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This is the one-mole, or equivalently molar volume, form of the Van der Waals equation, an approximate equation of state for a real gas:Here $p$ is the gas pressure, $T$ its absolute temperature, $R$ the molar gas constant, and $V$ the volume occupied by one mole. If $V$ denotes molar volume explicitly, its units are ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$.

The positive parameter $b$ represents excluded volume arising from the finite size and short-range repulsion of the molecules. The volume available for translational motion is reduced from $V$ to $V-b$. This is an effective excluded volume, not simply the sum of the molecules' geometric volumes. Its molar units are ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$.

The positive parameter $a$ measures intermolecular attraction in a mean-field approximation. Attractions reduce the force exerted on the walls, lowering the measured pressure by $a/V^2$ relative to the repulsion-only expression. Thus $p+a/V^2$ is the pressure corrected for this attraction. The squared inverse-volume dependence reflects a contribution proportional to the square of the molecular number density. The molar units of $a$ are $\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$.

For an arbitrary amount $n$ measured in moles, with total container volume $V_{\mathrm{tot}}$, the corresponding formula isAt low density, when $b\ll V$ and $a/V^2\ll RT/V$, it reduces to the ideal gas law. The model captures qualitative attraction, repulsion and liquid-gas behavior, but does not supply an exact equation of state for every material.

Electromagnetic waves drive mobile charges in a conducting medium. Their current changes both the phase and attenuation of the wave. A description in terms of a real, frequency-independent refractive index is therefore generally insufficient: the material response and the propagation constant are complex and frequency dependent.

Use phasors proportional to $e^{i(kz-\omega t)}$, with $k=k^{\prime }+i{k}^{\prime \prime }$ and $k^{\prime \prime }\ge 0$ for a passive medium. In a homogeneous, isotropic, linear material, let ${ϵ}_b$ represent the permittivity associated with bound-charge electric polarization, $\mu$ the magnetic permeability and $\sigma (\omega )$ the mobile-charge conductivity. The macroscopic Maxwell equations giveFor a transverse plane wave, taking another curl yieldsThis division of the response avoids counting mobile electrons both in ${ϵ}_b$ and in $\sigma$. Define the complex refractive index $\tilde{n}=ck/\omega =n+i\kappa$. For a nonmagnetic medium, ${\tilde{n}}^2={ϵ}_{\mathrm{eff}}/{ϵ}_0$. Its real part determines the phase variation, while the positive optical extinction coefficient $\kappa$ gives amplitude attenuation $e^{-\omega \kappa z/c}$ and intensity attenuation $e^{-2\omega \kappa z/c}$. The amplitude penetration length is

A simple microscopic explanation comes from the Drude model. Let electrons of charge $-e$, mass $m$ and number density $n_e$ move against an effectively stationary positive background. Collisions relax their mean velocity at rate $\nu$:ConsequentlyWith a vacuum-like bound-charge background, this givesIts imaginary part is positive for positive $\omega ,\nu$, consistent with dissipation under the chosen phasor convention. Real materials can require an effective electron mass, a nontrivial bound-charge response, or additional absorption associated with the electronic band structure.

A plasma is a conducting ionized medium with mobile electrons and ions, approximately neutral on large scales because of plasma screening. For the electron-frequency transverse waves considered here, the heavy ions can often be treated as stationary. In a dilute weakly collisional plasma, $\nu \ll \omega$, soFor $\omega >{\omega }_p$, $k$ is real in this collisionless model and the wave propagates. Its phase velocity is $v_{\mathrm{ph}}=c/n>c$, while its group velocity is $v_g=cn<c$; their product is $c^2$. A phase crest exceeding $c$ does not imply superluminal information transport. For $\omega <{\omega }_p$, the refractive-index square is negative andThe transmitted field is evanescent rather than a traveling bulk wave. A sufficiently thick collisionless plasma reflects such radiation without Ohmic absorption. Collisions make the response complex and permit absorption. This accounts for a plasma cutoff and, qualitatively, the reflection of some radio frequencies by ionized atmospheric layers. The simple model describes transverse waves; plasmas also support longitudinal collective modes.

