Unité 5 : Thermodynamique, rayonnement, oscillations et cosmologie
Pearson Edexcel · International A-Level · Physics Sujet 5 32:51 Narration en anglais · Sous-titres anglais + 中文 incrustés
Chapitres
Transcript
An ice pack absorbs energy without necessarily warming. A star emits energy while its nuclear fuel changes.
冰袋吸收能量时不一定升温，恒星发射能量的同时核燃料发生变化。
This lesson connects those ideas through nine skill blocks.
本课分九个技能模块连接这些现象。
For each calculation, name the model, write the equation, keep units and test the claim.
计算时先说明模型，再写方程、保留单位并检验题目中的判断。
The full reference contains the details; practise on the skill sheet before the corresponding authentic set.
完整讲义提供细节，先完成技能练习，再做对应真题。
Two samples can have the same temperature and different internal energies because their amounts or states differ.
两份物质温度相同，内能仍可因质量或状态不同而不同。
Internal energy includes the interactions between molecules as well as their random motion.
内能包括分子无规则运动和分子间相互作用的能量。
During melting, transferred energy changes molecular arrangements even while temperature stays constant.
熔化过程中温度可以不变，能量用于改变分子排列。
The motion of a whole falling ice block is a separate kinetic energy.
整块冰下落的动能要另算。
On a slope, temperature changes without a change of state. On a plateau, energy is still entering but the state changes at approximately constant temperature.
斜线段表示温度改变而物态不变；平台段仍有能量输入，但物态在近似恒温下改变。
At constant heater power and negligible losses, a steeper slope means a smaller product of mass and specific heat capacity.
恒定功率且热损失可忽略时，斜率越大，质量与比热容的乘积越小。
Do not interpret a flat line as no energy transfer.
水平线不表示能量停止传递。
First warm the solid by ten kelvin. Then melt the whole mass at zero degrees Celsius.
先把固体升温十开尔文，再在零摄氏度熔化全部冰。
Write separate expressions before adding, because specific heat capacity and specific latent heat describe different processes.
先分别列式再相加，因为比热容和比潜热对应不同过程。
Attempt the calculation before advancing.
翻页前先独立计算。
The warming energy is eight point four kilojoules, while melting requires one hundred and thirty-six kilojoules. Their sum is about one hundred and forty-four kilojoules.
升温需要八点四千焦，熔化需要一百三十六千焦，总计约一百四十四千焦。
A ten-degree temperature rise is a ten-kelvin difference, but absolute gas temperatures later must use kelvin.
十摄氏度的温差等于十开尔文，但后面气体定律中的绝对温度必须用开尔文。
The ice absorbs energy from its surroundings rather than sending cold energy outward.
冰吸收周围能量，并不是向外释放冷量。
A one-kilowatt heater used for six hundred seconds supplies six hundred kilojoules.
一千瓦加热器工作六百秒输入六百千焦。
Evaporating point two kilograms of water at the stated latent heat uses four hundred and fifty-two kilojoules, giving about seventy-five percent efficiency.
按给定汽化比潜热，蒸发零点二千克水需要四百五十二千焦，效率约百分之七十五。
In a kettle comparison, count all loads and compare the same time interval.
比较水壶方案时要统计全部次数，并使用相同时间范围。
A daily standby loss is not the energy for one boiling cycle.
一天的待机损耗不能直接等同于一次烧水能量。
The output is the supply voltage multiplied by the lower resistance divided by the total resistance.
输出等于电源电压乘以下方电阻与总电阻之比。
Here the thermistor is the lower component. Heating lowers its resistance and lowers the output voltage.
图中热敏电阻在下方，升温使它的电阻下降，输出也下降。
Reversing the positions reverses the trend.
交换元件位置会反转趋势。
A divider supplies a temperature-dependent signal; a separate switching stage is needed for a thermostat.
分压器只提供随温度变化的信号，恒温控制还需要开关电路。
For the thermistor, use a fixed supply and resistor, a high-resistance voltmeter, and several temperatures after equilibrium.
热敏电阻实验应固定电源和定值电阻，使用高内阻电压表，在多个温度达到平衡后读数。
Compare warming and cooling to detect lag.
