📚 Year 7 Edexcel Physics: Interdisciplinary Integrated Question Training | 七年级爱德思物理：跨学科综合题型训练
Interdisciplinary questions in Year 7 Edexcel Physics help you connect physical principles with mathematics, biology, geography, history and other subjects. This article provides a wide range of sample problems, step-by-step methods and key ideas to strengthen your skills in applying physics across different real-world contexts.
七年级爱德思物理中的跨学科题目帮助你建立物理原理与数学、生物、地理、历史等学科的联系。本文提供了丰富的例题、分步解题方法和核心思路，帮助你在不同实际情境中应用物理知识，提升综合解题能力。
1. Physics and Mathematics: Speed, Distance, Time | 物理与数学：速度、距离、时间
The basic formula linking speed, distance and time is: average speed = total distance / total time. Remember to convert units so that they match, for example km to m, hours to seconds.
联系速度、距离和时间的基本公式为：平均速度 = 总距离 ÷ 总时间。解题时要注意统一单位，例如将千米换算成米，小时换算成秒。
Example: A cyclist rides 36 km in 2 hours. Find the average speed in km/h and m/s. Solution: speed = 36 km / 2 h = 18 km/h. To convert to m/s, multiply by (1000 m / 3600 s) or divide by 3.6: 18 / 3.6 = 5 m/s.
例题：一名骑自行车的人2小时骑行了36千米。求平均速度（以km/h和m/s表示）。解答：速度 = 36 km ÷ 2 h = 18 km/h。换算成m/s只需除以3.6：18 ÷ 3.6 = 5 m/s。
Distance–time graphs are common in exams. The slope (gradient) of the line gives the speed. A horizontal line means the object is stationary.
距离–时间图在考试中很常见。图中直线的斜率（坡度）代表速度；水平线表示物体静止不动。
2. Physics and Biology: Energy in Food Chains | 物理与生物：食物链中的能量传递
In ecosystems, energy is transferred from one trophic level to the next, but only about 10% of the energy is passed on. The rest is lost as heat, movement or waste, following the principle of energy conservation.
在生态系统中，能量从一个营养级传递到下一个，但只有大约10%的能量被传递。其余以热、运动或废物的形式散失，这符合能量守恒原理。
Example: A grass plant absorbs 5000 J of light energy from the Sun. A rabbit eats the grass and obtains 500 J. How much energy is lost between the grass and the rabbit? Solution: 5000 J – 500 J = 4500 J lost, mostly as heat and metabolic work.
例题：一株草从太阳吸收了5000 J的光能，一只兔子吃了草，获得了500 J的能量。草和兔子之间损失了多少能量？解答：5000 J – 500 J = 4500 J 损失，大部分以热量和代谢活动的形式散失。
Linking physics to biology helps us calculate efficiency: efficiency = (useful energy output / total energy input) × 100%. Here it is (500 J / 5000 J) × 100% = 10%.
把物理与生物联系起来可以帮助我们计算效率：效率 = （有用能量输出 ÷ 总能量输入）× 100%。本例中为（500 J ÷ 5000 J）× 100% = 10%。
3. Physics and Chemistry: Conduction and Circuits | 物理与化学：导电性与电路
Metals are good conductors of electricity because they have free electrons. The chemical reactivity of a metal may affect its conductivity, but in a circuit the main factors are resistance (R), current (I) and voltage (V).
金属是电的良导体，因为它们拥有自由电子。金属的化学反应活性可能会影响其导电性，但在电路中主要因素是电阻（R）、电流（I）和电压（V）。
Ohm’s law states V = I × R. When you build a test circuit with wires of different metals (copper, iron, nichrome), you can measure how resistance changes. Copper has low resistance, so it is widely used in household wiring.
欧姆定律指出 V = I × R。当你用不同金属丝（铜、铁、镍铬合金）搭建测试电路时，可以测量电阻如何变化。铜的电阻很低，因此广泛用于家用导线。
Cross-subject question: A copper wire of length 1 m has resistance 0.5 Ω. A nichrome wire of the same length and thickness has resistance 10 Ω. If a 3 V battery is connected to each wire separately, calculate the current in each case. Coinciding with chemistry, nichrome’s high resistance is due to its alloy structure.