In a metal, positive ions usually form a crystal lattice. Mobile conduction electrons occupy states described by the electronic band structure and normally form a degenerate Fermi gas. The conduction-electron density is usually much higher than in a dilute gas plasma. Thus its plasma frequency is commonly in the optical or ultraviolet range, rather than the lower radio-frequency values possible in dilute plasmas. Electron scattering from the lattice, defects and other excitations gives a finite relaxation rate. The distinction is not that every plasma is collisionless and every metal collisional: the controlling ratio for either material is $\omega /\nu$, together with its carrier density and bound-charge response.

At sufficiently low frequency, $\omega \ll \nu$, the metal's conductivity is approximately real, ${\sigma }_0=n_e{e}^2/(m\nu )$. In a good conductor the conduction current dominates the displacement current, ${\sigma }_0\gg \omega {ϵ}_b$. The wave equation then givesFor example, a tangential electric field varies asThe current $\mathbf{J}={\sigma }_0\mathbf{E}$ is therefore concentrated within a few skin depths of the surface: this is the skin effect. As frequency increases within this regime, $\delta \propto {\omega }^{-1/2}$. In a wire whose radius greatly exceeds $\delta$, only a thin outer region carries appreciable alternating current, increasing the effective resistance. In a conducting shield, the same penetration suppresses the field transmitted through a sufficiently thick wall.

The Poynting theorem identifies the loss as Joule heating. The cycle-averaged heating density for a complex conductivity isThe wave impedance is $Z=\mu \omega /k$. For incidence from vacuum at normal incidence, the electric reflection amplitude is $r=(Z-Z_0)/(Z+Z_0)$, where $Z_0=\sqrt{{\mu }_0/{ϵ}_0}$. A good conductor has small $|Z|$ and therefore strong reflection, while the part of the incident power entering it is dissipated over the skin depth.

At high enough frequency that $\omega \gg \nu$, a metal's free-electron response also approaches the collisionless plasma form. The low-frequency formula $\delta \propto {\omega }^{-1/2}$ must not simply be extrapolated into that regime. Real optical behavior additionally depends on the electronic band structure and transitions between bands. Both plasmas and metals are therefore described by the same electromagnetic response framework, but their relevant densities, scattering rates and frequency regimes differ.

Use the pipe-diameter convention for the Reynolds number, with mean velocity $\bar{v}$, density $\rho$ and dynamic viscosity $\mu$:The Reynolds number compares inertial and viscous effects in the fluid flow. A conventional estimate for the onset of turbulence in a pipe is ${\mathrm{Re}}_{\mathrm{crit}}\simeq 2000$, givingHence the usual onset estimate is $\boxed{D\gtrsim 1.6\mathrm{mm}}$. Using the also common transition criterion ${\mathrm{Re}}_{\mathrm{crit}}=2300$ gives $1.84\mathrm{mm}$, whereas $\mathrm{Re}\gtrsim 4000$, used for established turbulence, gives $D\gtrsim 3.2\mathrm{mm}$. There is no universal sharp minimum: inlet disturbances and pipe conditions affect the transition from laminar flow, so the threshold convention must accompany the answer.

Substitution into the wave equation gives ${\partial }_x^2\psi +{\partial }_y^2\psi =-(k_x^2+k_y^2)\psi$ and ${\partial }_t^2\psi =-{\omega }^2\psi$. Therefore ${\omega }^2=c^2(k_x^2+k_y^2)$. The fixed edges at zero are automatically satisfied by the sine factors; the opposite edges requireNeither integer may vanish, because that would make the entire displacement zero.

The element's mass is $\rho dxdy$, and its speed is $-\omega A\sin (k_{x}x)\sin (k_{y}y)\sin (\omega t)$. Squaring and multiplying by half its mass gives the requested energy density. The two sine-squared spatial integrals are $a/2$ and $b/2$, soThere is no dissipation. The total mechanical energy equals the maximum kinetic energy, givingEquivalently, integrate the small-slope tension energy $\rho c^2|\nabla \psi {|}^2/2$ and use the dispersion relation.

