Unit 6: Keterampilan Praktis Fisika II
Pearson Edexcel · International A-Level · Physics Topik 6 16:22 Narasi bahasa Inggris · Subtitle bahasa Inggris + 中文 disematkan langsung
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Two students measure the same damped pendulum.
两名学生测量同一个阻尼摆。
One writes down amplitudes. The other processes them with logarithms, draws one straight line, and reads a decay constant and an initial amplitude from it.
一人只记下振幅，另一人对数据取对数、画一条直线，并从线上读出衰减常数和初始振幅。
That second step, from a reading to a result, is what Unit 6 tests.
从读数到结果的这一步，正是第六单元考查的内容。
The paper is eighty minutes, fifty marks, and at least twenty of them are Level-2 mathematics.
考试为80分钟、50分，其中至少20分为二级数学。
It is about the experiments you met in Units four and five; doing real practical work remains the foundation.
内容围绕第四、五单元的实验；真实动手实验仍是基础。
Unit three asked you to measure well, with half a resolution and half a range. Unit six builds on that with three new tools.
第三单元要求测得准：半分度值、半极差。
You will compound percentage uncertainties through a calculation. You will linearise models with natural logs and base-ten logs. And you will turn a result with its uncertainty into a justified conclusion about accuracy.
第六单元在此基础上增加三个工具：把百分比不确定度通过计算进行复合；用自然对数和常用对数把模型线性化；把带不确定度的结果转化为关于准确度的有依据结论。
Sheets six point one to six point four follow the paper's own flow.
6.1到6.4四张练习沿试卷自身的流程展开。
A plan that never names its graph is unfinished.
没有写明图像的计划是不完整的。
Work backwards from the model. Write the predicted relation, rearrange it into a straight-line form, and state what will prove it, the expected gradient or a line through the origin. Only then choose the readings.
要从模型倒推：先写出预测关系式，整理成直线形式，再说明什么能证明它——期望的斜率，或过原点的直线。
The six-mark devise-a-method questions expect the graph inside the answer.
然后再选读数：至少五个分布均匀的设置。 六分的“设计方案”题要求答案中包含图像。
Keep this table in mind. An exponential straightens with natural logs; a power law straightens with base-ten logs; an inverse relation straightens against one over the variable; and the mass on a spring gives period squared against mass.
记住这张表：指数关系用自然对数变直；幂律关系用常用对数变直；反比关系对自变量取倒数后变直；弹簧振子是周期平方对质量作图。
In every case the gradient and the intercept carry the physical quantities you want.
每种情形中，斜率和截距都承载着你要求的物理量。
The zero-pressure point cannot be reached with water baths, but a straight line can be extended to it.
零压点无法用水浴达到，但直线可以外推到那里。
Record paired pressure and temperature readings from ice to boiling water, at least five of them, stirring and reading the thermometer perpendicular.
从冰水混合物到沸水至少记录五组压强与温度读数，搅拌液体并垂直读数。
Plot pressure against temperature, draw the best-fit line, and extrapolate to zero pressure. The temperature intercept estimates absolute zero.
作压强—温度图，画最佳拟合直线，外推到零压强，温度轴上的截距就是绝对零点的估计值。
The answer inherits the scatter of the fitted line, which is the price of a point you cannot measure directly.
这个答案继承了拟合线的离散程度，这是测量不到的点必须付出的代价。
Pause the video and write your own method.
暂停视频，自己写出方案。
A light sensor sits a fixed distance from a filament bulb, and the model says power equals a constant times the fourth power of the sensor reading.
光传感器位于灯泡前方固定距离处，模型为功率与传感器读数的四次方成正比。
Name two control variables with the physical reason each matters, then choose your graph and state what would prove the model.
写出两个需要控制的变量及各自的物理原因，再选择作图方式并说明什么能证明该模型。
Take up to a minute.
限时一分钟。
Here is the answer.
这是答案。
The bulb current and supply change the brightness, so keep the electrical setting steady while you take each reading of current and voltage, and calculate the power.
灯泡电流和电源改变亮度，所以每取一组电流电压读数时电学条件要保持稳定，再计算功率。
Distance and alignment are controlled because the received light depends on them, and background light adds to the sensor reading.
距离和角度要控制，因为接收到的光取决于它们；背景光会叠加到传感器读数上。
Then lg P against lg X should be straight with gradient four.
然后作lg P对lg X图，斜率为4即证实模型。
Log graphs are not only for exponentials; the specification's own example is a gas pressure-volume power law.
对数图像不只用于指数关系，考试说明中自己的例子就是气体压强—体积的幂律关系。
Meter placement decides whether the circuit measures the investigation or something else.
电表的位置决定电路测量的是研究对象还是别的东西。
The ammeter goes in series, so the current flows through it. The voltmeter goes in parallel with the component under test, and only that component.
