נושא 1: אנרגיה
AQA · GCSE · Physics נושא 1 29:58 קריאת קול באנגלית · תרגום אנגלי + סינית שרוף בתוך הסרטון
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תרגום טקסט
Try the motor starter before continuing.
先尝试电动机问题。
Identify what power measures, use the equation, and give the unit. Explain the difference between a rate and an energy store.
说明功率表示什么，使用公式并写出单位，再区分能量转移速率与能量储存。
Power is energy transferred per second.
功率是每秒转移的能量。
Divide sixty joules by three seconds to obtain twenty watts.
六十焦耳除以三秒得到二十瓦。
A supplied equation still requires choosing the quantities and interpreting the result.
即使提供公式，也需要选择正确的量并解释结果。
Use the examples to recognise energy stores.
利用例子识别八种能量储存。
Heating, mechanical work and electrical work describe transfers, not additional stores.
加热、机械做功和电流做功描述能量转移，而不是额外的储存类型。
A chosen system may contain several stores.
选定的系统可以包含多种能量储存。
Name the increasing store in each case: heated water, raised water, and stretched cords.
分别说出热水、被提升的水以及拉伸绳索中增加的能量储存。
Use a store name rather than the transfer word heat or electricity.
使用储存类型名称，而不是用热或电来代替。
Heating water increases its thermal store.
加热水使其内能增加。
Raising water increases the gravitational potential store of the water and Earth system.
提升水使水与地球系统的重力势能增加。
Stretching the cords increases their elastic potential store.
拉伸绳索使其弹性势能增加。
Track the decreasing and increasing stores.
追踪减少和增加的能量储存。
For an upward-moving object, kinetic decreases and gravitational potential increases.
物体上升时动能减少、重力势能增加。
During braking and collisions, kinetic energy is transferred to thermal stores.
制动和碰撞把动能转移到内能。
A battery-driven vehicle transfers energy from chemical to kinetic stores; a kettle transfers energy electrically to the water.
电池驱动车辆把化学能转移为动能；电热水壶通过电流做功把能量转移给水。
Start with the energy balance for the motor. Calculate what is dissipated, then use proportional arrow widths.
先写出电动机的能量平衡，计算耗散的能量，再按比例画箭头宽度。
Explain where the energy went rather than saying that it disappeared.
解释能量转移到了哪里，而不是说能量消失。
One hundred joules enters; eighty joules is useful, so twenty joules is dissipated to thermal stores.
输入一百焦耳，其中八十焦耳为有用能量，所以二十焦耳耗散到内能。
The arrow shaft widths have the ratio five to four to one. Arrowhead size and arrow length do not encode energy.
箭头主体宽度之比为五比四比一；箭头尖端大小和长度不表示能量大小。
Choose the equation from the store that changes.
根据发生变化的能量储存选公式。
Kinetic energy depends on speed squared. Elastic potential energy depends on extension squared, provided the spring remains proportional. Gravitational potential energy gained depends on the height increase.
动能与速度平方成正比；在比例关系成立时，弹性势能与伸长量平方成正比；重力势能的增加量取决于上升高度。
Use metres, kilograms and the stated gravitational field strength.
使用米、千克和题目给出的重力场强度。
Try the three parts before the answer.
先独立完成三问。
Distinguish total length from extension, convert centimetres to metres, then use the elastic energy equation.
区分总长度与伸长量，把厘米换成米，再使用弹性势能公式。
For the last part, use the square rather than doubling the energy.
最后一问注意平方关系，不能直接把能量加倍。
The extension is twelve centimetres, or zero point one two metres.
伸长量为十二厘米，也就是零点一二米。
Substituting gives zero point three six joules.
代入后得到零点三六焦耳。
Doubling extension gives four times the stored energy, one point four four joules.
伸长量加倍时，储能变为四倍，即一点四四焦耳。
This statement depends on remaining in the proportional range.
这个结论以仍处于比例范围内为前提。
Try the energy calculation and then rearrange the kinetic energy equation for speed.
先算弹性势能，再把动能公式变形求速度。
Use the ideal transfer assumption given in the question. Finally explain how thermal transfers change the result.
使用题目给出的理想转移假设，最后解释热能转移会怎样改变结果。
The spring initially stores zero point three six joules.
