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Let $m_p=0.250\mathrm{kg}$ and let $m_w$ be the hanging mass. Taking up-slope and downward respectively as positive,Eliminate the common tension to obtainHere $\theta =10^{\circ }$, $\mu =0.1$ and $a=1.60\mathrm{m}{\mathrm{s}}^{-2}$. The pulley changes the direction of the string force; in the ideal model it does not change its magnitude.

Prefer molecular rotational spectroscopy or rovibrational spectroscopy, because nuclear masses enter their reduced mass directly. For an isotopologue with nuclear reduced mass ${\mu }_N$,Isotope substitution therefore produces appreciable shifts and distinct sets of lines, assuming the same electronic potential and approximately unchanged bond length and force constant. A suitable heteronuclear molecule can provide electric-dipole rotational/vibrational spectra; other molecules may require Raman spectroscopy.

Atomic electronic energies instead depend on the electron-nucleus reduced mass $m_{e}M/(m_e+M)$, whose fractional change is suppressed by $m_e/M$. High-resolution atomic spectroscopy can still detect isotope shifts, including nuclear-size and many-electron mass effects; it is not intrinsically incapable of identifying isotopes. Molecular spectra generally offer the stronger mass sensitivity for this purpose.

The transverse stiffness of a stretched spring follows because a spring of length $2l$ develops transverse force $T_0\Delta y/(2l)$ to first order. Its extension changes only at second order, so the leading transverse stiffness comes from its existing tension, not from that extension change. Here $T_0/(2l)=k/2$, givingThe mode vectors are again $(1,1)$ and $(1,-1)$. Treating the springs as unstretched would incorrectly remove this linear restoring force.

A normal mode is a small motion in which all coordinates oscillate at one common angular frequency with fixed relative amplitudes. For conservative linearized motion, these are eigenvectors of the mass-weighted stiffness matrix.

All three springs are initially stretched from $l$ to $2l$, giving equal tension $T_0=kl$ and zero resultant force on each mass. This prestress is essential for the transverse modes found below.

For the final mixed-axis trajectory of a prestressed spring pair, the initial displacements excite only the symmetric mode in each direction. Therefore, to first order in $ϵ/l$,The second mass has the same displacements about $4l$. Over the first transverse period $T_y=2\pi \sqrt{2m/k}$, write $s=t/T_y$:The trajectory is not closed after $T_y$, because the frequency ratio is $\sqrt{2}$. It ends at approximately $(-0.858,1)$ in these scaled coordinates rather than $(1,1)$.

For a normal from medium 1 to 2, the magnetic boundary conditions areWithout a free surface current, tangential $H$ is continuous. To obtain the magnetization of a permeable sphere of radius $R$ and permeability ${\mu }_0(1+\chi )$, choose magnetic scalar potentialsTangential continuity gives $H_i=H_0-C/R^3$, and normal $B$ continuity gives $(1+\chi )H_i=H_0+2C/R^3$. HenceThe fields are uniform inside. For negative susceptibility, magnetization is antiparallel to the applied field; the scalar formula retains its negative sign.

For a small weakly magnetic body, the induced magnetic energy is $U_m=-\chi V{B}^2/(2{\mu }_0)$. The factor one-half accounts for building the induced moment. The vertical force is $F_z=\chi VB{B}^{\prime }/{\mu }_0$. Balancing gravity therefore givesA diamagnet needs a field decreasing in magnitude with height for an upward force. For diamagnetic levitation in an inverse-height field,The force falls as $z^{-3}$, giving a restoring vertical derivative at equilibrium. Full lateral stability would require the transverse field geometry, which is not specified.

Use local east, north and upward axes. At the equator, Earth's angular velocity points north. To first order in its magnitude ${\Omega }_E$, the falling object has $v_z=-gt$, so the Coriolis acceleration $2\mathbf{v}\times {\mathbf{\Omega }}_E$ points east and has magnitude $2{\Omega }_{E}gt$. Integrating from zero initial eastward velocity and displacement givesFor $h=830\mathrm{m}$, $g\simeq 9.81\mathrm{m}{\mathrm{s}}^{-2}$ and ${\Omega }_E\simeq 7.29\times 10^{-5}{\mathrm{s}}^{-1}$,Local measured $g$ includes the stationary centrifugal correction. Corrections involving the acquired horizontal velocity or the height dependence of effective gravity are beyond this leading estimate. If the printed centrifugal term is read as $+\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{r})$, its sign is opposite the usual outward inertial force; it does not affect the retained first-order Coriolis result.