比较升温和降温可发现滞后。
For latent heat, measure mass change after conditions stabilise.
比潜热实验稳定后测质量变化。
Subtracting results at two heater powers removes approximately constant heat-loss power only if the phase-change temperature and losses stay comparable.
两个功率结果相减，只有在相变温度和散热近似相同的条件下才能消除散热功率。
Use low voltage and protect against hot water and surfaces.
使用低压电源并防止热水烫伤。
Ideal particles have negligible volume and no intermolecular forces except during elastic collisions.
理想气体分子体积可忽略，除弹性碰撞外不考虑分子间作用力。
Wall momentum changes create pressure.
分子撞壁改变动量而产生压强。
The proportionality laws require kelvin temperatures, absolute pressure and a fixed amount of gas.
比例关系要求开尔文温度、绝对压强和固定气体量。
Classical average translational kinetic energy tends towards zero at absolute zero; real gases condense before this extrapolation, so do not assert that every material loses every form of energy.
经典模型中平均平动动能趋于零，但真实气体在达到这种外推状态前会凝结，不能说任何物质的一切能量都归零。
The vessel is rigid and sealed, so volume and molecule number stay fixed.
容器密闭且刚性，因此体积和分子数都不变。
First calculate the molecule number using the initial state. Then use the pressure-temperature ratio without recalculating volume.
先用初始状态求分子数，再用压强温度比求末压强，不需要重新计算体积。
Write the assumptions before obtaining both answers.
先写出适用条件再计算两个结果。
The initial pressure-volume product is two hundred joules.
初始压强与体积的乘积为二百焦。
Dividing by Boltzmann constant \times three hundred kelvin gives four point eight three \times ten to the twenty-two molecules.
除以玻尔兹曼常数与三百开尔文的乘积，得到四点八三乘十的二十二次方个分子。
The final temperature is twenty percent larger on the absolute scale, so the final pressure is one point two \times ten to the five pascals.
绝对温度增加百分之二十，末压强为一点二乘十的五次方帕。
If molecular mass is supplied, gas mass is molecule number \times molecular mass.
若给出单个分子质量，总质量等于分子数乘以单个分子质量。
Consider a cube of side l and a molecule with velocity component c x.
考虑边长为l的立方体，分子沿x方向的速度分量为c x。
An elastic collision reverses that component, transferring twice its component momentum to the wall.
弹性碰撞使该分量反向，向墙传递两倍的分量动量。
The molecule returns after travelling twice the cube length.
分子运动两个边长后再次撞击同一面墙。
Divide momentum transfer by return time to find average force. Summing over molecules and dividing by wall area produces pressure.
动量变化除以返回时间得到平均力，再对所有分子求和并除以墙面积得到压强。
Isotropic random motion distributes mean squared velocity equally among three perpendicular directions. That supplies the factor of one third in the pressure relation.
各向同性的无规则运动使三个垂直方向的速度平方平均值相同，因此压强关系中有三分之一。
Equate it to the ideal-gas expression and cancel molecule number to obtain average translational kinetic energy.
令它与理想气体方程相等并约去分子数，得到平均平动动能关系。
This derivation explains the relation instead of treating it as an isolated formula.
推导能解释公式的来源，而不是孤立背诵。
Root mean square speed is the square root of the mean of squared speeds. It is neither mean speed nor its square.
方均根速率是速率平方平均值的平方根，既不是平均速率，也不是平均速率的平方。
Lighter particles move faster on average at the same temperature because mean translational energy is the same.
同温下平均平动动能相同，所以较轻分子的方均根速率较大。
Molecules have a distribution of speeds; heating does not give every molecule an identical speed.
分子速率具有分布，加热并不会让每个分子都具有相同速率。
Trap a fixed amount of gas and record several pressure-volume pairs.
封闭固定量气体，测量多组压强和体积。
Slow compression and waiting allow heat to escape so temperature returns to the surroundings.
缓慢压缩并等待，可以使温度重新接近环境温度。
Include the connecting tube volume and prevent leakage.
计入连接管内体积并防止漏气。
Pressure against reciprocal volume should be linear through the origin within uncertainty; pressure against volume is curved.
