. . . temperature?¹

. This is where $\sigma(U)$ has a maximum, and therefore zero slope. The inverse of zero is $\pm \infty$. There is no difference between $\tau = +\infty$ and $\tau = -\infty$.

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. . . ``hottest''?²

At (n=N). The slope of $\sigma(U)$ has its steepest negative value there, meaning that entropy is gained by losing U.

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. . . . ³

At that U the temperature $\tau$ has its smallest negative value.

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. . . answers]⁴

(i) ( $U = {3\over3} N \tau$)

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. . . molecules?⁵

The average velocity is proportional to the square root of temperature and to the inverse square root of the mass of the particle, so the ratio of average velocities of He and O₂ is the square root of the ratio of the mass of an oxygen molecule to that of a helium atom: ${\displaystyle \sqrt{m_{{\rm O}_2} \over m_{\rm He}} = \sqrt{32/4} = \sqrt{8} = }$ .

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. . . Explain. ⁶

The gravitational and electrostatic forces have the same dependence on the distance between the electron and proton, so they will never be equal. Their ratio will always be $2.3 \times 10^{39}$.

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. . . Explain. ⁷

. There is no net charge inside the sphere, so by Gauss' Law there is no net flux of $\vec{\mbox{\boldmath$E$\unboldmath }}$ out through the closed surface. If the net flux is zero, its average over the surface is too.

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. . . sentences:⁸

 
The electric field inside the slab is zero.
The electric field outside but very close to the surface of the slab is normal (perpendicular) to the surface of the slab.
The electric field outside but very close to the surface of the slab has a magnitude $E = Q/2\epsilon_0 A$.
The electric field at a distance $\sqrt{A}$ from the surface of the slab cannot be determined from the information given.
 

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. . . inert.⁹

The gravitational potential energy of a buckyball is $\varepsilon(h) = mgh$ where h is the altitude above sea level. If our ``system'' is just the altitude of the buckyball, then in thermal equilibrium the BOLTZMANN DISTRIBUTION implies that the probability of finding one buckyball at height h is proportional to $\exp(-\varepsilon/\tau)$ where $\tau \equiv k_{_{\rm B}} T$. Since all buckyballs share the same probability distribution, the density D(h) of buckyballs has this same distribution. We don't need to know the normalization to find the ratio $D(h)/D(0) = \exp(-mgh/k_{_{\rm B}}T)$. This ratio will be e^-1 when $mgh=k_{_{\rm B}}T$ or when $h=k_{_{\rm B}}T/mg$. Plugging in the numbers gives  .

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. . . )?¹⁰

as for any isotropic charge distribution viewed from outside.

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. . . $r \le R$)?¹¹

Here it is a little more complicated. First we need to know the constant of proportionality a in $\rho(r) = a \, r$: since ${\displaystyle Q = \int_0^R \rho(r) \times 4\pi r^2 dr = 4\pi a \int_0^R r^3 dr = \pi a R^4 }$, we have ${\displaystyle a = {Q \over \pi R^4} }$. Now imagine a spherical Gaussian surface at r<R. The total charge within that closed surface is ${\displaystyle Q_{\rm encl} = \int_0^r \rho(r') \times 4\pi r'^2 dr' = 4\pi a \int_0^r r'^3 dr' = \pi a r^4 = {\not{\pi} Q r^4 \over \not{\pi} R^4} }$ and the integral of $\vec{\mbox{\boldmath$E$\unboldmath }} \cdot \hat{\mbox{\boldmath$n$\unboldmath }} \, dA$ over that surface is $4 \pi r^2 E$, so Gauss' Law gives ${\displaystyle 4 \pi r^2 \epsilon_0 E = {Q r^4 \over R^4} }$ or .

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