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Elongation $\Delta L = \frac{FL}{AY} = \frac{FL}{\pi(d/2)^2 Y} \propto \frac{L}{d^2}$.
- New elongation $\Delta L' \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{1}{2} \Delta L$
- $\Delta L' = \frac{1}{2} \times 0.04 = 0.02\text{ m}$.
New radius \(= 1.1r\), so new surface area \(= 4\pi(1.1r)^2 = 4\pi \times 1.21r^2\).
Increase \(= 21\%\).
Who wrote 'The Prince', a work focused on political realism and power?
Niccolò Machiavelli wrote 'The Prince' as a guide for rulers to maintain power.
Reduction: $Mn^{2+} + 2e^{-} \rightarrow Mn$ ($E^{0}_{red} = -1.18\text{ V}$)
Oxidation: $2Mn^{2+} \rightarrow 2Mn^{3+} + 2e^{-}$ ($E^{0}_{ox} = -1.51\text{ V}$)
$E^{0}_{cell} = E^{0}_{cathode} - E^{0}_{anode} = -1.18 - 1.51 = -2.69\text{ V}$. Since $E^{0}_{cell}$ is negative, the reaction is non-spontaneous.
Identify the curve from the family defined by $(x^2-y^2)\,dx + 2xy\,dy = 0$ that passes through the point $(1,1)$.
Rewrite: $\frac{dy}{dx} = \frac{y^2-x^2}{2xy}$. This is a homogeneous ODE. Let $y=vx$:
$\frac{1-v^2}{2v}dx + x\,dv/... $
After solving, the general solution is $x^2 + y^2 = Cy$ (circles with centres on the $y$-axis).
Substituting $(1,1)$: $1+1=C\cdot1 \Rightarrow C=2$, giving $x^2+y^2=2y$, i.e. $x^2+(y-1)^2=1$.
This is a circle centred on the $y$-axis at $(0,1)$.
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