电流表串联接入，让电流流过它；电压表只并联在待测元件两端——若跨接两个元件，读数就不属于任何一个。
A variable resistor in series with a fixed supply gives you the spread of readings.
可变电阻与固定电源串联可提供分布的读数。
When a two-position switch moves a charged capacitor into a new circuit, draw the voltmeter across the one capacitor the question names.
当双掷开关把充好电的电容接入新电路时，电压表要画在题目点名的那只电容两端。
Every improvement or precaution names its mechanism.
每项改进或防护都要说明机理。
Hot water and immersion heaters burn, so move glass with tongs and clamp the heater so its leads cannot topple the beaker.
热水和浸入式加热器会烫伤，用夹钳移动玻璃器皿，并固定加热器以免导线碰倒烧杯。
Electrolytic capacitors have polarity, a working voltage, and stored charge: connect correctly, stay below the rating, discharge before handling.
电解电容有极性、额定电压和储存的电荷：正确接线、低于额定值、处理前先放电。
A data logger with probes records paired readings at the same moment, removes parallax, and its probes usually resolve better than a liquid thermometer.
带探头的数据采集器能同时记录成对读数、消除视差，分辨率通常优于液体温度计。
Say which of these your investigation gains, because more accurate alone names no mechanism.
要说明你的实验到底获得哪一项好处，因为只说“更准确”并没有指出机理。
A damped pendulum's turning point moves quickly past the rule, and you watch it at an angle.
阻尼摆的转折点很快掠过刻度尺，而且你是斜着看的。
Claiming the amplitude to the nearest millimetre is claiming precision the eye cannot deliver.
把振幅记到毫米，就是声称眼睛达不到的精度。
The nearest five millimetres is defensible.
记到最近的5毫米才站得住脚。
This argument works both ways: criticise a recording resolution that is too fine, and be ready to justify a coarse one.
这个论证是双向的：既能批评过细的记录精度，也要能为较粗的精度辩护。
Timing many oscillations beats timing one, because your reaction time is nearly a constant absolute uncertainty: spread over ten swings it becomes a tenth in the period.
计时多个振动优于单个，因为反应时间近似是固定的绝对不确定度：分摊到十次振动后，每个周期只占十分之一。
Put the timing marker at the centre of the oscillation, where the mass moves fastest and the moment of passing is clearest.
把计时标记放在振动中心，那里质量运动最快，经过时刻最清晰。
Start after several oscillations so the motion has settled, record repeats of the total time, and only then divide.
等振动稳定后再开始计时，记录总时间的重复值，最后再除以次数。
Pause here.
在这里暂停。
One row of times reads one point one, one point zero six, one point zero eight seconds. A later row holds two point six one beside one point nine five and one point nine one.
某一行时间读数为1.1、1.06、1.08秒；后一行在1.95和1.91旁边出现2.61。
Find two separate faults and say what the honest response to each is.
找出两个独立的问题，并分别说明正确的处理方式。
Take up to a minute.
限时一分钟。
The one point one entry does not use the two decimal places its timer resolves, so the column mixes precisions: record to the instrument's resolution every time.
1.1这个读数没有保留秒表两位小数的分辨率，使该列精度混杂：每次都应按仪器分辨率记录。
The two point six one entry does not fit the pattern of one point nine five and one point nine one: identify it as a possible anomaly, check the method and repeat it. Never silently delete the least convenient value.
2.61与同行1.95、1.91的模式不符：应标记为可能的异常值，检查方法并重测，绝不能悄悄删掉最不方便的数据。
And every column heading carries its unit once, in the heading.
另外，每个表头只写一次单位。
Measure a coil or cylinder diameter at several orientations and average: the object is not perfectly round, and the average attacks that random variation.
沿几个方位测直径再取平均：物体不是正圆，平均能减小这种随机变化。
Close the jaws first and read the zero: an offset shifts every reading the same way, a systematic error that repeating never removes; subtract it.
先闭合卡尺读零点：零点偏移会把每个读数整体移动，这是重复测量消不掉的系统误差，应予减去。
Use the micrometer ratchet so the jaws tighten the same amount each time.
使用千分尺棘轮，保证每次夹紧力一致。
Justifying an instrument is quantitative.
论证仪器选择要靠计算。
A micrometer resolves one hundredth of a millimetre, so a single reading of one point two one millimetres carries an uncertainty of half a division, five thousandths of a millimetre. As a percentage that is zero point four per cent.
千分尺的分度值是0.01毫米，一次1.21毫米的读数带有半分度即0.005毫米的不确定度，换算成百分比是0.4%。
The small percentage is the justification. Saying the instrument is accurate is not.
