 Vo MN 0=8( PU O139 JX &x XX S=4.64 3 z(tl39) Db ME 326