压强对体积倒数应在误差范围内呈过原点直线，压强对体积本身则是曲线。
Rapid compression changes temperature and invalidates the intended test.
快速压缩会升温，破坏实验条件。
Binding energy is the energy required to separate a nucleus into free nucleons.
结合能是把原子核完全分离成自由核子所需的能量。
Its bound mass is smaller than the total mass of those free particles.
束缚态原子核的质量小于自由核子的总质量。
Keep all masses in one unit and account for electrons consistently if atomic rather than nuclear masses are supplied.
质量单位必须一致；若使用原子质量而非核质量，要一致处理电子。
The speed of light here is c zero, distinct from specific heat capacity.
这里的c零表示光速，与比热容不同。
Convert the mass deficit to kilograms before using mass-energy equivalence.
先把质量亏损换成千克，再使用质能关系。
Then convert joules to mega-electronvolts and divide by the number of nucleons.
接着把焦耳换为兆电子伏，并除以核子数。
Do not confuse the nucleus total binding energy with binding energy per nucleon.
不要把总结合能与每核子结合能混为一谈。
The mass deficit is four point nine eight \times ten to the minus twenty-nine kilograms.
质量亏损为四点九八乘十的负二十九次方千克。
Multiplying by the square of light speed gives four point four eight \times ten to the minus twelve joules.
乘光速平方得到四点四八乘十的负十二次方焦耳。
Divide by four nucleons and the joules-per-mega-electronvolt conversion to get seven mega-electronvolts per nucleon.
再除以四个核子和兆电子伏换算系数，得到每核子七兆电子伏。
Check each conversion separately.
分别检查每一步换算。
Light nuclei can fuse towards the peak of the binding-energy curve, while very heavy nuclei can split towards that peak.
轻核聚变和重核裂变都可能使产物向结合能曲线峰值靠近。
Products with greater total binding energy have smaller total rest mass. The energy difference appears in other forms; energy is conserved.
产物的总结合能增加时，总静质量减小，差额转化为其他形式的能量，总能量仍守恒。
For a numerical reaction, compare total masses rather than subtracting two per-nucleon values without their nucleon counts.
计算反应释能应比较总质量，不能不计核子数就直接相减两个每核子结合能。
Positive nuclei repel electrically.
带正电的原子核相互排斥。
High temperature helps nuclei approach closely enough for fusion to occur.
高温使核具有足够动能接近到能发生聚变的距离。
High density and confinement increase the number of encounters and help sustain energy release.
高密度和约束提高碰撞机会并有助于持续释能。
Naming pressure alone does not explain both roles. A good answer connects each condition to its physical effect.
只写高压不能解释两个不同作用，完整答案应把每个条件与具体物理效果相连。
Alpha particles leave thick tracks because they strongly ionise, and are relatively straight because their mass is large compared with beta particles. Those explanations are different.
阿尔法径迹粗是因为电离作用强，较直则与它比贝塔粒子重有关，两个解释不能互换。
A paper-thickness monitor needs radiation partly transmitted and partly absorbed.
纸张测厚需要辐射部分透过、部分吸收。
Gamma attenuation through lead reduces intensity progressively rather than stopping every photon at one precise thickness.
铅对伽马射线是逐步衰减，并不存在使所有光子突然停止的精确厚度。
Balance the top nucleon numbers and bottom charge numbers independently.
分别核对上方核子数和下方电荷数。
In beta-minus decay, a neutron becomes a proton, an electron and an electron antineutrino. A beta particle is not an electron previously orbiting the nucleus.
贝塔负衰变中，中子转变为质子、电子和电子反中微子，电子并不是原来绕核运动的电子。
Energy and momentum conservation still apply even when the two printed rows balance.
两个数字行配平后仍必须满足能量和动量守恒。
Only the source contribution follows this inverse-square distance model.
只有测试源的贡献服从这里的距离平方反比模型。
Separate it from background, change that contribution, then put the background back.
先扣除本底，再改变源的贡献，最后加回本底。
Decide which part of the measured count is actually affected before doing arithmetic.
计算前先判断测得计数中的哪一部分随距离改变。
Subtract fourteen from sixty-two to obtain forty-eight counts per minute from the source.
