Theoretical questions for the resumption of the national physical olympics competition for secondary school students in 2026 and detailed answer instructions for the examination: eight questions in total, with a full score of 160 and an examination time of 180 minutes. The solution needs to be written with the necessary text description, the core formula and the extrapolation step, with only the final result not to be divided; all physical constants use competitive standard values, gravitational acceleration \\(g=9. 8\mathrm{m/s^), planck constant \\(h=6. 63\times10^-34}\mathrm{j\cdots}, vacuum light speed \(c=3\times10^28\mathrm{m/s}), vacuum media constant \(\varepsilon 0=8. 85\) times10}\mathrm{c^2/(c\cdotm^)} i. (20 minutes)Mechanical: the re-movement of the connector question is illustrated by the placement of a long plank of mass (m=4mathrm{kg}\) on the smooth-level desktop, a light block fixed on the left side of the plank, a strength factor connected between the plate and the small piece of mass \(m=mathrm{kg}\(k=1) light spring (mathrm{n/m}\), which is initially long. Now gives small blocks a horizontal, right-to-right, instantaneous velocity \\ (v 0=10 \\mathrm{m/s}, during which springs remain within elastic limits, with no friction between the plank and the desktop and ignore air resistance. (1) the speed of the first compression of the spring to the shortest possible degree, shared by the planks and blocks; (2) the maximum compression of the springs; and (3) the speed of the plate and the planks upon the first return of the analysis blocks to the long spring positions. Detailed answers (1) the direction of the common velocity solver system is free from external forces and the momentum is constant. The spring is compressed to the shortest time, with equal speed and a common speed of \\(v \\). \ (mv 0=(m+m)v\) for data: \ (times10=(4+1)v\), solved (\bardsymbol{v=2\mathrm{m/s}}). (2) the maximum compression solvency system is mechanically sustainable and the difference between the initial and co-velocity kinetic energy of the block equals the maximum elasticity of the spring. \(\deltae k=\dfrac}mv 0^-\dfrac{2}additional energy loss:\(e p=\dfrac{2}(k k=dfrac}\times=100-\dfrac}\times\times=50-10=40 \mathrm{j}) resilient energy formula:\(e p=\dfrac{2}}kx^m\) (\dfrac^) ({kx^m}) (\m\m\m\m\m\m) (\m}\m) (3) the two velocitys of the original long position seek the recovery of the spring from the original long process, and the system is constant in kinetic and mechanical and is a fully elastic collision model. The block speed is \ (v ) and the plank speed is \\ (v 2\) and the right is the right direction. \ (mv 0=mv 1+mv 2\) mechanical persistence: \ (\dfrac}mv 0^m/s}(mv 1^dfrac{}mv 1^ \dfrac}}mv 2^)contrusive collision speed formula: \ (v 1=m\dfrac{m}m}m}m}dfrac{m}4}m/s} (\times10=-6\mathrm{m/s\) (negative sign indicates left)\(v m{m{m\m\m\m\m\m} }m\m\m\m\m} }m\m}m}m}dfrac{m}m}m/s} (orient) to the right \bold=4mathrm{m{m}m} \m}m\s}m\bold\m\m\m\m\m\m\m}m\m\m\m\m\m\m\m\m ii. (20 minutes) mechanics: an unknown planet mass of \ (m\), radius of \ (r\), a negligible detector at altitude from the planet's surface \ (h=r\) round orbits around planets at round speed. The gravitational acceleration of the surface of the planet is known to be (g ) without consideration for planetary rotation effects or for gravity of other celestial bodies. (1) speed and cycle of operation of the detector in circular orbit; (2) duration of operation of the detector in an elliptical orbit in an elliptical