The transverse-wave impedance of a string is $Z=\sqrt{T\mu }$ for common tension $T$. The density ratio four gives $Z_2/Z_1=2$. Continuity of displacement and transverse force gives transmitted amplitude ratio $t=2Z_1/(Z_1+Z_2)=2/3$. The power transmission coefficient includes the impedance ratio:The reflected fraction is $1/9$, checking energy conservation. Squaring the displacement amplitude alone would give the wrong transmitted power.

In a central potential, $J=m{r}^2\dot{\theta }$ is conserved. Substituting it into $U=m({\dot{r}}^2+r^2{\dot{\theta }}^2)/2+V(r)$ givesWhile the string stays taut, $V=k(r-L{)}^2/2$. Circular force balance gives $m{r}_0{\omega }_0^2=k(r_0-L)$, hence ${\omega }_0^2=k(r_0-L)/(m{r}_0)$. A radial impulse changes no angular momentum. The radial oscillation of an elastic-string orbit follows by expanding the same fixed-$J$ effective potential about its minimum:soFor an outward impulse, one convenient linear approximation is $r=r_0+ϵ\sin \Omega t$ and $\theta ={\omega }_{0}t+[2{\omega }_0ϵ/(r_0\Omega )](\cos \Omega t-1)$. The angular modulation follows from $\dot{\theta }=J/(m{r}^2)$ and is needed for a consistent trajectory sketch.

Configurational entropy counts the arrangements compatible with a macroscopic configuration: $S=k_B\ln \Omega$. Let $x=L/(Na)$, $N_{\downarrow }=N(1+x)/2$ and $N_{\uparrow }=N(1-x)/2$. The multiplicity is $N!/(N_{\downarrow }!N_{\uparrow }!)$. Stirling's approximation therefore givesWith downward $L$ positive, the gravitational potential energy is $-mgL$. Minimizing the Helmholtz free energy $F=-mgL-TS$ gives $\mathrm{artanh}x=mga/(k_{B}T)$, henceRaising temperature reduces the extension: the entropic preference for many up/down arrangements opposes alignment by the load. For small loads, $L\simeq N{a}^{2}mg/(k_{B}T)$.

For the positive-feedback entropic chain with replacement force $f(L)=f_{0}L/a$, $f_0>0$, the conservative potential is $U(L)=-{\int }_0^{L}f(s)ds=-f_0{L}^2/(2a)$. Using $-f(L)L$ would overcount it by a factor two. The stationary equation becomesExpansion about $x=0$ givesFor $T>T_c$ the stable extension is zero. Below $T_c$, nonzero minima appear, with $x\simeq \pm \sqrt{3(T_c-T)/T}$ close to the transition and $|x|\to 1$ as $T\to 0$. Restricting to downward extensions selects the positive branch. The onset is continuous in this model, with a nonanalytic mean-field change at $T_c$; a finite jump in $L$ does not follow from the printed force law, despite the question's discontinuity wording. A finite chain has rounded fluctuations rather than a literal thermodynamic singularity.

This is the same self-consistency equation as the zero-field mean-field treatment of the Ising model: the segment orientation corresponds to a spin, extension to magnetization, and the extension-dependent force to the self-consistent molecular field. The competition between alignment energy and binary configurational entropy explains the shared temperature dependence.

In the $S_z$ basis $\alpha ,\beta$, $S_x=(\hslash /2)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Its normalized eigenstates are $(\alpha +\beta )/\sqrt{2}$ and $(\alpha -\beta )/\sqrt{2}$, with eigenvalues $+\hslash /2$ and $-\hslash /2$. Expanding the initial state in this basis and evolving the two phases gives, with $\Omega =\lambda B$,The sign of $\Omega$ fixes rotation sense; the probability is insensitive to that sign.