这个很小的百分比才是选择它的理由，而“它很准确”不是。
The examiner's graph checklist is short. Enough points, spread over more than half of each axis, on scales that are easy to subdivide; multiples of three or seven are not. Points plotted to within half a small square. A thin continuous best-fit line that balances the scatter, and a check of whether the points are actually straight before drawing one. Criticising is not listing everything you know; point at the actual defect.
阅卷人的图像检查清单很短：数据点足够多，铺满每根坐标轴的一半以上；刻度易于细分——3或7的倍数不行；描点误差不超过半小格；一条细而连续的最佳拟合线，并且画线前先确认点确实呈直线。 批评不是罗列所有知识，而是指出真实存在的缺陷。
The Unit three building blocks still stand. A single reading carries half the instrument's resolution. Repeated readings carry half the range, and the reading furthest from the mean is an accepted alternative.
第三单元的基础依然成立：单次读数的不确定度取仪器分度值的一半；重复读数取极差的一半，离均值最远的读数也可作为替代。
Percentages are the working currency because doubling a length leaves its percentage unchanged.
百分比是通用语言，因为长度加倍其百分比不变。
Quote them to one or two significant figures.
百分比保留一到两位有效数字。
Three rules compound uncertainties. A quantity raised to a power has its percentage uncertainty multiplied by that power, so an area doubles its length's percentage. Products and quotients add their percentage uncertainties, so the density of a cube triples the side's percentage and adds the mass's. Sums and differences add absolute uncertainties instead, which is why subtracting two similar measurements leaves a large percentage. Combine the rules as needed.
复合不确定度有三条规则：量取幂时，百分比不确定度乘以幂次，面积就是长度百分比的两倍；相乘相除时，百分比不确定度相加，立方体的密度是边长百分比的三倍加质量的百分比；相加减时，绝对不确定度相加，这就是两个相近读数相减后百分比很大的原因。 需要时可组合使用这些规则。
Here is the classic A2 compounding.
这是经典的A2复合题。
A sphere rolls a distance s down a ramp of height difference delta h in time t, and the model rearranges to g equals fourteen s squared over five t squared delta h.
小球沿斜面滑过距离s、下落高度差Δh、用时t，模型整理得g等于14s²除以5t²Δh。
The distance is squared, the time is squared, and the height difference appears once, so the percentage in g adds twice the s percentage, twice the t percentage, and the height percentage. That gives zero point two two, plus one point seven eight, plus four point seven six: six point eight per cent.
s取平方、t取平方、Δh取一次，所以g的百分比等于2倍s百分比加2倍t百分比加Δh百分比：0.22加1.78加4.76，得6.8%。
The height difference dominates, so improving the timing would barely change the result.
高度差是主导项，改进计时几乎不改变结果。
Compounding is also an improvement finder.
复合计算也能指出改进方向。
Pause and compound it yourself.
暂停并自己复合计算。
The diameter terms are D equals fifty-nine point four millimetres plus or minus zero point two, and d equals thirteen point nine plus or minus zero point one.
D为59.4±0.2毫米，d为13.9±0.1毫米。
Show that the percentage uncertainty in D squared over two d squared is about two per cent.
证明D²/(2d²)的百分比不确定度约为2%。
Take up to a minute.
限时一分钟。
The diameters carry zero point three four and zero point seven two per cent.
两个直径的百分比分别为0.34%和0.72%。
Squaring each term doubles its percentage, and the quotient adds them: twice each, giving two point one per cent.
平方使百分比翻倍，相除再相加：各乘二后得2.1%。
Give compound percentages to a minimum of two significant figures, because one digit here would hide the difference the question is testing.
复合百分比至少保留两位有效数字，因为一位数会掩盖题目要考查的差别。
With a percentage uncertainty, turn the result into an interval and compare. Nine point six with six point eight per cent spans eight point nine to ten point three, and the accepted nine point eight one lies inside, so the measurement is accurate to the precision claimed. Without an uncertainty, compare percentages instead: a difference below about five per cent suggests accuracy.
有了百分比不确定度，就把结果化为区间再比较：9.6±6.8%跨越8.9到10.3，公认值9.81在其中，所以测量在此精度下是准确的。
And state the honest strength: a value of zero point two seven six plus or minus six per cent contains the steel value zero point two six five, so the spring could be steel; the interval does not prove it is.
若没有不确定度，就比百分比：差值小于约5%可认为准确。 结论要诚实：0.276±6%包含钢的0.265，说明弹簧“可能”是钢制的，区间并不能证明它一定是。
A damped pendulum's amplitude follows A equals A nought e to the minus lambda n.
阻尼摆的振幅满足A=A₀e^(−λn)。
Take the natural log of both sides and you get ln A equals ln A nought minus lambda n. That is the straight-line form: plot ln A against n, and the gradient is minus lambda while the intercept is ln A nought.