弹簧最初储存零点三六焦耳。
For this ideal model that becomes the plane’s kinetic energy.
在这个理想模型中，这些能量全部变成飞机的动能。
Rearrange kinetic energy before substituting: speed is the square root of twice the energy divided by mass.
先把动能公式变形：速度等于二倍动能除以质量后开平方。
The ideal result is six metres per second, not a guarantee that the real plane achieves it.
理想结果是每秒六米，不代表真实飞机一定达到这个速度。
Apply the chain to the actual exam data.
把能量计算链应用于真实考题的数据。
Calculate one cord and then both, before using gravitational potential energy.
先算一根绳，再算两根绳的总能量，随后使用重力势能公式。
Explain a specific mechanism for the lower real height.
最后说明真实高度较低的具体原因。
One cord stores twenty-three thousand five hundred and twenty joules.
一根绳储存两万三千五百二十焦耳。
Add the energy of both cords to get forty-seven thousand and forty joules.
两根绳的能量相加得到四万七千零四十焦耳。
In the ideal model this becomes gravitational potential energy at the highest point.
在理想模型的最高点，这些能量变成重力势能。
Divide by mass times gravitational field strength to obtain a twenty metre height gain.
除以质量与重力场强度的乘积，得到二十米的上升高度。
Do not double the extension: that would quadruple the energy of one cord.
不要把伸长量加倍，那会使一根绳的能量变为四倍。
Link air resistance to the smaller height: because of air resistance, not all of the elastic energy becomes gravitational potential energy.
把空气阻力与高度降低联系起来：由于空气阻力，弹性势能没有全部转化为重力势能。
The wasted energy has not disappeared: some is transferred to the surroundings, so the gain in gravitational potential energy is smaller, and the height is lower.
被浪费的能量并没有消失：一部分转移到周围环境中，所以重力势能的增加量变小，高度也就降低。
Specific heat capacity is the energy needed to raise one kilogram of a substance by one degree Celsius.
比热容表示让一千克某种物质升温一摄氏度所需的能量。
The equation uses the temperature change, not the final temperature.
公式使用温度变化，而不是最终温度。
With equal mass and equal energy input, a material with a greater specific heat capacity has a smaller temperature rise.
在质量相同、吸收能量相同的条件下，比热容较大的物质温升较小。
This does not by itself establish how quickly a spoon heats in soup.
仅凭比热容不能判断汤里的勺子升温多快。
Try the conversion, temperature change and rearrangement before continuing.
先完成单位换算、温度变化和公式变形。
The energy stated is gained by the block, so this question does not require a correction for transfer to the surroundings.
题目给出的是金属块获得的能量，因此这里不需要另行扣除传给环境的能量。
Twenty-six kilojoules is twenty-six thousand joules.
二十六千焦等于两万六千焦。
The rise is twenty-eight degrees Celsius.
温升为二十八摄氏度。
Divide energy by mass times temperature change to obtain about four hundred and sixty joules per kilogram per degree Celsius.
能量除以质量与温升的乘积，得到约四百六十焦每千克每摄氏度。
Substituting the final temperature would overstate the temperature change and make the calculated value too small.
若把最终温度当作温升，就会把温升算大，使比热容结果偏小。
This exam question changes the unknown and gives specific heat capacity in kilojoules.
这道考题改变了未知量，并用千焦给出比热容。
Convert consistently, rearrange, then substitute.
先统一单位，再变形、代入。
Calculate a temperature change, not a final reading.
求的是温度变化，不是最终温度。
Converting the specific heat capacity gives one thousand and ten joules per kilogram per degree Celsius.
换算后的比热容是一千零十焦每千克每摄氏度。
Divide zero point zero one three zero joules by the product of mass and specific heat capacity. The temperature change is about four hundred and ninety-five degrees Celsius.
把零点零一三零焦除以质量与比热容的乘积，得到约四百九十五摄氏度的温升。
The initial temperature is still needed to calculate the final reading.
若要计算最终温度，还需要知道初温。
The ammeter is in series with the heater. The voltmeter measures potential difference across the heater.
电流表与加热器串联，电压表测量加热器两端电压。
Measure the block’s mass with a balance and time with a stopwatch.
用天平测质量、秒表测时间。
Electrical energy supplied equals voltage times current times time when voltage and current remain constant. It is not automatically all gained by the block.
