📚 Year 7 CIE Engineering: Cross-curricular Problem-Solving Training | 七年级 CIE 工程：跨学科综合题型训练
In Year 7 CIE Engineering, you are expected to apply knowledge from different subjects — such as physics, mathematics and materials science — to solve real-world problems. This article provides integrated practice questions and step-by-step explanations to sharpen your critical thinking and exam skills.
在七年级 CIE 工程课程中，你需要综合运用物理、数学和材料科学等多学科知识来解决实际问题。本文提供跨学科综合练习题和逐步解析，帮助你提高批判性思维和应试能力。
1. Understanding Forces in Engineering | 理解工程中的力
the motion or shape of an object. In engineering, we measure force in newtons (N). Newton's Second Law states that force equals mass times acceleration (F = m a).
力是能够改变物体运动状态或形状的推或拉。在工程中，力的单位是牛顿（N）。牛顿第二定律指出，力等于质量乘以加速度（F = m a）。
ommon question asks you to find the weight of an object on Earth: weight = mass × 10 N/kg.
在设计桥梁等结构时，工程师必须计算作用在梁上的力，以防止弯曲或断裂。常见的问题是求物体在地球上的重量：重量 = 质量 × 10 N/kg。
Example: A steel beam has a mass of 500 kg. Calculate its weight.
例题：一根钢梁质量为 500 kg。计算它的重量。
Weight = 500 kg × 10 N/kg = 5000 N.
重量 = 500 kg × 10 N/kg = 5000 N。
2. Simple Machines and Mechanical Advantage | 简单机械与机械优势
tio of output force to input force. For a lever, MA = effort arm length ÷ load arm length.
杠杆、滑轮和斜面等简单机械可以帮助我们放大作用力。机械优势 (MA) 是输出力与输入力的比值。对于杠杆，MA = 施力臂长度 ÷ 抗力臂长度。
Example: A crowbar has an effort arm of 1.2 m and a load arm of 0.3 m. Find MA.
例题：一根撬棍的施力臂长 1.2 m，抗力臂长 0.3 m。求机械优势。
MA = 1.2 ÷ 0.3 = 4.
MA = 1.2 ÷ 0.3 = 4。
This means the lever multiplies your force by 4. Such calculations blend maths with design decisions.
这意味着杠杆能将你的力放大 4 倍。这类计算将数学与设计决策结合在一起。
3. Electrical Circuits and Ohm's Law | 电路与欧姆定律
In electronics, Ohm' rrent (I) in amperes, and resistance (R) in ohms (Ω). It is essential for designing circuits and selecting components.
在电子学中，欧姆定律（V = I × R）将电压（V）、电流（I，单位安培）和电阻（R，单位欧姆 Ω）联系起来。它是设计电路和选择元件的关键。
Example: A resistor of 220 Ω has a current of 0.05 A flowing through it. What is the voltage across it?
例题：一个 220 Ω 的电阻上有 0.05 A 的电流流过。它两端的电压是多少？
V = 0.05 A × 220 Ω = 11 V.
V = 0.05 A × 220 Ω = 11 V。
problems often involve series and parallel connections, which combine science and arithmetic.
你还需要会变换公式：R = V ÷ I 和 I = V ÷ R。电路问题常涉及串联和并联，综合了科学和算术。
4. Material Properties and Selection | 材料特性与选材
Engineers choose materials based on properties like strength, hardness, density, thermal conductivity and electrical conductivity. A material property table is often given in exams.
工程师根据强度、硬度、密度、导热性和导电性等特性来选择材料。考试中常会给出材料特性表。
For instance, you may be asked to select a material for a cooking pot handle: it must be a thermal insulator and have a high melting point. Wood or certain plastics are common choices.
例如，你可能会被要求为锅柄选材：它必须是热绝缘体，并具有高熔点。木材或某些塑料是常见选择。
This requires cross-curricular thinking: linking science knowledge about heat transfer with practical constraints like cost and weight.
这需要跨学科思维：将热传导的科学知识与成本和重量等实际限制联系起来。
Below is a sample table for quick reference:
下表为快速参考示例：
| Material | Density (g/cm³) | Thermal Conductivity | Tensile Strength (MPa) |
|---|---|---|---|
| Steel | 7.8 | High | 400 |
| Aluminium | 2.7 | High | 200 |
| Wood (pine) | 0.5 | Low | 40 |
5. Measurement and Unit Conversion | 测量与单位转换
it conversions include 1 m = 100 cm, 1 cm³ = 1 mL, 1 tonne = 1000 kg, and 1 hour = 3600 s.