六十二减十四得到源贡献每分钟四十八次。
Doubling distance divides that by four, giving twelve. Add the unchanged background to obtain twenty-six.
距离加倍后除以四，得到十二次，再加回本底十四次，总计二十六次。
Count rate is proportional to activity only for fixed detection efficiency and geometry; a detector count is not automatically the source activity in becquerels.
只有探测效率和几何条件固定时，计数率才与活度成正比，探测计数不能直接当作贝可勒尔活度。
Spontaneous means no external trigger is required. Random means that an individual nucleus decay time cannot be predicted.
自发表示不需要外界触发，随机表示不能预测某一个核何时衰变。
A large population still follows an exponential pattern statistically.
大量原子核仍在统计上服从指数规律。
Activity counts decays per second; the decay constant is probability per unit time.
活度是每秒衰变次数，衰变常数是单位时间的衰变概率。
Use seconds if activity is required in becquerels.
求贝可勒尔活度时要使用秒。
Taking the logarithm of the half-life condition gives \lambda equals log two divided by half-life.
对半衰期条件取对数，得到衰变常数等于二的自然对数除以半衰期。
Log corrected activity against time has negative gradient \lambda.
校正活度的对数对时间作图，斜率为负的衰变常数。
Do not take the logarithm of source-plus-background readings: their nonzero plateau changes the model.
不能对含本底的读数直接套用，因为非零平台改变了模型。
Read several half-life intervals when using a graph, rather than relying on one noisy crossing.
读图时可测多个半衰期间隔，避免只用一个噪声较大的交点。
Two percent means point zero two of the initial activity, not the number two.
百分之二表示初值的零点零二，而不是二。
Use the same time unit in the half-life, decay constant and final time.
半衰期、衰变常数和最终时间必须使用一致单位。
If a radioactive power is requested, multiply activity by energy released per decay in joules.
若求放射性功率，活度乘以每次衰变释放的焦耳能量。
Useful electrical power can be less than released nuclear power.
有效电功率可能低于核释能功率。
Keep geometry unchanged, subtract background and account for fluctuations by repeating or extending counting time.
固定几何位置，扣除本底，通过重复或延长计时处理计数波动。
Three half-value thicknesses transmit one eighth, not one third.
三个半值厚度透过八分之一，不是三分之一。
Thickness attenuation and time decay use different constants.
厚度衰减与时间衰变使用不同常数。
Near-background readings have large relative uncertainty.
接近本底的读数相对不确定度很大。
Follow supervised source handling, minimise exposure, use shielding and wash after handling lead.
按教师要求操作辐射源，缩短暴露时间、使用屏蔽并在接触铅后洗手。
The resultant acceleration is proportional to displacement and directed towards the fixed equilibrium.
合加速度与位移成正比，方向始终指向固定平衡位置。
For a vertical spring, weight is already balanced at the loaded equilibrium position; the extra restoring resultant is minus spring constant \times displacement from there.
竖直弹簧在负载平衡位置已经平衡重力，从该处偏移后的合回复力为负的劲度系数乘位移。
A repeated motion or a force pointing to a centre alone does not establish this proportionality.
仅有周期性或指向某个中心的力，都不足以证明简谐运动。
At positive maximum displacement, velocity is zero and acceleration is most negative.
在正最大位移处，速度为零，加速度为负最大值。
At equilibrium, acceleration is zero and speed is greatest.
平衡位置加速度为零而速率最大。
Velocity is the gradient of the displacement-time graph, so read its sign as well as its magnitude.
速度是位移时间图的斜率，要同时读出符号和大小。
The velocity curve is a quarter period out of phase with displacement; acceleration is opposite in phase.
速度与位移相差四分之一周期，加速度与位移反相。
Read the full period rather than half a cycle.
读取完整周期而不是半周期。
Convert it to angular frequency first, then use amplitude \times angular frequency for maximum speed and amplitude \times its square for maximum acceleration.
先换成角频率，再用振幅乘角频率求最大速率，用振幅乘角频率平方求最大加速度。
State the positions so the numbers remain connected to the motion.
说明最大值出现的位置，使数值与运动联系起来。
Two \pi divided by half a second gives about twelve point six radians per second.