在电压和电流不变时，供给电能等于电压乘电流乘时间，但这些能量不一定全部被金属块吸收。
Connect each reading to the quantity it determines. Mass comes from the balance; energy from voltage, current and time; temperature change from the thermometer.
把每个读数与所求量联系起来：天平给出质量，电压、电流和时间给出供能，温度计给出温度变化。
The practical handbook uses temperature against energy supplied and warns about initial curvature.
实验手册要求画温度对供给能量图，并提醒初始部分可能弯曲。
A straight-region gradient can determine the material value, but heat loss still affects the estimate.
可用直线段斜率求材料比热容，但散热仍会影响估计。
Work out both values before predicting the direction.
先分别求出两个数值，再判断误差方向。
The example specifies an underestimated temperature change.
这里明确给出温升被低估。
A vague statement about poor thermometer contact does not by itself establish which change was recorded.
仅说温度计接触不好，不能自动确定记录的温度变化是偏大还是偏小。
The actual data give five hundred joules per kilogram per degree Celsius; the underestimated rise gives six hundred.
实际数据得到五百焦每千克每摄氏度；低估温升时算出六百。
A smaller denominator makes the calculated value larger.
分母变小会使结果变大。
Heat loss can also produce an overestimate when all electrical input is treated as energy gained by the block.
若把全部电能当作金属块吸收的能量，散热也会导致比热容被高估。
Explain the measured quantities rather than memorising that poor contact must give one particular direction.
应根据测量量解释，不能死记接触不良必定产生某个方向的误差。
Both motors lift the same mass through the same height.
两台电动机把相同质量提升到相同高度。
Try the energy and power calculations, then distinguish transfer rate from efficiency.
先计算能量和功率，再区分转移速率与效率。
There is no information about electrical input.
题目没有给出输入电能。
Each motor transfers four hundred joules usefully.
两台电动机的有用能量转移都为四百焦耳。
Dividing by two seconds gives two hundred watts for A; dividing by four seconds gives one hundred watts for B.
A 用两秒，功率二百瓦；B 用四秒，功率一百瓦。
The faster motor has twice the useful power, but this does not establish its efficiency.
较快的电动机有用功率加倍，但这不能确定其效率。
Calculate the power associated with gaining height.
计算与高度增加有关的功率。
This is an average over the stated time.
这是在给定时间内的平均值。
Distinguish the useful gain from other transfers in the athlete’s body.
区分有用势能增加和人体内其他能量转移。
Convert one hundred and seventy-five centimetres to one point seven five metres. The gravitational potential gain is one thousand and twenty-nine joules.
一百七十五厘米换成一点七五米，重力势能增加一千零二十九焦耳。
Dividing by one point four seconds gives seven hundred and thirty-five watts of average useful power.
除以一点四秒，得到七百三十五瓦的平均有用功率。
Heating and other transfers are additional to this useful gain.
加热等其他转移不包含在这个有用势能增加中。
Here the unknown is energy rather than power.
这次未知量是能量而不是功率。
Convert megawatts to watts and multiply by time in seconds.
把兆瓦换成瓦，再乘以秒数。
The question gives output, with no total input for an efficiency calculation.
题目只给出输出，没有总输入，不能据此算效率。
Five hundred megawatts is five hundred million watts.
五百兆瓦等于五亿瓦。
Multiplying by three thousand six hundred seconds gives one point eight times ten to the twelve joules.
乘以三千六百秒，得到一点八乘十的十二次方焦耳。
Watts alone do not label a quantity as useful output: read what the question says.
瓦这个单位本身并不表示有用输出，必须看题目描述。
Choose the system before accounting for energy.
先选定系统再做能量核算。
Find the missing kinetic energy during the fall, then follow the energy through impact.
求下落时缺少的动能，再追踪碰撞后的能量。
In this model the chosen whole system exchanges no energy with its outside.
在这个模型中，选定的整个系统不与外界交换能量。
At the start, the gravitational potential contribution is twenty joules.
开始时重力势能为二十焦耳。
During the fall, five joules of gravitational potential plus fifteen joules of kinetic energy still makes twenty.
下落途中，五焦耳重力势能加十五焦耳动能，总量仍为二十焦耳。
Once the ball and floor are at rest, the initial energy has been redistributed to thermal stores in this simplified model.
