Sujet 6 : Ondes
AQA · GCSE · Physics Sujet 6 22:33 Narration en anglais · Sous-titres anglais + 中文 incrustés
Chapitres
Transcript
Build a comparison using paired statements.
用成对陈述比较声音与可见光。
Sheets 6.1–6.5 develop the models and calculations.
练习6.1至6.5逐步建立模型和计算方法。
Sound in air is a longitudinal, mechanical wave; visible light is a transverse, electromagnetic wave.
空气中的声波是纵波、机械波；可见光是横波、电磁波。
Sound needs particles to travel through, but light can cross a vacuum.
声音需要粒子才能传播，而光可以穿过真空。
Sound travels at about three hundred and thirty metres per second here, light at about three hundred million.
这里声速约为每秒三百三十米，光速约为每秒三亿米。
Both transfer energy, and both can reflect and refract.
二者都传递能量，都能反射和折射。
Sound speed depends on conditions, so use the value given.
声速与环境条件有关，所以要用题目给定的数值。
The longitudinal marks show particle positions, not a transverse displacement curve.
纵波图中的线表示粒子位置，不是横向位移曲线。
Compare vibration with travel direction.
比较振动方向与传播方向。
Particles oscillate around equilibrium. They move without travelling across the pond with the wave.
粒子在平衡位置附近振动，但不会随波一起穿过池塘。
Sound likewise passes through vibrating air particles; air need not flow from the source to the listener.
同样，声音通过振动的空气粒子传播，不需要空气从声源流到听者。
Amplitude is not peak-to-trough distance. Wavelength is not an arbitrary gap.
振幅不是波峰到波谷的距离，波长也不是任意两点之间的距离。
The period is the reciprocal of frequency.
周期等于频率的倒数。
Higher frequency means a shorter cycle.
频率越高，周期越短。
Keep the time unit in the final answer.
最终答案应写时间单位。
Equation-sheet support depends on the exam series.
公式表内容取决于考试年份。
A kilohertz is a thousand hertz.
千赫兹等于一千赫兹。
Use consistent units before substitution.
代入前应统一单位。
Distance per second divided by cycles per second gives distance per cycle.
每秒距离除以每秒周期数，得到每个周期对应的距离。
Use a calibrated surface scale.
使用已校准的水面标尺。
Correct projected measurements for magnification.
投影测量应校正放大倍数。
First crest to eleventh crest gives ten cycles.
从第一波峰到第十一波峰是十个周期。
Ten crests span only nine intervals.
十个波峰之间只有九个间隔。
Keep unrounded intermediate values.
保留中间计算值。
Constant speed is a condition supplied by this question.
波速不变是本题给定的条件。
Use two loops per wavelength and read frequency from the generator.
每两个波腹段对应一个波长，频率从信号发生器读取。
For echo measurement of sound speed, use twice the wall distance.
回声法测声速时，路程是到墙距离的两倍。
Electronic timing reduces reaction-time uncertainty.
电子计时可减小反应时间误差。
With a stationary source and boundary, wavelength changes with speed while frequency stays fixed.
对于静止声源和界面，波长随波速改变，频率保持不变。
This is sound entering water. Do not confuse it with light entering glass.
这里是声音进入水中，不要与光进入玻璃混淆。
Draw the normal at the incidence point. Incident and reflected rays lie on opposite sides.
在入射点画法线，入射光和反射光位于法线两侧。
Distinguish reflection angle from angle to the surface.
区分反射角与光线和镜面的夹角。
Rough surfaces do not break the reflection law; their local orientations differ.
粗糙表面不违反反射定律，只是各位置的法线方向不同。
Describe a glass-block method using angles of incidence from ten to sixty degrees.
描述一个用玻璃砖进行的实验，入射角从十度到六十度。
Put the tracing and measuring in a logical order, locate the normal, use ten degree steps, and keep the ray inside the block separate from the ray that emerges.
按合理的顺序安排描画和测量步骤，确定法线，以十度为间隔，并把玻璃砖内的光线与射出的光线区分开。
A mirror-only method does not answer this task.
只用镜子的方法不能回答这道题。
Pause and plan it.
暂停并构思。
Place the block on paper, trace its outline, and shine a narrow ray into it.
把玻璃砖放在纸上，描出它的轮廓，让一束细光射入。
Mark where the ray enters and leaves, then remove the block and join the points to show the ray inside.
标出光线射入和射出的位置，然后拿走玻璃砖，把这些点连起来，画出砖内的光线。
Draw the normal where the ray enters, and measure the angles of incidence and refraction from it.
在入射点画法线，从法线量出入射角和折射角。
Repeat at ten, twenty, thirty, forty, fifty and sixty degrees, with the same block.
用同一块玻璃砖，在十、二十、三十、四十、五十和六十度时重复。
Fine marks help locate the rays, and repeats improve reliability.
细小的标记有助于确定光线位置，重复测量可以提高可靠性。
These processes can happen together; a boundary need not reflect all incident energy.
这些过程可以同时发生，界面不一定反射全部入射能量。
Follow vibrations through the particles and ear structures.
沿空气粒子和耳部结构描述振动传递。
Ultrasound is defined by frequency.
超声波按频率定义。
Hearing limits vary between people.
听觉范围因人而异。
A microphone also converts vibration into an electrical signal.
麦克风也能把振动转化为电信号。
The time includes downward and upward travel.
时间包括向下和向上传播。
Identify the complete path before calculating depth.
计算深度前应确定完整路程。
Round the original value once.