二派除以半秒得到约十二点六弧度每秒。
Multiplying by the amplitude gives point three seven seven metres per second.
乘振幅得到零点三七七米每秒。
Squaring angular frequency before multiplying gives four point seven four metres per second squared.
角频率先平方再乘振幅得到四点七四米每二次方秒。
Maximum speed and maximum acceleration do not occur at the same instant.
最大速率与最大加速度不是同时出现。
The spring period depends on mass and stiffness; the small-angle pendulum period depends on length and gravity. State the small-angle condition.
弹簧周期取决于质量和劲度系数，小角度单摆周期取决于摆长和重力加速度，并要说明小角度条件。
In undamped spring motion, kinetic and potential energy interchange while total energy stays constant.
无阻尼弹簧振动中，动能和势能相互转化，总能量不变。
The kinetic-energy-versus-displacement graph is a downward parabola with zero at both endpoints, not a sinusoidal time graph.
动能对位移图是端点为零的开口向下抛物线，不是时间轴上的正弦曲线。
Use several known masses and identify resonance by varying the driving frequency carefully.
使用多个已知质量，细调驱动频率找共振。
Convert each measured frequency to period squared and plot against mass. Use that calibration to infer the unknown mass.
把频率换算成周期平方，再对质量作图，用标定曲线推断未知质量。
Apparatus such as a spring and holder can contribute effective mass, producing an intercept; do not assume every graph must pass through zero.
弹簧和支架等可贡献有效质量而产生截距，不要默认所有图都必须过原点。
Repeat measurements and avoid large amplitudes outside the intended model.
重复测量并避免振幅过大而偏离模型。
A free oscillator moves at its natural frequency after release.
自由振动在释放后以固有频率运动。
A periodically driven oscillator settles to the driving frequency.
周期驱动下，稳定振动跟随驱动频率。
Damping removes energy from the oscillation, often into thermal stores.
阻尼使振动能量转移到其他储能，常为内能。
Under otherwise comparable conditions, a forced response becomes large near a natural frequency; that is resonance.
在其他条件相同时，驱动频率接近固有频率会产生较大振幅，即共振。
These terms describe different aspects and can apply together.
这些词描述不同方面，可以同时适用。
Each point represents a steady amplitude at a chosen driving frequency.
每个点对应一个驱动频率下的稳定振幅。
Extra damping usually reduces the peak and broadens the response, so less energy accumulates in a narrow frequency band.
增加阻尼通常降低峰值并加宽响应，减少某个狭窄频段中的能量积累。
A real damped amplitude maximum can be slightly shifted from the undamped natural frequency.
真实阻尼系统的振幅峰值可能略偏离无阻尼固有频率。
Do not confuse increasing driving frequency with automatically increasing amplitude.
驱动频率增大并不意味着振幅必然增大。
Build a causal chain rather than listing resonance, energy and damping as separate keywords.
不要把共振、能量和阻尼作为互不相连的关键词堆砌。
Start with the driving force, connect frequency matching to efficient energy transfer, then connect damper work to energy loss and a smaller steady amplitude.
先说明周期驱动力，再把频率接近与有效能量传递相连，最后把阻尼器做功、能量损失和较小稳定振幅相连。
Attempt a linked explanation before advancing.
翻页前先写完整因果解释。
People walking supply a periodic driving force. Near a natural frequency, efficient transfer can build up bridge oscillation energy and amplitude.
行人提供周期驱动力，接近固有频率时能量有效传入桥梁，振动能和振幅增大。
A damper opposes motion and does work, transferring energy out of the oscillation. For a comparable drive, the steady amplitude is reduced.
阻尼器阻碍运动并做功，把振动能转移出去，在相似驱动条件下减小稳定振幅。
Do not claim that a heavier oscillator always damps more slowly from one paper-pendulum example.
不能从某个纸摆例子推导所有重振子都衰减更慢。
Field strength is force per unit test mass and is a vector. Potential is energy per unit mass and is a scalar.
重力场强度是单位测试质量所受的力，是矢量；引力势是单位质量的势能，是标量。
For a spherical source, radius is measured from its centre.
球对称源的距离从中心量起。
Moving outward weakens the force while potential becomes less negative.