球和地面静止后，在这个简化模型中，初始能量转移到了内能储存。
Sound can carry energy within the chosen surroundings before dissipating.
声音可以先在系统内传递能量，随后耗散。
Ball alone is a different system that can exchange energy with other objects.
若只把球作为系统，它就能与其他物体交换能量。
Choose a method that addresses the transfer mechanism.
选择能针对转移机制的方法。
For the wall comparison keep dimensions and temperature difference fixed.
比较墙体时，应保持尺寸和温差相同。
Think about the competing effects if both thickness and conductivity change.
如果厚度和热导率同时改变，要考虑它们可能产生相反影响。
Lubrication acts on friction, while insulation slows thermal transfer.
润滑针对摩擦，保温材料减慢热传递。
Conductivity and thickness comparisons require the other conditions to remain the same.
比较热导率或厚度时，其他条件应保持相同。
If both change in opposing directions, qualitative labels do not establish the ranking.
若两项变化产生相反作用，仅凭定性描述不能确定快慢。
Plan the material comparison separately from the thickness comparison.
把材料比较与厚度比较分开设计。
State the measured temperature change over a fixed time and identify what must stay the same.
说明在固定时间内测量温度变化，并指出哪些条件应相同。
Repeats help you judge variation.
重复测量有助于判断结果的变化范围。
Use measured thickness when comparing different materials: equal layer counts need not mean equal thicknesses.
比较不同材料时要测量厚度：层数相同不一定表示厚度相同。
For one material, layer count can vary thickness.
对于同一种材料，可以通过改变层数改变厚度。
Keep the water, apparatus and starting conditions consistent, then compare cooling over the same time and assess repeat variation.
保持水量、装置和初始条件一致，再比较相同时间内的降温，并检查重复结果的差异。
These points are real readings from a cooling practical.
这些点是一次冷却实验的真实读数。
The lines just join the recorded readings; they are not a fitted physical law.
连线只是把记录的读数连起来，并不是拟合出的物理规律。
Compare the numbers of newspaper layers at the same time, and check the starting temperatures: the run with no layers starts at eighty-five degrees, and the two- and six-layer runs at eighty-six.
比较同一时刻不同报纸层数的结果，并检查初始温度：没有包报纸的一组从八十五摄氏度开始，两层和六层的从八十六摄氏度开始。
Calculate initial minus final temperature for each run.
分别用初温减末温。
Give evidence for the thickness comparison and identify the difference in starting conditions.
用数据支持厚度比较，并指出初始条件的差异。
The data compare newspaper thickness, not different insulating materials.
这些数据比较的是报纸厚度，不是不同保温材料。
The falls are twenty-eight, twenty-four and twenty degrees Celsius.
降温分别为二十八、二十四和二十摄氏度。
Six layers cool four degrees less than two layers over the same time, from the same starting temperature.
在初温相同、时间相同的情况下，六层比两层少降四摄氏度。
The zero-layer run starts one degree cooler.
零层实验的初温低一摄氏度。
A temperature-change calculation is useful but does not make unequal starting conditions identical.
计算温度变化很有用，但不能让不同的初始条件变得相同。
Do not rank the materials from the final readings alone.
不要只按最终读数给材料排名。
Calculate the changes, then ask whether material was the only changed factor.
先计算温度变化，再检查是否只改变了材料这一项。
Explain how a fairer investigation would separate the effects.
说明怎样通过更公平的实验区分各因素的作用。
Cotton falls fifteen degrees and foil ten, but thickness and starting temperature also change. These results do not isolate the material effect.
棉花降十五摄氏度，铝箔降十摄氏度，但厚度和初温也不同，因此结果不能单独说明材料效应。
Match the other conditions, measure cooling over equal times and compare repeats before drawing a supported conclusion.
应统一其他条件，在相同时间内记录降温，并比较重复结果，再作出有依据的结论。
The intended task is lifting, so useful output is mechanical power transferred to the load.
目标是提升重物，因此有用输出是传给负载的机械功率。
Calculate both ratios and express the difference in percentage points.
分别计算比值，用百分点表示差距。
Heating the surroundings is not useful output for this task.