orbit with a perigee in the original orbit and an apogee in the surface of the planet; detailed answers (1) round orbital velocity is equal to gravitational force of 10,000 (mg 0=g\dfrac{mm}{r^2), \ (gm=g 0r^2). \(r=r+h=2r\), gravity provides the center:\(g\dfrac{m^(2r}}=m\dfrac{v^2}{2r}\(gm=g 0r^2)\, and solves the orbital \(boldsymbol{v=sqrt}dfrac{g 0r}{). By a periodic formula\(t=dfrac{2\pir}}, \ (r=2r\), \v=sqrt{g 0r}}, \boldsymbol{t=4\sqrt{2r}}). (2) a semi-long axis of an elliptical orbit proportional to the perigee deceleration \(a=\dfrac{2r+r}{2}=1. 5r\). An elliptical orbital perigee is \\ (v \) and an apogee (planet surface) is \\ (v \). According to kepler's second law (angular kinetic): \ (cdot2r=v 2\cdotr\), \ (v 2=2v ). \(-\dfrac{um}{2r}}dfrac}1^-\dfrac}{r}+dfrac}\mv 2^\) is replaced by:\(-\dfrac}2\), \(v 0r^2\), \ (v 2=2v dfrac}\(-\dfrac{g 0r^2}{2r}}}dfrac}{2}v 1^2=g 0r+dfrac}\(2}\times4v 1^2\) and \(-\v 1=sqrt{\g\dfrac}6}}). \(k=\dfrac{v}=\dfrac{\sqrt{g 0r}}{\sqrt{\dfrac{g 0r}}}=bardsymbol{\dfrac}}}) (3) the elliptical orbit cycle was introduced by capeler's third law: \ (\dfrac{t'2}=dfrac{a^3}}} (r=2r\), \ (t=4\sqrt{dfrac{2r}), \ (t't\cdot\sqrt}{g}), \ (t'cdot\sqrift{\ (\dfrac{3r}{2r}{2}\right}3}}}4\sqrt{\dfrac{2}}}cdot\sqrt{dfra{27{64}}) \f) and \boldsymbol{t=g=dfrac{3\sqirt}} }sqc}f}d{\f}f}). Iii. (20 minutes) thermals: a closed insulated tank is vertically placed for the desired change of gas state, with a mass (m=2 \\mathrm{kg}), a trans-section \ (s=10 \\mathrm{cm^2}) smooth insulation piston, which divides the cylinder into two upper and lower parts, with a vacuum at the top, and an ideal gas of a certain quality at the bottom. Initial piston static, gas column length \ (l 1=mathrm{cm}), temperature \ (t 1=3) the sand is now slowly folded on the piston, the total mass of the sand \ (m=8mathrm{kg}\), and the length of the gas column becomes \ (l 2=12mathrm{cm} after stabilization). Atmospheric pressure \\ (p 0=times10^5mathrm{pa}\, regardless of piston thickness,\(g=9. 8\mathrm{m/s^2}}) (1) the pressure of the gas in its initial state; (2) the temperature of the gas in its final state; and (3) the amount of energy in the gas during that process。detailed answers (1) initial piston pressure balance, upper vacuum without pressure, gas pressure supporting piston gravity: \ (p 1=\dfrac{mg}{s} add data: \ (s=10^-3}mathrm{m^2), \ (p 1=\dfrac{2\times9. 8/} {10^-3}}=196\times10^4\mathrm{pa}) is the initial pressure \\(\bardsymbol{p 1=1. 96\times10^4\mathrm{pa}}). (2) the end-state pressure \\ (p 2=\dfrac{(m+m)g}{dfrac{10\times9. 8}{10^-3}}=9. 8\times10^4\mathrm{pa}) after the temperature of the last-state gas is folded with fine sand. Anticipated gas state equation: \\ (\dfrac{p 1v ) =dfrac{p 2 v}t 2} \ (v 1=l 1s\), \ (v 2=l 2s\), the volume margin can be approximated. (\dfrac{1. 96\times10^4\times2) (3) thermal insulation of the gas in which the gas can vary from the cylinder to the piston, and no heat exchange between the gas and the outside world, \ (q=xiii). According to the first law of thermodynamics. External action on gas is equivalent to a reduction in pistons and fine sand gravity (slow process, kinetic energy change of 0). Pistol drops height \\(\deltah=l 1-l 2=8mathrm{cm}=0. 08\mathrm{m}). External performance (w=overline{f}\cdot\deltah\), average pressure \((overline{f}=dfrac{mg+(m+m)g}{2}=49\mathrm{\). \ (w=49\times0. 08=3. 92\mathrm{j}) so the gas can increase \ (\boldsymbol(deltau= 3. 