For $b=4a/5$, the two lowest modes are $(p,q)=(1,1)$ and $(2,1)$:The fundamental has one antinode at $(a/2,b/2)$ and no interior nodal line. The next mode has a node along $x=a/2$ and opposite-sign antinodes at $(a/4,b/2)$ and $(3a/4,b/2)$. Every boundary is a node.

The square fundamental has $f=c\sqrt{2}/(2\beta a)$. Equating it to $f_{11}$ givesAt the same peak displacement amplitude and frequency, time-averaged kinetic energy is proportional to membrane area. ThereforeThe reciprocal ratio is $41/40$ if the rectangle is placed first.

A gravitational field is force per unit test mass, $\mathbf{g}=\mathbf{F}/m$. Newton's law of universal gravitation gives the field of a point mass M:The potential is potential energy per unit mass, with zero at infinity for an isolated mass. Fields and potentials superpose. The potential difference fixes the work per unit mass, and a conservative gravitational field has zero circulation around a closed path.

For a continuous mass distribution, the gravitational Gauss law is $\oint \mathbf{g}\cdot d\mathbf{S}=-4\pi G{M}_{\mathrm{enc}}$. A spherical body therefore has the same exterior field as a point mass at its centre. Inside a uniform sphere of density $\rho$,The field is continuous at its surface and vanishes at its centre. An empty spherical shell has zero interior field but a nonzero constant interior potential. Near a small region of Earth's surface one may approximate $\mathbf{g}$ as uniform and the potential change as $g\Delta h$. Farther away its spatial gradient produces tidal relative accelerations. Escape to infinity from radius R requires speed $\sqrt{2GM/R}$ when rotation and other bodies are neglected.

In vacuum the electric and magnetic field energy densities areFor example, charging a capacitor requires $U={\int }_0^Q(q/C)dq=Q^2/(2C)=C{V}^2/2$. A parallel-plate capacitor with $C={ϵ}_{0}A/d$ and $E=V/d$ then gives $U={ϵ}_0{E}^{2}Ad/2$, identifying the electric field energy density. Similarly an inductor stores $L{I}^2/2$; applying this to a long solenoid yields $B^2/(2{\mu }_0)$ times its interior volume.

Electromagnetic energy is transported by the Poynting vector $\mathbf{S}=\mathbf{E}\times \mathbf{B}/{\mu }_0$. Poynting's theorem,expresses local energy conservation and transfer to matter. In a vacuum plane wave $E=cB$, so the two energy densities are equal and the energy flux has magnitude $c(u_E+u_B)$. For nondispersive linear media the corresponding reversible expressions are $\mathbf{E}\cdot \mathbf{D}/2$ and $\mathbf{B}\cdot \mathbf{H}/2$; they must not be applied unchanged to arbitrary dispersive or hysteretic materials.

An inverse-square field spreads through area proportional to squared distance. More generally the flux contribution of a point source is proportional to the solid angle subtended by a surface element; the oriented solid angles sum to $4\pi$ for an enclosed source and zero for an exterior one. Superposition extends this to a mass distribution.

Newton's law of universal gravitation gives $\mathbf{g}=-GM\hat{\mathbf{r}}/r^2$ for a point mass. A surrounding sphere therefore has flux $-4\pi GM$, soFor the tunnel, approximate its water as an infinite uniform cylinder with $R=3.5\mathrm{m}$ and density $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$. The bob is at radial distance $r=R+1.5\mathrm{m}=5.0\mathrm{m}$. A coaxial Gaussian surface givesFor the plumb-line deflection from cylindrical gravity, horizontal equilibrium gives $d/L\simeq |g_r|/g$. Draining removes the attraction toward the tunnel, so the bob shifts away byThe unchanged rock and ordinary gravitational field cancel in the before/after comparison; end effects and mechanical disturbances are neglected.