对这个任务来说，加热环境不是有用输出。
Device A is zero point eight five efficient, or eighty-five percent. Device B is zero point seven five, or seventy-five percent.
A 的效率是零点八五，即百分之八十五；B 是零点七五，即百分之七十五。
The difference is ten percentage points.
两者相差十个百分点。
Efficiency has no unit because the same units cancel in the ratio.
比值中的同类单位相消，因此效率没有单位。
Rearrange before substituting.
先变形再代入。
Find useful light power, then use the power balance for the remainder.
求出有用光功率，再用功率平衡求剩余部分。
Distinguish the decimal zero point eight five from eighty-five percent.
区分小数零点八五与百分之八十五。
Multiplying the decimal efficiency by four watts gives three point four watts of useful light power.
小数效率乘以四瓦，得到三点四瓦有用光功率。
The remainder is zero point six watts transferred in other ways.
剩余零点六瓦以其他方式转移。
These outputs add to the input; the efficiency is eighty-five percent, not eighty-five as a decimal.
输出之和等于输入；效率是百分之八十五，而不是小数八十五。
Calculate useful fractions and useful energies for both actual methods. Then decide whether the higher-efficiency method necessarily provides the greater useful energy for the stated mass. Include the installation context.
先计算两种实际方法的有用比例和有用能量，再判断效率较高的方法是否一定能为给定质量提供更多有用能量，并考虑安装地点。
Method A gives twenty thousand one hundred and sixty kilojoules usefully; B gives three hundred and sixty-seven point five.
A 的有用能量为两万零一百六十千焦，B 为三百六十七点五千焦。
B has the greater useful fraction, seventy-five percent rather than sixty, but A has much more useful energy per hundred kilograms.
B 的有用比例较高，为百分之七十五而不是百分之六十，但 A 每一百千克水提供的有用能量多得多。
Include the stated location restriction and the need to insulate hot water.
比较时还要考虑题给地点限制和热水保温需求。
Match the improvement to the intended task.
把改进方法与目标任务对应起来。
Explain the energy pathway, then compare useful fractions at fixed input or required input at fixed useful output.
说明能量转移途径，再分别比较固定输入时的有用比例，以及固定有用输出时所需输入。
Keep those comparisons distinct.
不要混淆这两种比较。
Lubrication and insulation address different unwanted transfers.
润滑和保温针对不同的不需要的转移。
With fixed input, keeping more energy useful increases the ratio.
输入固定时，有用能量更多，效率比值就更高。
With fixed useful output, greater efficiency reduces the required input.
有用输出固定时，效率越高，所需输入越少。
Useful depends on the task; thermal energy is not universally wasted.
有用与否取决于任务，热能并不总是浪费的。
Separate replenishment from availability and environmental impact.
区分资源的补充、供电是否可用，以及环境影响。
Give a reason for each classification.
每种分类都要给出理由。
Pause to answer before checking the next slide.
先暂停作答，再核对下一页。
Wind is renewable but variable.
风能可再生，但输出会变化。
Nuclear fuel is non-renewable, yet generation need not depend on wind.
核燃料不可再生，但发电不必依赖风。
Bio-fuel can be renewed, but burning it still releases carbon dioxide.
生物燃料能得到补充，但燃烧仍释放二氧化碳。
No classification removes the need to examine impacts.
任何分类都不能代替对环境影响的分析。
Fossil fuels, nuclear fuel and bio-fuel have different uses and impacts.
化石燃料、核燃料和生物燃料的用途及影响不同。
Do not treat a power station as always available.
不要把电站看作始终可用。
The carbon dioxide comparison here concerns operation, not manufacture, mining or construction.
这里的二氧化碳比较针对运行阶段，不包含制造、采矿或建设。
All six resources are renewable.
这六种资源都可再生。
Predictability is not continuous availability: the timing of tides can be known while output still varies.
可预测并不等于持续可用：潮汐时间可预测，但输出仍变化。
Reservoir storage helps match demand only while enough water is available.
水库只有在水量足够时才能帮助满足需求。
Descriptions of generating machinery are outside this topic requirement.
本节不要求描述发电机械的过程。
Compare named effects rather than saying renewable means harmless.
应比较具体影响，而不是说可再生就无害。
Some effects are environmental, such as habitat change; others are technical or economic, such as suitable sites and cost.