In the Stern-Gerlach experiment, a collimated atomic beam passes through a strongly nonuniform magnetic field. The force component $F_z\simeq {\mu }_z{\partial }_z{B}_z$ separates magnetic-moment projections spatially. Silver atoms provide an effective spin-half moment, yielding two resolved components rather than a continuous distribution of deflections. To test the rotation above, use a first analyzer and aperture to prepare $+z$, pass that selected beam through a controlled approximately uniform $x$-directed field, and analyze it with a second $z$-oriented Stern-Gerlach magnet. Varying field strength or interaction time changes the two output populations as the calculated cosine and sine squares. A narrow velocity distribution and controlled field-entry geometry preserve the contrast by reducing variations of interaction time and unwanted spin evolution.

Zero-point energy is the ground-state energy above the classical minimum, remaining even at zero temperature. For a harmonic oscillator it is $\hslash \omega /2$; simultaneous vanishing position and momentum fluctuations is forbidden by the uncertainty principle.

A concrete test is the Casimir interaction: changing conducting boundaries changes electromagnetic ground-state modes and produces a measurable force. Ideal parallel plates have pressureA torsion-pendulum experiment measured the attraction of closely spaced conducting surfaces, testing this boundary-dependent interaction.^[ref] Such measurements test changes in ground-state energy, not an arbitrary absolute additive vacuum energy; real-material and thermal corrections must be included when comparing precision data.

The method of images replaces conductors by fictitious charges outside the physical solution domain. Choose images so the real-plus-image potential obeys the specified conductor boundary values. The electrostatic uniqueness theorem then identifies that potential in the physical domain; fields inside the replaced conductor are not physical predictions.

Put the cylinder axis at the origin, real line at $(2a,0)$ and image at $(a/2,0)$. Let $r_1,r_2$ be distances from these lines. The exterior potential can be writtenOn the surface, $r_1^2=a^2(5-4\cos \theta )$ and $r_2^2=r_1^2/4$, so $\varphi =0$ everywhere as required. The additive constant is essential for a grounded cylinder; the two-dimensional potential need not also be zero at infinity.

The physical line feels the image field at separation $3a/2$. Its force per unit length is attractive:Measure $\theta$ from the radius toward the real line, as in the original diagram. The radial field just outside the surface isSince the interior field is zero,Integrating $a\sigma (\theta )$ around the circumference gives total induced charge $-\lambda$ per unit length.

Black-body radiation is the equilibrium electromagnetic radiation associated with a perfect absorber, whose spectrum is fixed by its temperature and the available modes. Photons can be emitted and absorbed, so their equilibrium chemical potential is zero. The model here retains two transverse polarizations in an effectively one-dimensional cavity; the resulting pressure is a one-dimensional mechanical force, not a three-dimensional pressure in pascals.

Newton's law of universal gravitation gives the force $-GmM\hat{\mathbf{r}}/r^2$ on a test mass a distance $r$ from a point source. Gauss's law for gravity statesThe inverse-square point field has the stated flux through a surrounding sphere. Superposition extends it to distributed sources; the divergence theorem relates the integral and local forms. Symmetry is still needed to turn a flux into a pointwise field value.

For the gravitational field of an infinite uniform cylinder, a coaxial cylindrical Gaussian surface has no end-cap flux and constant inward field on its curved surface. The enclosed mass per unit length is $\pi \rho r^2$ inside and $\pi \rho R^2$ outside. HenceIts magnitude rises linearly from zero, reaches $2\pi G\rho R$ at the surface, then decreases as $1/r$. The field is continuous at $R$, while its radial derivative changes because the density has a sharp edge.

Use signed distance $y$ along the diameter hole. A rod displaced by $x$ occupies $y\in [x-R,x+R]$, so its force is $F(x)=\lambda {\int }_{x-R}^{x+R}{g}_y(y)dy$. At equilibrium symmetry makes $F(0)=0$. Differentiating the moving endpoints givesThus the leading restoring force and oscillation frequency areFor an explicit positive-displacement check, put $C=2\pi G\rho$ and integrate the inside and outside pieces for $0<x<2R$:Its first-order term is $-2C\lambda Rx$, agreeing with the endpoint argument. The result is the diameter-rod oscillation in cylindrical gravity; it assumes a sufficiently long source cylinder and a narrow hole whose removed mass can be neglected.

The electric displacement field is $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}$. A homogeneous linear isotropic dielectric has $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$, so $\mathbf{D}={ϵ}_0{ϵ}_r\mathbf{E}$. At a free-charge-free interface the normal component of D is continuous, while electrostatics requires continuity of tangential E.