跨学科问题：一根长1 m的铜导线电阻为0.5 Ω，同长度同粗细的镍铬合金丝电阻为10 Ω。若分别接在3 V电池上，计算每种情况的电流。与化学联系起来看，镍铬合金的高电阻源于其合金结构。
Solution: For copper, I = V / R = 3 V / 0.5 Ω = 6 A. For nichrome, I = 3 V / 10 Ω = 0.3 A.
解答：铜导线中 I = 3 V ÷ 0.5 Ω = 6 A；镍铬合金中 I = 3 V ÷ 10 Ω = 0.3 A。
4. Physics and Geography: Solar Radiation and the Earth’s Temperature | 物理与地理：太阳辐射与地球温度
The Earth’s temperature is determined by the balance between incoming solar radiation and outgoing heat. Ice and snow have a high albedo, reflecting most sunlight; dark oceans absorb more energy. This connects physics (radiation) with geography (climate zones).
地球温度由入射太阳辐射与输出热量之间的平衡决定。冰雪具有高反照率，反射大部分阳光；深色海洋则吸收更多能量。这将物理（辐射）与地理（气候带）联系起来。
Integrated problem: If the average solar power reaching the Arctic region is 200 W/m² and the ice reflects 80%, how much power per square metre is actually absorbed? Solution: Absorbed power = 200 W/m² × (1 – 0.8) = 200 × 0.2 = 40 W/m².
综合题：如果到达北极地区的平均太阳功率为200 W/m²，冰面反射了80%，那么每平方米实际吸收的功率是多少？解答：吸收功率 = 200 W/m² × (1 – 0.8) = 200 × 0.2 = 40 W/m²。
When the ice melts due to global warming, the exposed dark water absorbs far more energy, creating a feedback loop. This shows how physics principles help explain geographical phenomena.
当全球变暖导致冰层融化时，暴露的深色海水会吸收更多的能量，形成正反馈循环。这体现了物理原理如何解释地理现象。
5. Physics and History: Archimedes’ Principle and Density | 物理与历史：阿基米德原理与密度
According to the story, Archimedes discovered how to test the purity of a golden crown by measuring its volume through water displacement and calculating its density. Density = mass / volume.
相传阿基米德发现可以通过排水法测量王冠的体积，再计算其密度，从而检验王冠的纯度。密度 = 质量 ÷ 体积。
Historical problem: A crown has a mass of 1.5 kg. When submerged in a container completely filled with water, it causes 100 cm³ of water to spill out. Is the crown pure gold? (Density of pure gold is 19.3 g/cm³.)
历史问题：一顶王冠质量为1.5 kg，把它沉入装满水的容器后，溢出了100 cm³的水。这顶王冠是纯金的吗？（纯金的密度为19.3 g/cm³）
Solution: First convert mass to grams: 1.5 kg = 1500 g. Density of crown = 1500 g / 100 cm³ = 15 g/cm³, which is less than 19.3 g/cm³, so the crown is not pure gold — it may contain lighter metals. This cross-links physics and history through experimental archaeology.
解答：首先将质量换算为克：1.5 kg = 1500 g。王冠密度 = 1500 g ÷ 100 cm³ = 15 g/cm³，小于19.3 g/cm³，因此王冠不纯，可能掺有轻金属。这道题通过实验考古学将物理和历史联系起来。
6. Physics and PE: Forces and Motion in Sports | 物理与体育：运动中的力与运动
In long jump, an athlete takes off with a certain horizontal speed and spends some time in the air. The horizontal distance travelled is given by distance = horizontal speed × time of flight. This ignores air resistance.
在跳远中，运动员以一定的水平速度起跳，并在空中停留一段时间。水平方向移动的距离可用 距离 = 水平速度 × 腾空时间 计算，这里忽略空气阻力。
Sports science problem: A long jumper has a horizontal take-off speed of 5 m/s and a flight time of 0.8 s. How far does she jump? Solution: d = 5 m/s × 0.8 s = 4.0 m.
运动科学问题：一名跳远运动员的水平起跳速度为5 m/s，腾空时间0.8 s。她能跳多远？解答：d = 5 m/s × 0.8 s = 4.0 m。
Another example links force and acceleration. If a hockey player applies a force of 12 N to a ball of mass 0.4 kg, what is the ball’s acceleration? Using F = m × a, a = F / m = 12 N / 0.4 kg = 30 m/s². This combines PE data with Newton’s second law.