栖息地变化属于环境影响，选址及成本则涉及技术或经济因素。
Manufacturing and construction also matter when judging the whole life of a system.
评价整个系统寿命期时，制造和建设同样重要。
A nuclear power station generated electricity on ninety-two percent of the days in a three hundred and sixty-five day year.
一座核电站在一年三百六十五天中，有百分之九十二的天数在发电。
Calculate how many days it generated.
计算它发电的天数。
Then two extensions: what does this tell you about a claim that nuclear stations are always available?
再思考两个拓展问题：这对“核电站总是可用”这一说法说明了什么？
And can a time percentage alone tell you the energy output or the efficiency?
仅凭时间百分比，能否知道能量输出或效率？
Pause and try it.
暂停并试着回答。
Nought point nine two times three hundred and sixty-five gives three hundred and thirty-five point eight days, about three hundred and thirty-six; either three hundred and thirty-five or three hundred and thirty-six is accepted.
零点九二乘以三百六十五，得三百三十五点八天，约三百三十六天；答三百三十五或三百三十六都可以。
These data concern one station's operating time, not its energy efficiency.
这些数据只涉及一座电站的运行时间，而不是它的能量效率。
These are made-up data, not a real country's record, and only two categories make up each total.
这些是虚构的数据，不是某个国家的真实记录，每期的总量只由这两类组成。
A terawatt-hour is a unit of energy.
太瓦时是能量单位。
Calculate the total energy in each period and the fossil-fuel percentage.
计算每期的总能量和化石燃料所占的百分比。
Then decide: does a smaller share mean a smaller amount?
然后判断：所占比例变小，是否意味着用量变少？
Pause and try it.
暂停并试着计算。
The total grows from one hundred to one hundred and fifty terawatt-hours.
总量从一百增至一百五十太瓦时。
Fossil share falls from eighty to sixty percent, but its amount rises by ten terawatt-hours.
化石燃料占比从百分之八十降到六十，但用量增加十太瓦时。
Distinguish percentage points from energy amounts, and do not infer total emissions from share alone.
区分百分点和能量用量，不能仅凭占比推断总排放。
Use the stated night-time need and the supplied facts.
利用夜间供电需求和题目给出的事实。
There is no price or storage capacity given, so a complete design choice needs more evidence.
题目没有价格或储能容量，因此完整方案需要更多证据。
Connect a social or economic constraint to this clinic.
把社会或经济限制与这家诊所联系起来。
A useful conclusion states its conditions.
有用的结论要说明条件。
Solar plus adequate storage is a possible plan, not a guarantee.
太阳能配足够储能是一种可能方案，不是保证。
Science can quantify demand and impacts; funding, planning and fair access also affect what is chosen.
科学可以量化需求和影响，但资金、规划和公平获得服务同样影响选择。
There is no universal final sentence that guarantees marks.
没有一句通用结尾能保证得分。
Attempt all parts without looking back.
先不翻前页，独立完成各部分。
The cart starts from rest, so its final kinetic energy is four joules.
小车从静止开始，因此最终动能为四焦耳。
Use the named equations, substitute with units, and identify the remaining transfer.
写出公式，带单位代入，并说明剩余能量的转移。
The final speed is four metres per second, efficiency eighty percent, and mean input power two point five watts.
最终速度为每秒四米，效率为百分之八十，平均输入功率为二点五瓦。
One joule is transferred by heating.
一焦耳通过加热转移。
Energy is conserved even though less of it is useful for the chosen task.
即使对该任务有用的能量减少，能量仍然守恒。
Diagnose the reasoning, not just a missing word.
诊断推理错误，而不只是补一个词。
For case A use energy divided by mass and temperature rise.
A 用能量除以质量和温升。
For B identify the confounding variable.
B 找出同时变化的变量。
For C distinguish replenishment from availability.
C 区分资源补充和供电可用性。
Check your explanation against each cause.
按每个原因核对解释。
Overcounting energy raises the calculated specific heat capacity in the stated case.
在给定条件下，高估吸收的能量会使计算的比热容偏高。
A controlled cooling comparison changes one intended variable.
公平冷却比较只改变预定变量。
A renewable resource may still need storage or backup to serve night-time demand.
可再生资源仍可能需要储能或备用供应来满足夜间需求。