At half immersion the conductor centre lies in the flat air-liquid interface. Set $\varphi =C/r$ in both media, with potential zero at infinity. The sphere is an equipotential, and the potential is continuous across the liquid plane. On that plane the radial E field is tangential, so tangential E agrees while both normal D components vanish. Each half-space also satisfies the source-free potential equation. The fields outside the sphere areGauss's law for D on a sphere gives $Q=2\pi {ϵ}_0(1+{ϵ}_r)C$. ThereforeThe half-immersed conducting sphere in a dielectric has radial E lines with equal field strength at equal radius on either side, whereas the D flux density is ${ϵ}_r$ times larger in the liquid. The corresponding free surface charge is larger on the lower hemisphere. The unchanged field direction at this particular interface follows from its tangential geometry, not from a claim that electric fields never refract at dielectric interfaces.

The outward electrostatic pressure is $p_e=\mathbf{E}\cdot \mathbf{D}/2$. Integrating its vertical component over the two hemispheres uses ${\int }_{\mathrm{upper}}\cos \theta dA=\pi a^2$ and the negative of that integral below. ThusThe force is downward for ${ϵ}_r>1$. Initially, buoyancy on the uncharged sphere gives ${\rho }_{\ell }gf(4\pi a^3/3)=Mg$. At half immersion the upward buoyancy is therefore $Mg/(2f)$, leaving a required downward electric force $Mg(1/(2f)-1)$. Equating the two givesThis uses the ideal flat-interface, linear-dielectric model with surface tension and interface deformation neglected. If ${ϵ}_r=1$, there is no electric force of this kind, consistently preventing this mechanism from pulling an initially less-than-half-immersed sphere down to the specified level.

A magnetic dipole is the leading magnetic source moment of a localized steady current distribution. For a small planar loop, $\mathbf{m}=I\mathbf{A}$ with area orientation set by the right-hand rule. A fixed permanent moment in a slowly varying external field has energy $U=-\mathbf{m}\cdot \mathbf{B}$ and force $\mathbf{F}=\nabla (\mathbf{m}\cdot \mathbf{B})$.

For the given purely azimuthal vector potential, the spherical curl givesThese are the magnetic dipole field components outside the localized source.

The conducting shell has $Q=4\pi {ϵ}_{0}aV$ and surface charge density $\sigma =Q/(4\pi a^2)$. A latitude strip of area $2\pi a^2\sin \theta d\theta$ carries current $dI=\sigma \omega a^2\sin \theta d\theta$. Its loop area is $\pi a^2{\sin }^2\theta$, givingIn vector form the magnetic moment of a rotating charged spherical shell isThe positive-z expression in the question takes $\mathbf{\omega }=\omega \hat{\mathbf{z}}$. “Clockwise” needs a viewing direction: this rotation looks clockwise when viewed looking along $+z$ from below, and counterclockwise when viewed from the positive-z side. Looking from the other end reverses the verbal label, not the vector relation.

Since the positive charge and the loop current circulate in the same stated sense, their dipole moments are parallel when judged from a common viewpoint. At $h\gg R$, the loop's axial field magnitude is $B\simeq {\mu }_{0}I{R}^2/(2h^3)$. With the sphere small compared with that variation scale,The magnetic force is attractive, pulling the sphere toward the loop. Reversing only one circulation would reverse that magnetic force. The calculation treats the rotating charge as a prescribed moment and uses the stated separated length scales; it does not add unrelated induction or electric interactions to the dipole model.

Ridge push is a horizontal pressure force integrated with depth, equivalently a difference in gravitational potential energy per horizontal area. It is not the weight of a slab multiplied by the surface slope at one chosen point.

For an explicit compensated-column derivation, let $y$ measure depth below the ridge seafloor; the old seafloor is at $y=h$. Set $b={\rho }_m\alpha \Delta T$. The hot column has greater pressure. Its pressure excess over the old column isThe second line follows by integrating $\delta \rho (s)=b(1-s/a)$ upwards from the compensation level. The two expressions agree at $y=h$ because $({\rho }_m-{\rho }_w)h=ba/2$. Integrating this positive pressure excess gives the ridge-push force in a plate model, directed away from the ridge:Consistently retaining only first order in thermal expansion givesIf the topographic term is retained within the same linear-density column model, it adds $1.52\times 10^{11}\mathrm{N}{\mathrm{m}}^{-1}$, giving $F\simeq 2.75\times 10^{12}\mathrm{N}{\mathrm{m}}^{-1}$. This is a useful small correction, not a claim that all other second-order material effects have been included. Both values use the intended hot initial plate, not the inconsistent printed zero-age profile.