For the tidal water loading of a spring balance, approximate the extra water at high tide as an infinite plane slab of thickness $h=8\mathrm{m}$, with the suspended mass above it. The gravitational field of an infinite plane sheet, derived with a pillbox Gaussian surface, gives added downward field $\Delta g=2\pi G\rho h$. With initial spring extension $x_0=mg/k=1\mathrm{m}$,The extension increases at high tide. These estimates use $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and the stated water-density approximation; salinity, finite estuary geometry and terrestrial tidal deformation alter a real measurement.

Using metres and SI viscosity, the pressure difference is $\Delta P=200$ Pa and the central flow estimate isIf the quoted uncertainties are independent standard uncertainties, the pressure-difference uncertainty is ${\sigma }_{\Delta P}=\sqrt{10^2+10^2}=14.14$ Pa. First-order propagation givesHenceThe radius sensitivity of Poiseuille-flow uncertainty comes from the fourth-power dependence on radius. If the quoted errors instead mean independent worst-case bounds, the first-order absolute fractional bound is $4(0.1/2.3)+0.3/8.9+0.01/0.50+20/200=0.3276$, giving approximately $\pm 1.62\times 10^{-6}$ m³/s. That conservative bound and the standard uncertainty result are different conventions, not interchangeable errors. The supplied flow model presumes a Newtonian fluid and laminar flow that is fully developed; no density is given for an independent Reynolds-number check.

An impulse is $\mathbf{J}=\int \mathbf{F}dt$, giving $\Delta \mathbf{P}=\mathbf{J}$. For a short localized impulse at displacement $\mathbf{r}$ from the centre of mass, the angular impulse is $\Delta {\mathbf{L}}_{\mathrm{CM}}=\mathbf{r}\times \mathbf{J}$; for a changing lever arm use its instantaneous value inside the integral. For a uniform solid sphere with volume density ${\rho }_V$, direct integration about the $z$ axis givesTake forward translation $u$ and forward-rolling clockwise spin $\Omega$ as positive. Neglect the table's tangential impulse during the much shorter cue impact. Immediately afterwards,The bottom contact point has slip velocity $s_0=u_0-a{\Omega }_0$. The cue height for immediate rolling is therefore $d=2a/5$.

For the impulse-driven slipping-to-rolling transition of a sphere at an arbitrary admissible impact height, let $J_f$ be the signed friction impulse accumulated before slipping ends. Translation and spin then obey $u_f=u_0+J_f/M$ and ${\Omega }_f={\Omega }_0-a{J}_f/I$. Impose $u_f=a{\Omega }_f$ and eliminate $J_f$:This expression handles both sliding directions. If $d<2a/5$, the contact point slips forward and friction slows the centre; if $d>2a/5$, excess spin makes it slip backward and friction accelerates the centre. friction still dissipates total mechanical energy. For ${\mu }_d>0$, $\dot{s}=-(7/2){\mu }_{d}g\mathrm{sgn}s$, so the slipping time is $2|s_0|/(7{\mu }_{d}g)$. Its duration depends on friction, while the final speed does not. With zero friction no slipping-to-rolling transition occurs unless rolling was present from the outset.

Far from its ends, approximate the body by an infinite uniform cylinder. By cylindrical symmetry, the Newtonian gravitational field is radial. Applying Gauss's law for gravity to a coaxial cylindrical surface of radius $s$ and length $\ell$ outside the material givesAt the surface, $R=50\mathrm{m}$, soThe gravitational acceleration points inward toward the axis. Its small magnitude reflects both the small size and the modest density. The infinite-cylinder approximation uses the specified distance from both ends; a spherical approximation would discard the relevant geometry. No rotational contribution is included, since no rotation rate is supplied.

The moment of inertia about a specified axis is $I=\int s^{2}dm$, where $s$ is perpendicular distance from the axis. The parallel axis theorem states $I=I_{\mathrm{cm}}+m{d}^2$ for two parallel axes separated by $d$, one passing through the centre of mass of a body of mass $m$.