92\mathrm{j}}). Iv. (20 minutes) electromagnetics: in the vacuum of electrostatic field power and power field strength problems, there is a flat-banded circle, circle radius \\ (r=0. 2\mathrm{m}, charge \\(q=2\times10^-6}\mathrm{c}). (1) the strength and power of the field at the centre of the circle; (2) the strength of the field at the centre of the circle at a distance from the centre of the circle \\(x=0. 2\mathrm{m}\; and (3) the work required at least for external strength (not counting the gravity of the particles) if a mass \\(m=times10^-3}mathrm{kg}\), a small tether particle with charge \q=times10^-8}mathrm{c}\) is taken from an infinite distance to the axis \\(x\). Detailed answers (1) the power of the centre is equal to the power of the power, and the amount of the charge is offset by the strength of the field vector at the centre, so the centre is strong (e 0=1). The power is superheavy to the scale, and the circle power is: \\(\varphi 0=\dfrac{c\cdot\dfrac{q}\r} \cdot\dfrac{ \r} \ \varphi=9\times\9\times\dfrac{2\times10^6}0}}9\times10^4\mathrm{v}). (2) \\(x\) field-strength formula on the field with a circle:\(e=dfrac{ 4 \ \ \ \ q ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } } \ \ \ \ \ } } } } } } \ \ \ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^. Simplified calculations: \ (e=9\times10^9\times\dfrac{2\times10^6}\times0. 2}(16\sqrt{2}\times10^3}\approx\boldsymbol{1. 59\times10^6\mathrm{n/c}}) (3) exogenous power works to solve zero power and field power works at zero. Axes \\(x) at power: \\(varphi x=\dfrac1}cdot\dfrac{sqrt{r^2+x^}\approx6. 36\times10^4mathrm{v}). The particle moves slowly, the kinetic energy changes to 0, by the kinetic energy theorem: \ (w extra+w electric=) \ (w extra=q\varphi x\). \(\bardsymbol{w \approx6. 36\times10^4}\mathrm{j}}) v. (20) electromagnetics: constant current and jolt problems such as circuits, power dynamics (e=24\mathrm{v}), internal resistance \ (r=2\omega\), nominal resistance \ (r 1=6\omega\), \ (r 2=12\omega\), maximum resistance value for slide transformers \ (r 3=18\omega\). The circuits are stable and the meters are ideal. (1) when the slider slider is placed at the leftmost end, the circuit end voltage and the total power of the power of the power source; (2) the slider slider at the rightmost end, the jolt heat power \ (r 2\); and (3) the position of the slider, the maximum power available to the slider. Detailed answer (1) to the left end of the slider ((r 3=1) at this time (r 2\) is short-circuited, total circuit block \ (r total=r+r 1=8\omega\). Total current \\ (i=\dfrac{e} {r general} =3mathrm{a}, circuit-end voltage\ (u=e-ir=24-6=\bardsymbol{18\mathrm{v}}) power \ (p total=ei=24\times3=\bardsymbol{72\mathrm{w}}) (2) the right-hand end of the slider ((r 3=18\omega\)\(r 2\) is linked to (r 3\) and is electrically restrained\(r 23}=\dfrac{12\times18}{12+18}=7. 2/omega\). \ (r total=r+r 1+r 23}=15. 2/omega\), \ (i=dfrac{24}{15}\mathrm{a}\) \(u 23}=ir 23}\approx11. 37\mathrm{v}}) \(r 2) power\(p 2=\dfrac{u 23}}r}2}approx\bardsymbol{10. 8\mathrm{w}}) (3) the maximum power of the sliding transformer is based on the equivalent power source method, which is the same as new internal resistance. Equivalent electric motion\ (e et al.=\dfrac{r 2} {r 1+r 2}e=mathrm{v}\), equivalent internal resistance\ (r et al.=r+\dfrac{r 1r 2}{r 1+r 2}=6\omega\). When \\(r 3=r et al=6\omega\), the smoother power is maximum. Maximum power \ (p max} \dfrac{e {2 } {4r } } } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)。