向外移动时引力减弱，但引力势变得较不负。
A height above the surface is not the radial distance in either equation.
表面高度不是方程中的中心距离。
An isolated spherical mass and a point charge both produce radial fields with inverse-square strengths outside the source. Gravity attracts a test mass.
孤立球形质量和点电荷在源外都产生径向的平方反比场。
Electric field direction is defined using a positive test charge, so a negative charge feels the opposite force.
引力吸引测试质量；电场方向按正测试电荷定义，负电荷受力方向相反。
Similar distance laws do not make the interactions identical.
距离规律相似，不表示相互作用完全相同。
The distance doubles, so treating gravity as a constant surface value is unsuitable.
距离加倍，不能把重力加速度视为恒定的表面值。
Write negative gravitational potential energy at both radii, then subtract initial from final.
分别写两个半径处的负引力势能，再用末值减初值。
Predict the sign before calculating: moving outward should increase potential energy towards zero.
计算前先判断符号，向外移动使势能向零增加。
Final potential energy is less negative than initial potential energy, so the change is positive.
末势能比初势能较不负，所以变化为正。
The difference in reciprocal radii gives about three point one three \times ten to the ten joules.
两个半径倒数之差给出约三点一三乘十的十次方焦耳。
Kinetic energy may also change during launch or an orbital transfer.
发射或轨道转移还可能改变动能。
Do not use a mistaken intermediate scheme sign as a new physical rule.
不能把评分标准中的中间符号错误当成新的物理规律。
The gravitational force is the centripetal resultant, not an extra force added to it.
引力本身提供向心合力，不能再额外加一个向心力。
Set the two expressions equal and cancel the orbiting mass. Using circumference divided by speed gives the period relation.
令两种表达式相等并约去卫星质量，用周长除以速率得到周期关系。
A measured period and orbital radius can then give central mass.
已知周期和轨道半径可以求中心天体质量。
Check whether a supplied distance is radius or diameter.
要检查题目给的是半径还是直径。
For the reference values, the orbital radius is four point two three \times ten to the seven metres.
按讲义数值得轨道半径四点二三乘十的七次方米。
A satellite appears fixed over one point only if its circular orbit is equatorial, moves in the same direction and matches the rotation period.
相对地面固定还需要赤道圆轨道、同向运动和相同转动周期。
Matching period on an inclined orbit is insufficient.
倾斜轨道即使周期相同也不够。
Subtract the planet radius only if the question asks for altitude.
只有题目求离地高度时才减去行星半径。
A black body is an ideal absorber and thermal emitter.
黑体是理想吸收体和热辐射体。
Raising its temperature moves the wavelength-spectrum peak towards shorter wavelengths and increases total emitted power per unit surface area.
升温使波长谱峰移向短波，并增加单位表面积的总辐射功率。
A frequency-spectrum peak cannot generally be converted into the wavelength-spectrum peak simply by using c equals f \lambda, because the spectral density definition changes.
频率谱的峰通常不能直接用光速等于频率乘波长换成波长谱的峰，因为谱密度定义改变了。
Read the question convention carefully.
要仔细阅读题目采用的定义。
Luminosity is total emitted power in watts. Received intensity is power per receiving area.
光度是以瓦为单位的总发射功率，接收强度是单位接收面积的功率。
The star surface sets emitted power, while the sphere at observer distance distributes that power.
恒星表面决定发射功率，观察者距离处的球面决定功率如何分散。
Substituting observer distance into the star surface area confuses the two stages.
不能把观察距离代入恒星表面积。
Absorption and departures from a black body can limit the model.
吸收及非理想黑体行为都会限制模型。
This is a three-stage chain: wavelength peak to temperature, temperature and stellar surface to luminosity, then luminosity and observer distance to intensity.
这是三步链条：波长峰求温度，温度和恒星表面求光度，光度和观察距离求强度。
Convert nanometres to metres first. Keep unrounded temperature for the fourth power.
先把纳米换为米，四次方计算保留未舍入温度。
Attempt the chain before advancing.
翻页前独立完成这条链。
Wien law gives about five thousand kelvin.