It is easiest to calculate the eccentric disc by subtraction. Let $m_f$ be the mass of the unperforated disc and $m_h$ the removed mass. Uniform thickness and density give $m_h=m_f/4$ and $m_f-m_h=M$, hence $m_f=4M/3$ and $m_h=M/3$. The hole is tangent to the rim, so its centre, and therefore the shaft axis, lies $R/2$ from the original disc centre. About the shaft axis,The shaft contributes $I_{\mathrm{shaft}}=M(R/2{)}^2/2=M{R}^2/8$. ThusThe motor's torque is $P/\omega$ at nonzero angular velocity, while the resisting torque is $-\gamma \omega$. The rotational equation of motion and its equivalent energy balance areAt the steady state, input power equals dissipated power, givingFor the acceleration, use $u={\omega }^2$. Then $\dot{u}+(2\gamma /I)u=2P/I$ with $u(0)=0$, soThe approach to ${\omega }_0$ takes infinite time. This constant-power rotational acceleration idealizes the start: $P/\omega$ diverges at rest, although the energy balance has a well-defined limiting solution. A real motor has finite starting torque.

The central rod occupies the holes in the discs, so the disc masses must be annular rather than counted as solid discs overlapping the rod. Define $C=L_1{\rho }_1=2L_2{\rho }_2$. ThenThe axial moment of inertia isThus $I=\frac{1}{2}M{r}_2^2$, exactly the axial inertia of a uniform disc of outer radius $r_2$ and mass $M$. The cancellation uses the given density–length relation.

$V(x)$ is the particle's potential energy as a function of position, and $E$ is its total energy eigenvalue in this time-independent Schrödinger equation. Solving it with the boundary conditions determines the allowed stationary states and energy levels.

The wavefunction is a probability amplitude: the Born rule gives position probability density $|\psi (x){|}^2$, normalized to unit integral for one particle. $\psi$ itself need not be real or positive and is not a classical matter-density profile. The full stationary-state time dependence is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$, leaving $|\Psi {|}^2$ time independent.

The position probability density isand zero outside. It has zeros at $0,L/2,L$ and equal maxima $2/L$ at $L/4,3L/4$. The two lobes have equal integrated probability. The vertical axis is a density with units of inverse length, not a dimensionless probability at a point.

The density is symmetric about $L/2$: $p(L-x)=p(x)$. Denote the expectation by $\mu ={\int }_0^{L}xp(x)dx$. Substitution $u=L-x$ givesThus $⟨x⟩=L/2$. An expectation need not coincide with a high-density position: the centre is a node of this particular state, but the two equally weighted lobes balance about it.

For a symmetric finite well supporting this excited bound state, the wavefunction and its derivative match continuously at each finite potential step. Outside, where $V_{\mathrm{wall}}>E$, the solution decays exponentially withThe density therefore extends into the classically forbidden region instead of dropping to zero at the walls. Its interior lobes and normalization change, and the energy is lower than for the corresponding infinite-wall state at the same well bottom. Symmetry preserves the central node for the first excited state. The diagram shows an explicitly illustrative barrier height, not an extra numerical value supplied by the question.

The far fields obey $E_{\theta }=c{B}_{\varphi }$. The radial component of their Poynting vector is $E_{\theta }{B}_{\varphi }/{\mu }_0$. Averaging ${\ddot{p}}^2=p_0^2{\omega }^4{\cos }^2\omega (t-r/c)$ and integrating $\int {\sin }^2\theta d\Omega =8\pi /3$ givesFor a small isotropically polarizable particle, $p_0=\alpha E_0$ and incident intensity is $I={ϵ}_{0}c{E}_0^2/2$. Thus the total Rayleigh scattering cross-section isEach incident linear polarization gives the same integrated power; averaging unpolarized light does not add another factor of two. The scalar real polarizability approximation neglects absorption and molecular anisotropy.