vi. (20 minutes) electromagnetic induction: placement of a smooth parallel metal track level with a sufficiently long dynamic and energy problem, length of the track \ (l=0. 5\mathrm{m}\), fixed-value resistance to the left end of the track \ (r=4\omega\) and absence of electrical resistance to the track. \(b=0. 8 \mathrm{t}\, a metal bar of mass\(m=0. 2 \mathrm{kg}), a resistance\(r=omega\) is placed vertically in the guidance track, giving the metal bar horizontally to the right at the beginning \(v 0 = 10 \mathrm{m/s}, and the metal bar is not removed from the guide. (1) insisting electric motion and currents at the beginning of the metal bar; (2) analysing the metal bar's state of motion with a view to its final stabilization; (3) valuing the total process for electro-reception (r \) resulting from jj. Detailed answers to (1) initial induction electric motion and mobile electric motion \ (e=blv 0=0. 8\times0. 5\times10=\bardsymbol{4\mathrm{v}}) sensor current \\(i=\dfrac{e{r+r}=\dfrac{4}{5}=bardsymbol{0. 8\mathrm{a}}) (2) the motion and stable speed bar is subject to a left-to-left ampellation motion, while the reduced speed is associated with electrical currents, ablution is synchronized, and speed reduction straight-line motion is performed. When the speed is reduced to zero, the induction current is zero, the ampere is lost and the movement is stable. So finally, the speed is stable. (3) the total kinetic energy of the fixed-value resistance jj during the whole process is converted to circuit jj, total jj (q = total \dfrac{2}mv ^0^dfrac{2}\times 0. 2\times100=10\mathrm{). A positive ratio of serial circuit joule fever to electrical resistance, \ (q r=q general\cdot\dfrac{r+r}=10\times\dfrac{4}}} \bardsymbol{8\mathrm{j}\) vii. (20 minutes) opticals: refraction and intervention of a transparent medium to visible light \(n=1. 5\), visible wavelength range in a vacuum \ (mathrm{nm}\sim760\mathrm{nm}). (1) the critical angle at which the light is directed from the medium to the vacuum; (2) the wavelength range of the visible light in the medium; (3) the thin film of the medium, which is evenly thick, is placed in the air, thin film thickness \\ (d=9 milm{nm}\), a single-colour visible light vertically into the thin film, and the wavelength of the single light in the vacuum is requested if the reflection light is enhanced. Detailed answers (1) to the full reflection threshold angle from the critical angle formula \\ (\sinc=\dfrac{n}\), to (n=1. 5\), to (\sinc=\dfrac{2}{3}\). Threshold angle \\(\bardsymbol{c=arcsin\dfrac{2}\approx41. 81^ \circ}). (2) the medium's medium wavelength range light is wavelength in the medium \\ (\lambda=\dfrac/2003/lambda ), \\ (\lambda themselves) is vacuum wavelength. Shortest wavelength: \\(\lambda min}=dfrac{40 }1. 5}\approx266. 7\mathrm{nm}), maximum wavelength: \ (\lambda {max}=dfrac{76 {1. 5}\approx506. M{nm}) is the medium medium wavelength range \\(\bardsymbol{266. Ammathrm{nm}\sim506. Mmathrm{nm}). (3) there is a half-wave loss of reflections of the upper and lower surfaces of the wavelengths of reflecting light interference, with no additional pv differentials (\delta=2nd\). Interventions to strengthen conditions: \\(k=1,2,3\cdots\), \(k=0,\dfrac{2nd}{dfrac{2700}k}\mathrm{nm}\) \\(k=4\) when filtering the visible wavelength range:\ (\lambda 0=6\mathrm{nm}); \ (k=5\) when filtering:\ (\lambda 0=540\mathrm{nm}); \(k=6\) when filtering:\ (\lambda 0=450\mathrm{nm}). Eligible wavelengths are \\(\bardsymbol{450\mathrm{nm}, 54/1mathrm{nm}, 6\mathrm{nm}). Viii. (20 minutes) modern physics: light