对原始结果直接舍入一次。
Rounding 0.1248 to 0.125 and then to 0.13 gives a double-rounding error.
先把0.1248舍为0.125，再舍为0.13会造成重复舍入错误。
One absent signal is not proof of a liquid core.
一个信号缺失不能证明液态地核。
Interpret the spatial pattern using known wave properties and an Earth model.
应根据已知波性质和地球模型解释空间分布。
Use consistent units.
统一单位。
Time-gap interpretation uses different known P- and S-wave speeds.
到达时间差的解释利用P波和S波不同的已知速度。
The teacher check connects to the sheet model; it is not an extra claim about this question’s marks.
教师检查联系练习中的模型，不额外声称这是原题的评分要求。
The spectrum is continuous; visible light is one small part.
电磁波谱是连续的，可见光只占很小部分。
All EM waves are transverse.
所有电磁波都是横波。
Distinguish production, shared wave properties and biological effects.
区分产生方式、共同的波性质和生物效应。
Increasing frequency is not a universal hazard scale.
频率增大不是统一的危险程度标尺。
Exposure, absorption and wave type matter.
风险还取决于暴露、吸收和波的类型。
Explain why the property suits the use.
解释相关性质为何适合该用途。
Ultrasound images internal boundaries, not just the skin.
超声成像利用内部界面，不只是皮肤表面。
A gigahertz is a billion hertz.
吉赫兹等于十亿赫兹。
Show the prefix conversion.
应写出单位前缀换算。
Compare this with the 200 MHz radio worked example in the worksheet.
与练习中的200 MHz无线电波例题比较。
Wavefronts stay continuous at the boundary.
波面在界面处保持连续。
Their spacing shrinks in glass while arrival rate stays fixed.
进入玻璃后间距变小，但到达频率不变。
Slowing does not always change direction. The incidence angle matters.
减速并不总意味着改变方向，入射角也很重要。
This counterexample prevents the false rule that every speed change bends a ray.
这个反例避免“每次波速变化都会使光线偏折”的错误规则。
Equal temperature and geometry are essential.
温度和测量几何条件必须相同。
A detector reading is not a temperature rise.
探测器读数不是温升。
Use surface properties in the infrared range.
使用表面在红外范围的性质。
Visible colour alone does not determine infrared emissivity.
仅凭可见颜色不能确定红外发射率。
Separate supplied arithmetic from claims about typical clinical doses.
区分给定数据的计算与真实临床典型剂量的陈述。
For ultraviolet, explain skin ageing and cancer risk; for X-rays and gamma, connect ionisation to gene damage.
紫外线可能导致皮肤老化并增加癌症风险；X射线和伽马射线可通过电离损伤基因。
The parallel ray refracts through the far focus; the centre ray goes straight.
平行光线折射后经过另一侧焦点，过光心的光线沿直线传播。
Locate their actual intersection.
用实际交点确定像的位置。
Parallel rays strike a lens with its principal focus marked F.
平行光射向一个透镜，图中标出了主焦点F。
Name the lens, and complete the refracted rays: they must converge at the focus.
说出透镜的名称，并画完折射光线：它们必须会聚在焦点上。
Then think about how finding the image of an object is different: that uses rays from one point on the object.
再想一想：作物体的像有什么不同？ 那要用从物体上同一点发出的光线。
Pause and try it.
暂停并试着画。
An object’s image is not always at the principal focus.
物体的像并不总在主焦点处。
For a real object, a concave lens gives a virtual upright diminished image.
凹透镜对实物成正立缩小虚像。
A convex lens with an object inside f gives a virtual upright magnified image.
凸透镜在物体位于焦距内时成正立放大虚像。
The unit cancels only when both lengths use the same unit.
只有两长度单位相同时，单位才能约去。
Use the supplied product relation in the worksheet’s two-lens extension; it is not assumed syllabus recall.
练习中的双透镜拓展题提供乘积关系，不要求作为考纲知识记忆。
A filter after an object changes what reaches the eye, not the illumination.
放在物体之后的滤光片改变到达眼睛的光，不改变照明。
These are ideal models under white light.
这是白光照明下的理想模型。
Real surfaces may reflect wider wavelength ranges.
真实表面可能反射更宽的波长范围。
A perfect black body absorbs all incident radiation and is the best possible emitter.
理想黑体吸收所有入射辐射，也是最好的发射体。
Compare peak position and area under the distribution.
比较峰值位置和曲线下面积。
A halved peak wavelength does not imply intensity merely doubles.
峰值波长减半不意味着强度仅加倍。
Absorption equals emission at steady temperature in this radiation-only model.
在这个仅考虑辐射的稳态模型中，吸收等于发射。
Real objects also conduct and convect.
真实物体还有传导和对流。
The conclusion follows from the complete energy balance.
结论来自完整的能量平衡。
Reflected sunlight was not absorbed.
被反射的太阳光没有被吸收。
Less reflection can increase absorption and initially cause warming.
反射减少可增加吸收，最初导致升温。
Match mass, heat capacity and initial temperature.
应保持质量、热容和初温相同。
Visible colour alone is not evidence of infrared emissivity.
可见颜色本身不能说明红外发射率。
Do not compare cooling after unequal warming without considering the starting temperatures.
如果升温不同，比较冷却时必须考虑初温差异。
Explain the reference line, complete path, measurement interval and model conditions.
说明参考线、完整路程、测量间隔和模型条件。
Practise independently with sheets 6.1–6.5 and explain why each model applies.
用练习6.1至6.5独立训练，并解释每个模型为何适用。