For Rayleigh attenuation from refractivity in dilute air, polarization density is $\mathbf{P}=N\alpha \mathbf{E}$. Consequently ${ϵ}_r=1+N\alpha /{ϵ}_0$ and, with ${\mu }_r\simeq 1$ and $n-1\ll 1$,The ideal-gas number density at the supplied pressure and temperature is $N=P/(k_{B}T)=2.47\times 10^{25}{\mathrm{m}}^{-3}$, yielding $\alpha =2.15\times 10^{-40}\mathrm{C}{\mathrm{m}}^2{\mathrm{V}}^{-1}$. For an explicit representative visible wavelength $\lambda =550\mathrm{nm}$,Direct-beam intensity follows $I(x)=I(0)e^{-x/L_{\mathrm{att}}}$. Equivalently $L_{\mathrm{att}}=3N{\lambda }^4/[32{\pi }^3(n-1{)}^2]$. Its strong wavelength dependence gives roughly $21$–$200\mathrm{km}$ across $400$–$700\mathrm{nm}$ if refractivity is held fixed. This is a uniform-density dry-air scattering estimate, not absorption or transmission through a vertically stratified atmosphere; aerosols and molecular depolarization corrections are outside the given model.

Here $\mathbf{H}$ is magnetic field intensity, $\mathbf{J}$ free electric current density, and $\mathbf{D}$ electric displacement. In vacuum, $\mathbf{B}={\mu }_0\mathbf{H}$ and $\mathbf{D}={ϵ}_0\mathbf{E}$. Integrating the equation gives Ampère's circuital law with displacement current; in a steady state the displacement term vanishes.

Take $z$ upward and define $V={\varphi }_{\mathrm{upper}}-{\varphi }_{\mathrm{lower}}=V_0\sin \omega t$. Neglect fringing and use the low-frequency thin-capacitor approximation. ThenInside the gap, a circular Ampère loop encloses only displacement current:Because $\hat{\mathbf{z}}\times \hat{\mathbf{\varphi }}=-\hat{\mathbf{r}}$, the Poynting vector isIt points inward while the electrical energy is increasing and outward while it decreases. The net power entering the curved side isThis is the leading quasistatic energy balance. The time-averaged field energies areFor the axisymmetric high-frequency capacitor field when $\omega a/c$ is not small, induction and radial propagation prevent a spatially uniform $E_z$. In an ideal axially uniform axisymmetric interior model, Faraday's and Ampère's equations giveThe solution regular at the axis and normalized to the central voltage is approximatelyAt $\omega =\pi c/a$, the radial profile is $J_0(\pi r/a)$: its first zero is at $r\simeq 0.765a$, and the edge has opposite phase, $J_0(\pi )\simeq -0.304$. This is an illustrative thin-disc interior profile; the exact driven high-frequency field also depends on the feed and edge/radiation boundary conditions. Continuing the uniform quasistatic field would be inconsistent.

For the potential of an inverse-radius spherical charge density, assume there is no additional point charge at the origin and choose $\varphi (\infty )=0$. The enclosed charge is $Q(r)=4\pi A{\int }_0^r{r}^{\prime }d{r}^{\prime }=2\pi A{r}^2$ for $r<a$, and $Q(a)=2\pi A{a}^2$ outside. Gauss's law then givesIntegrating the field inward from infinity,The potential and its first radial derivative are continuous at $a$, since there is no surface charge there. The integrable density singularity has a finite potential limit $\varphi (0^{+})=Aa/{ϵ}_0$.

For the focal length of a rotating liquid mirror, consider steady rigid rotation at angular frequency $\Omega =2\pi \nu$. The free surface is an equipotential of $gz-{\Omega }^2{r}^2/2$. HenceA paraboloidal mirror has $z-z_0=r^2/(4f)$, givingThe numerical estimate uses the conventional standard acceleration $g_n=9.80665\mathrm{m}{\mathrm{s}}^{-2}$, rather than an asserted local measurement or a recovered handbook entry.^[ref] The mercury density cancels; the ideal surface shape follows from gravity and rotation.