The depth-averaged added horizontal compressive stress is $\bar{\sigma }=F/a$. Resolving its traction on a plane dipping at angle $\delta$ gives the fault shear stress from a plate-integrated force,Keeping the previous topographic correction gives about $13.8\mathrm{MPa}$. This is the mean additional shear from ridge push, assuming the force is transmitted through thickness $a$ and resolving the nonisotropic stress increment. An isotropic lithostatic pressure adds normal traction but no shear. The force integral alone does not specify the local stress at each point on the fault or establish failure.

In vacuum, the source-free Maxwell equations arePut $\psi =\omega (t-z/c)$ and denote the constant electric amplitude by $E_0$. A corresponding traveling-wave magnetic field isBoth divergences vanish because both the electric field and the magnetic field depend only on $z,t$ and have zero $z$ component. Direct differentiation gives $\nabla \times \mathbf{E}=(\omega E_0/c)(\cos \psi ,\sin \psi ,0)=-{\partial }_t\mathbf{B}$ and $\nabla \times \mathbf{B}=(\omega E_0/c^2)(-\sin \psi ,\cos \psi ,0)=c^{-2}{\partial }_t\mathbf{E}$. A time-independent background magnetic field satisfying the homogeneous equations could be added; the formula is the wave component.

Introduce the magnetic vector potential and electric potential by $\mathbf{B}=\nabla \times \mathbf{A}$ and $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$. These identities automatically give the divergence equation for $\mathbf{B}$ and Faraday's law. For this wave choose $\varphi =0$ and $\mathbf{A}=(c/\omega )\mathbf{B}$; the derivatives just calculated verify both potential equations.

The electromagnetic force is the Lorentz force, $Q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. On the smooth constraint plane $z=z_0$, the constraint force is normal to the plane. Both $\mathbf{v}$ and $\mathbf{B}$ are tangential, so $\mathbf{v}\times \mathbf{B}$ is normal and is balanced by that constraint force; it does not affect the tangential motion. ThusA particular motion is ${\mathbf{r}}_{\parallel }={\mathbf{r}}_0-[Q{E}_0/(m{\omega }^2)](\cos \psi ,\sin \psi )$, a circle of radiusThe general planar motion also includes a constant drift velocity; the particle moves on a circle when that drift is zero. The constraint force can do no work because it is perpendicular to the velocity.

Orient $C$ counterclockwise around the solid body and let $\mathbf{n}$ point from the solid into the fluid. The pressure force is $-{\int }_{C}p\mathbf{n}ds$. Since $\mathbf{n}ds=(dy,-dx)$,For steady irrotational flow, Bernoulli's equation gives $p=p_0-\rho |w^{\prime }{|}^2/2$. The constant term integrates to zero. The boundary is a streamline, so $dw=w^{\prime }dz$ is real there, giving $w^{\prime }dz=\overline{w^{\prime }}d\overline{z}$. Thereforewhich is Blasius force theorem with the specified orientation.

The prescribed stream function is the imaginary part of $w_0(z)=\alpha (z-c{)}^3$. The zero-circulation solution from the Milne-Thomson circle theorem isOn $|z|=a$ the two terms are conjugates, so the boundary stream function is constant. The added term has no singularities in the exterior and does not change the prescribed growing far-field part. Additive constants are immaterial.

Differentiating givesThe coefficient of $z^{-1}$ in $(w^{\prime }{)}^2$ is $36{\alpha }^{2}c{a}^2(a^2+c^2)$. The residue theorem and Blasius force theorem giveThe zero-circulation convention is relevant: if an independently imposed circulation $\Gamma$ is allowed, adding $\Gamma \log z/(2\pi i)$ changes $F_y$ to $-3\rho \alpha \Gamma c^2$. Thus the claimed zero transverse force belongs to the standard circle-theorem construction, or to a single-valued exterior complex potential of a two-dimensional potential flow, which excludes that extra circulation.