两边取自然对数得ln A = ln A₀ − λn，这正是直线形式：作ln A对n图，斜率为−λ，截距为ln A₀。
The minus sign is part of the physics, not a slip; the line slopes down because the amplitude decays.
负号是物理本身，不是笔误——图线向下倾斜，因为振幅在衰减。
Convert both read-offs back into physics.
把两个读数换回物理量。
The gradient gives lambda directly, and since n counts oscillations, lambda has no unit; had the axis been time in seconds, lambda would carry per second.
斜率直接给出λ；由于n数的是振动次数，λ没有单位——若横轴是以秒计的时间，λ就带每秒。
The intercept two point two nine five is the logarithm of A nought in centimetres, because that is what the axis carried: e to the two point two nine five is nine point nine centimetres.
截距2.295是以厘米为单位的A₀的对数，因为轴上是ln(A/cm)：e的2.295次方等于9.9厘米。
Process every value to three decimal places and take the gradient from a large triangle on the line, not from data points off it.
所有对数值保留三位小数，用线上大三角形求斜率，而不是用偏离线的数据点。
Pause for one more.
再暂停一次。
Light passes through glass of thickness w, and the voltmeter reading follows V equals A e to the minus B w.
光穿过厚度为w的玻璃，电压表读数满足V=Ae^(−Bw)。
Explain how a graph of ln V against w determines B, and what the intercept would give you.
说明如何用ln V对w的图像求出B，以及截距能给出什么。
Take up to a minute.
限时一分钟。
Taking ln of the model gives ln V equals ln A minus B w, so the gradient is minus B and the intercept is the natural log of A. Both constants come from the one line.
对模型取对数得ln V = ln A − Bw，斜率为−B，截距是ln A，两个常数都来自同一条线。
For a power law use base-ten logarithms instead, where the gradient is the power and the intercept is lg k.
幂律关系改用常用对数：斜率是幂次，截距是lg k。
Keep the bases separate: ln for exponentials, lg for power laws, and the unit always divided out before the logarithm, so the axis reads lg of P over watts, never lg P with watts attached.
两种底数不要混用：指数用ln，幂律用lg；取对数前先把单位除掉，坐标轴写lg(P/W)，绝不写带单位的lg P。
A fitted line answers questions the raw table cannot.
拟合线能回答原始数据表回答不了的问题。
With B equal to zero point zero zero eight eight per millimetre, the thickness that dims the reading to sixty per cent follows from ln of zero point six zero equals minus B w: fifty-eight millimetres.
B为每毫米0.0088时，把读数降到60%所需厚度由ln 0.60 = −Bw得出：58毫米。
With two-millimetre filters that is twenty-nine, but twenty-nine filters leave the brightness marginally above sixty per cent, so thirty are needed.
以2.0毫米的玻璃片计是29片——但29片时亮度仍略高于60%，所以需要30片。
Round up when the requirement is at most.
要求“至多”时要向上取整。
Prediction questions test the direction of the final rounding as much as the algebra.
预测题既考代数，也考最后取整的方向。
The intercept often sits where there is no data, because the first reading was taken five oscillations in. So the intercept is only as good as the fitted line.
截距常常位于没有数据的地方，因为第一次读数在第5次振动之后。
Large scatter leaves the line's position uncertain and moves the intercept. A systematic error in judging amplitude shifts the whole line. And the model itself may not hold over the earliest swings, if the starting angle was large.
所以截距的可信度取决于拟合线：离散大会使线的位置不确定、截距漂移；判断振幅的系统误差会整体移动直线；若起始摆角较大，模型在最初几次振动中可能不成立。
Extrapolating beyond the data assumes the model still holds; that is the sentence that earns the mark.
向数据之外外推要假设模型仍成立——这句话就是得分点。
Say which reason applies to the experiment in front of you.
要说出哪条理由适用于眼前的实验。
Before you attempt a Unit six question, check that you can do all of this. Choose the processing graph before writing the method, and place both meters where the measurement is. Judge recording resolution, tables and graphs by naming the actual fault. Compound uncertainties for powers, products and differences, and let the largest term name the improvement. Linearise exponentials with ln and power laws with lg, convert gradients and intercepts back into physical quantities, and round predictions the right way. Sheets six point one to six point four train each skill in order.
做第六单元题目前，确认这些你都会：先选处理图像再写方案；两块电表接在测量所在的位置；用点名缺陷的方式评判记录精度、表格和图像；按幂、乘除、加减复合不确定度，并让最大项指出改进方向；指数用ln、幂律用lg线性化，把斜率和截距换回物理量，预测取整方向正确。 6.1到6.4四张练习依次训练这些技能。