另一个例子联系力和加速度。如果曲棍球运动员对质量为0.4 kg的球施加了12 N的力，球的加速度是多少？利用 F = m × a，a = F ÷ m = 12 N ÷ 0.4 kg = 30 m/s²。这结合了体育数据和牛顿第二定律。
7. Physics and Music: Sound, Frequency and Vibrations | 物理与音乐：声音、频率与振动
Musical notes are produced by vibrations. The pitch of a note depends on its frequency: a higher frequency means a higher pitch. String instruments show that frequency increases when the vibrating length decreases.
音符由振动产生。音调的高低取决于频率：频率越高，音调越高。弦乐器表明，振动长度越短，频率越高。
Musical-physics challenge: A guitar string of length 60 cm vibrates at a certain frequency f. If you press the string at the 30 cm mark and pluck only the shorter part, how does the frequency change? The frequency is inversely proportional to length, so halving the length doubles the frequency. This produces a note one octave higher.
音乐与物理挑战：一根长60 cm的吉他弦以某一频率 f 振动。如果按住30 cm处，仅拨动较短部分，频率如何变化？频率与长度成反比，因此长度减半，频率翻倍，产生的音高八度。
The human ear can detect frequencies between about 20 Hz and 20,000 Hz. Sound travels through air at roughly 340 m/s. These facts are essential for designing musical instruments and auditoriums, blending physics with music and technology.
人耳能察觉约20 Hz到20,000 Hz的声音。声音在空气中以大约340 m/s的速度传播。这些事实对设计乐器和音乐厅至关重要，将物理与音乐及技术融合。
8. Physics and Art: Light and Colour Mixing | 物理与艺术：光的颜色混合与颜料
In physics, white light is a mixture of all colours. The primary colours of light are red, green, and blue. When you mix them, you get different colours: red + green = yellow. This is additive colour mixing, used in screens and stage lighting.
在物理中，白光是一切颜色的混合。光的三原色是红、绿、蓝。将它们混合能得到不同颜色：红光 + 绿光 = 黄光。这就是加色混合，应用于屏幕和舞台灯光。
Art uses pigments, which work by subtractive mixing. The primary pigments are cyan, magenta and yellow. Mixing cyan and yellow paint produces green because cyan absorbs red and yellow absorbs blue, leaving green to be reflected.
艺术中使用颜料，遵循减色混合原理。颜料的三原色是青、品红、黄。将青色和黄色颜料混合得到绿色，因为青色吸收红光，黄色吸收蓝光，剩下绿光被反射出来。
Exam-style combined question: Shine a red spotlight and a green spotlight simultaneously on a white shirt. What colour is seen? Answer: yellow, because the shirt reflects both red and green light to the eye. Applying physics to art helps explain stage design and painting techniques.
考题风格综合题：将红色追光灯和绿色追光灯同时照射在一件白色衬衫上，会看到什么颜色？答：黄色，因为衬衫同时把红光和绿光反射到眼睛里。将物理运用到艺术中可以解释舞台设计和绘画技法。
9. Physics and Engineering: Simple Machines and Levers | 物理与工程：简单机械与杠杆
A lever is a simple machine that can magnify force. The law of the lever states: effort × effort arm = load × load arm. This is the moment balance condition.
杠杆是一种可以放大力的简单机械。杠杆定律为：动力 × 动力臂 = 阻力 × 阻力臂，这是力矩平衡条件。
Engineering problem: A worker uses a crowbar to lift a heavy rock. The rock (load) is 0.2 m from the pivot and weighs 800 N. The worker applies effort 1.6 m away from the pivot. What minimum effort is needed? Solution: effort × 1.6 m = 800 N × 0.2 m → effort = (800 N × 0.2 m) / 1.6 m = 100 N.
工程问题：工人用撬棍撬起重石头。石头（阻力）距支点0.2 m，重800 N。工人施力处距支点1.6 m。需要至少多大的力？解答：动力 × 1.6 m = 800 N × 0.2 m → 动力 = （800 N × 0.2 m）÷ 1.6 m = 100 N。
This principle helped ancient engineers build pyramids and it still applies in modern tools like scissors, wheelbarrows and nutcrackers. Calculating moments links physics with practical engineering and history.
这一原理帮助古代工程师建造了金字塔，至今仍应用于剪刀、手推车和胡桃夹子等现代工具。计算力矩将物理与实际工程和历史联系起来。
10. Physics and Environmental Science: Renewable Energy | 物理与环境科学：可再生能源
Solar panels convert sunlight directly into electricity. The power output depends on the surface area, the solar irradiance (power per square metre) and the panel efficiency.