维恩定律得到约五千开尔文。
Using the unrounded value with four \pi R squared and the Stefan-Boltzmann constant gives two point one eight \times ten to the twenty-six watts.
用未舍入温度、四派半径平方和斯特藩玻尔兹曼常数，得到二点一八乘十的二十六次方瓦。
Divide that luminosity by four \pi d squared to obtain about seven hundred and seventy watts per square metre. Only then compare with a reference intensity.
再除以四派观察距离平方，得到约每平方米七百七十瓦，之后才能与参考强度比较。
Observe a nearby star against distant background stars six months apart.
相隔六个月，以远方恒星为背景观察近星。
The full baseline spans two Earth-orbit radii, while the triangle using parallax angle p uses one radius.
完整基线为两个地球轨道半径，而使用视差角p的半三角形对应一个轨道半径。
In radians the small-angle relation is distance approximately orbit radius divided by p. In arcseconds the reciprocal gives parsecs.
p用弧度时，距离近似为轨道半径除以p；用角秒时，其倒数为秒差距。
Do not use Earth diameter as the orbital baseline.
不能把地球直径当作轨道基线。
A standard candle has known luminosity, not known apparent brightness.
标准烛光已知的是光度，不是视亮度。
Measure received intensity to infer distance.
测接收强度后推算距离。
For a Cepheid, average several complete periods, read the calibrated period-luminosity plot and convert solar units if required.
造父变星应对多个完整周期取平均，再读周期光度标定图，并按需要换算太阳光度单位。
A straight line on log axes does not prove a raw linear law.
双对数图的直线不能证明原始量是线性关系。
Parallax can fail because its angle becomes too small to measure accurately, even if the star is visible.
视差太小会难以精确测量，即使仍能看见恒星。
The main sequence runs from hot bright stars at upper left to cool faint stars at lower right.
主序带从左上方高温高光度延伸到右下方低温低光度。
Red giants are luminous despite low surface temperature because their emitting area is large. White dwarfs are faint despite high temperature because their surface area is small.
红巨星表面较冷但面积大，所以光度高；白矮星温度高但面积小，所以较暗。
Mark the Sun near one solar luminosity and about fifty-eight hundred kelvin, on the main sequence.
太阳位于约一倍太阳光度、五千八百开尔文处的主序带上。
An extended answer needs a causal sequence.
长答题需要因果顺序。
Explain what happens when core hydrogen runs low, why the core contracts and heats, and how shell burning and later helium fusion relate to the giant stage.
解释核心氢燃料减少后发生什么，核心为何收缩升温，以及壳层燃烧和后来的氦聚变如何联系到巨星阶段。
Then name the Sun-like remnant.
最后说明类太阳恒星的遗迹。
Do not give the Sun a supernova merely because massive stars can have one.
不能因为大质量恒星会超新星爆发，就让太阳也这样结束。
Core hydrogen depletion reduces fusion there, so gravity contracts and heats the core.
核心氢耗尽使该处聚变减弱，引力促使核心收缩升温。
Shell hydrogen burning and later helium fusion accompany expansion into a red giant. The outer layers are lost and a white dwarf remains without sustained fusion.
氢壳层燃烧和后续氦聚变伴随外层膨胀成为红巨星，外层散失后留下无持续聚变的白矮星。
More massive stars can fuse heavier nuclei, then undergo core collapse and a supernova.
更大质量恒星可聚变更重元素，最终核心坍缩并超新星爆发。
More fuel does not guarantee longer lifetime because their consumption rate is much greater.
燃料多不保证寿命长，因为消耗速率可能高得多。
For recession, successive wavefronts arrive farther apart; for approach they arrive closer together.
远离时相邻波面到达的间隔增大，接近时减小。
Compare the same identified spectral line so the change measures motion rather than a different atomic transition.
必须比较同一条已识别谱线，避免把不同原子跃迁误当作运动。
Pure transverse motion is not the recession speed in this simple model.
简单模型中的退行速度是视线方向分量，不是纯横向速度。
Absorption lines come from photons matching allowed atomic energy differences.
吸收线源于光子能量匹配允许的原子能级差。
The denominator is the rest wavelength.
分母使用静止波长。
Predict the sign before calculating: a longer observed wavelength indicates recession.