精确测量和单位换算在工程中至关重要。常见的换算有 1 m = 100 cm，1 cm³ = 1 mL，1 吨 = 1000 kg，1 小时 = 3600 秒。
Example: Convert a speed of 36 km/h into m/s.
例题：将速度 36 km/h 换算成 m/s。
36 km/h = 36 × 1000 m / 3600 s = 10 m/s.
36 km/h = 36 × 1000 m / 3600 s = 10 m/s。
When calculating density (mass ÷ volume), units must be consistent: if mass is in g and volume in cm³, density is in g/cm³.
计算密度（质量 ÷ 体积）时，单位必须统一：如果质量用 g，体积用 cm³，密度单位就是 g/cm³。
6. Work, Energy and Power Calculations | 功、能与功率的计算
, measured in joules (J). Power (P) is the rate of doing work: P = W ÷ time, in watts (W).
做功（W）= 力 × 在力的方向上移动的距离，单位是焦耳（J）。功率（P）是做功的快慢：P = W ÷ 时间，单位是瓦特（W）。
Example: A crane lifts a 2000 N load through a height of 15 m in 10 seconds. Calculate work done and power.
例题：一台起重机在 10 秒内将一个 2000 N 的重物举升 15 m。计算所做的功和功率。
Work = 2000 N × 15 m = 30 000 J. Power = 30 000 J ÷ 10 s = 3000 W.
功 = 2000 N × 15 m = 30 000 J。功率 = 30 000 J ÷ 10 s = 3000 W。
These topics combine maths and physical concepts to evaluate machine performance.
这些主题将数学和物理概念结合起来，以评估机器的性能。
7. Structures and Loads | 结构与载荷
Structures like towers and trusses are subjected to tension, compression, bending and torsion. Identifying these forces helps in choosing suitable materials and cross-sections.
塔和桁架等结构会受到拉伸、压缩、弯曲和扭转。识别这些力有助于选择合适的材料和截面形状。
Example: A beam supporting a weight experiences compression on the top and tension on the bottom. A hollow circular tube resists torsion better than a solid rod of the same mass.
例题：承受重量的梁，顶部受压缩，底部受拉伸。空心圆管比同质量的实心棒更能抵抗扭转。
Integrated problems might ask you to calculate the stress (force ÷ area) and compare it with the material's strength.
综合题可能要求你计算应力（力 ÷ 面积），并与材料强度进行比较。
0 N and has a cross-sectional area of 0.002 m², the stress is 500 000 Pa. Comparing this with the material's compressive strength determines safety.
应力 = F ÷ A。如果一根柱子承受 1000 N，截面积为 0.002 m²，则应力为 500 000 Pa。将其与材料的抗压强度比较，即可判断安全性。
8. Design and Problem Solving Process | 设计与问题解决过程
Engineering design follows a cycle: identify the problem, research, propose solutions, develop a prototype, test and evaluate. You must justify decisions with scientific and mathematical reasoning.
工程设计遵循一个循环：确定问题、调研、提出方案、开发原型、测试和评估。你必须用科学和数学推理来证明决策的合理性。
For example, designing a bridge may involve calculating load capacity, selecting materials based on strength-to-weight ratio, and estimating cost — all integrated skills.
例如，设计一座桥梁可能涉及计算承载能力、依据强度-重量比选择材料以及估算成本——这些都是综合技能。
Exam questions often present a scenario and ask you to evaluate competing designs using criteria like efficiency, sustainability and cost.
考试题目常会给出一个场景，要求你使用效率、可持续性和成本等标准来评价不同设计。
9. Data Interpretation and Graphs | 数据解读与图表
Engineers use graphs to display data from experiments. You must be able to read values, calculate gradients and areas under lines, and interpret relationships.
工程师用图表来展示实验数据。你必须能够读取数值、计算斜率和线下的面积，并解释变量之间的关系。
Example: A graph of force (N) vs. extension (m) for a spring has a straight line through the origin. The gradient gives the spring constant (k) in N/m.
例题：弹簧的力（N）与伸长量（m）关系图是一条过原点的直线。斜率给出弹簧常数 (k)，单位为 N/m。
Hooke's Law: F = k × x. This blends physics with graph skills.
胡克定律：F = k × x。这结合了物理与图表技能。
s (0.1 m, 5 N) and (0.3 m, 15 N), the gradient is (15 N - 5 N) ÷ (0.3 m - 0.1 m) = 50 N/m.
如果图上显示 (0.1 m, 5 N) 和 (0.3 m, 15 N)，则斜率为 (15 N - 5 N) ÷ (0.3 m - 0.1 m) = 50 N/m。
10. Integrated Practice Questions | 综合练习题
Here is a multi-step problem that combines several engineering topics.
以下是一个结合多个工程主题的多步骤问题。
Problem:
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