太阳能板直接将太阳光转化为电能。输出功率取决于表面积、太阳辐照度（每平方米的功率）和面板效率。
Environmental problem: On a clear day, solar irradiance is about 1000 W/m². A roof-mounted solar panel has an area of 1.8 m² and an efficiency of 18%. What is the usable electrical power output? Solution: incident power = 1000 W/m² × 1.8 m² = 1800 W; output = 1800 W × 0.18 = 324 W.
环境问题：晴天时太阳辐照度约为1000 W/m²。一块屋顶太阳能板面积1.8 m²，效率18%。可用的电功率是多少？解答：入射功率 = 1000 W/m² × 1.8 m² = 1800 W；输出 = 1800 W × 0.18 = 324 W。
Wind turbines also harness renewable energy. The power of the wind depends on wind speed and blade swept area. Discussing these topics combines physics with environmental decision-making, geography and even economics.
风力发电机也利用可再生能源。风力功率取决于风速和叶片扫风面积。讨论这些话题将物理与环境决策、地理甚至经济学结合起来。
11. Physics and Computing: Sensors and Data Logging | 物理与计算机：传感器与数据记录
Thermistors and light-dependent resistors (LDRs) are analogue sensors whose resistance changes with temperature or light. They are often used in potential divider circuits to send signals to a microcontroller for data logging.
热敏电阻和光敏电阻是模拟传感器，其电阻随温度或光照变化。它们常用于分压电路，将信号发送给微控制器进行数据记录。
Cross-field investigation: A thermistor has a resistance of 10 kΩ at 25°C and 5 kΩ at 40°C. It is placed in series with a fixed 10 kΩ resistor across a 6 V battery. At 40°C, what is the voltage across the fixed resistor? The circuit is a potential divider: V_out = V_in × (R_fixed / (R_thermistor + R_fixed)) = 6 V × (10 kΩ / (5 kΩ + 10 kΩ)) = 6 V × (10/15) = 4 V. This voltage can be read by a computer interface to monitor temperature.
跨学科探究：一个热敏电阻在25°C时阻值为10 kΩ，在40°C时阻值为5 kΩ。它与一个10 kΩ的固定电阻串联，接在6 V电池上。在40°C时，固定电阻两端的电压是多少？分压电路公式：V_out = V_in × (R_fixed / (R_thermistor + R_fixed)) = 6 V × (10 kΩ / (5 kΩ + 10 kΩ)) = 6 V × (10/15) = 4 V。计算机接口可以读取这个电压来监测温度。
This integration requires knowledge of Ohm’s law, sensor behaviour and basic programming logic, linking physics with computing and technology.
此类整合需要欧姆定律、传感器特性和基本编程逻辑的知识，将物理与计算机和技术结合起来。
12. Physics and Space: Gravity and Orbits | 物理与太空：重力和轨道
Gravity is the force of attraction between masses. On Earth, the gravitational field strength is about 10 N/kg (or 9.8 N/kg). On the Moon it is about 1.6 N/kg, roughly one-sixth of Earth’s. An object’s mass stays the same, but its weight changes.
重力是物体之间的吸引力。地球表面的重力场强约为10 N/kg（或9.8 N/kg）；月球上约为1.6 N/kg，大约是地球的六分之一。物体的质量不变，但重量会发生改变。
Space exploration problem: An astronaut with a mass of 72 kg travels to the Moon. What is her weight on Earth? On the Moon? (Use g = 10 N/kg for Earth, 1.6 N/kg for Moon.) Solution: weight on Earth = 72 kg × 10 N/kg = 720 N; weight on Moon = 72 kg × 1.6 N/kg = 115.2 N.
太空探索问题：一位质量为72 kg的宇航员登上月球。她在地球上重多少？在月球上呢？（地球 g 取10 N/kg，月球取1.6 N/kg。）解答：地球上重量 = 72 kg × 10 N/kg = 720 N；月球上重量 = 72 kg × 1.6 N/kg = 115.2 N。
Understanding gravity also helps explain why planets orbit the Sun. The Sun’s gravitational pull provides the centripetal force keeping Earth in a nearly circular orbit. This links physics with astronomy and the geography of the solar system.
理解重力也有助于解释为什么行星绕太阳公转。太阳的引力提供了使地球保持近似圆形轨道的向心力。这使物理与天文学以及太阳系的相关地理知识相连。
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