先判断符号，观察波长变长表示远离。
Use the low-speed approximation and do not extend it to arbitrary relativistic speeds.
使用低速近似，不能推广到任意相对论速度。
Attempt the numerical value and a sentence interpreting it.
先计算数值，再用一句话解释其意义。
The redshift is positive point zero one. Multiplying by light speed gives three million metres per second away along the line of sight.
红移为正零点零一，乘光速得到沿视线远离速度三百万米每秒。
The observed frequency is lower, so observed minus rest frequency divided by rest frequency is negative.
观察频率更低，所以观察频率减静止频率再除以静止频率为负。
At small shifts its magnitude approximately matches the redshift; its sign does not.
小位移下其绝对值近似等于红移，但符号不同。
Two opposite limbs do not each have the full observed wavelength separation as their individual shift.
两侧边缘的单独位移都不是观测谱线的全部间隔。
If their shifts are equal and opposite, each is half the separation.
若两侧位移等大反向，每侧位移是间隔的一半。
Use that individual shift to infer line-of-sight speed and then rotation period.
先求视线速度，再求自转周期。
The interpretation as equatorial speed requires the appropriate viewing geometry, not an unstated universal assumption.
把它当作赤道速度需要题目给定的观察几何条件，不能默认普遍成立。
Using one megaparsec as three point one \times ten to the twenty-two metres, convert seventy kilometres per second per megaparsec into two point two six \times ten to the minus eighteen per second.
一兆秒差距取三点一乘十的二十二次方米，将每兆秒差距每秒七十千米换为二点二六乘十的负十八次方每秒。
Its reciprocal is about four point four three \times ten to the seventeen seconds.
其倒数约四点四三乘十的十七次方秒。
Larger Hubble constant means smaller reciprocal time.
哈勃常数越大，倒数时间越小。
This is a timescale; treating it as age requires assumptions about past expansion.
这是时间尺度，把它解释为年龄需要对过去膨胀作假设。
Galaxy motions and rotation provide evidence for gravitating matter that does not emit enough detectable light to explain those effects.
星系运动和转动显示，可见物质不足以解释全部引力效应，由此推断暗物质。
Uncertainty in matter content and expansion history affects age and fate predictions. Modern models also include dark energy.
物质含量和膨胀历史的不确定性影响年龄与命运预测，现代模型还考虑暗能量。
One measurement of Hubble constant alone does not prove eternal expansion or future collapse, and expansion is not an explosion from a centre into pre-existing empty space.
单独一个哈勃常数测量不能证明永远膨胀或最终坍缩，宇宙膨胀也不是从中心向预先存在的空空间爆炸。
Return to the matching reference section when a practice error reveals a knowledge gap.
练习错误暴露知识缺口时，回到对应讲义章节。
Reattempt an equivalent skill question before returning to the authentic set.
再做同类技能题，之后回到真题。
Track whether an error came from a model, units, graph interpretation or explanation.
记录错误属于模型、单位、读图还是解释。
The reserved questions remain separate for later assessment; familiarity with a taught example is not evidence of independent mastery.
预留试题与教学材料分开，用于后续评估。 熟悉讲过的例子不能证明独立掌握。
Answer these without advancing.
先不翻页作答。
Connect constant-temperature phase change to molecular potential energy. Explain why background does not obey the test source distance or decay law.
把恒温相变与分子势能相连，解释本底为何不服从测试源的距离或衰变规律。
Finally read the acceleration-displacement slope as minus angular frequency squared, not minus angular frequency.
最后注意加速度位移图斜率是负角频率平方，而非负角频率。
Temperature tracks average random kinetic energy, so changing molecular arrangements can increase internal energy without a temperature rise.
温度反映平均无规则动能，因此分子排列改变可使内能增加而不升温。
Background must be separated before any ratio or logarithm based on the test source.
以测试源为依据作比值或取对数前，必须把本底分离。
The negative restoring slope of twenty-five per second squared gives angular frequency five radians per second.
负二十五每二次方秒的回复斜率对应五弧度每秒角频率。
Explain each answer, rather than quoting its keyword alone.
每个答案都要解释，不能只报关键词。