📚 Year 11 OCR Physics: Case Study Practice | Year 11 OCR 物理案例分析实战演练
Case study questions in OCR GCSE Physics require you to apply your knowledge to real-world scenarios. This article provides several worked examples covering key topics from energy to radioactivity, helping you sharpen problem-solving skills and exam technique.
在OCR GCSE物理中，案例分析题要求你将知识应用于实际情景。本文提供多个涵盖从能量到放射性等关键主题的解题示范，帮助你提升问题解决能力和考试技巧。
1. Solar Panel Efficiency Calculation | 太阳能电池板效率计算
A rooftop solar panel has an area of 2.5 m² and receives sunlight with an irradiance of 800 W/m². The panel produces 400 W of electrical power. Let's calculate its efficiency and discuss real-world factors.
某屋顶太阳能电池板面积为2.5 m²，接收到的太阳辐照度为800 W/m²，输出电功率为400 W。我们来计算其效率并讨论实际影响因素。
First, find the total incident power: Pin = irradiance × area = 800 W/m² × 2.5 m² = 2000 W. The useful output power Pout 400 W. Efficiency is given by the useful output divided by total input, multiplied by 100%.
首先，计算总入射功率：Pin = 辐照度 × 面积 = 800 W/m² × 2.5 m² = 2000 W。有用输出功率 Pout = 400 W。效率等于有用输出除以总输入再乘以100%。
Efficiency = (Pout / Pin) × 100% = (400 / 2000) × 100% = 20%
So the solar panel operates at 20% efficiency. In practice, factors such as dust on the surface, the angle of sunlight, and temperature rise can reduce this value. Cleaning panels and adjusting tilt angle can improve energy yield.
因此该太阳能电池板的效率为20%。实际使用中，表面灰尘、太阳光入射角以及温度升高等因素都会降低效率。清洁面板和调整倾角可以提升发电量。
2. Electric Vehicle Energy and Power | 电动汽车的能量与功率
An electric car of mass 1500 kg accelerates from rest to 20 m/s. We will find the kinetic energy gained and the electrical energy required from the battery, given a motor efficiency of 90%.
一辆质量为1500 kg的电动汽车从静止加速到20 m/s。我们来计算获得的动能，以及在电机效率为90%的情况下电池需要提供的电能。
The kinetic energy (KE) formula: KE = ½ m v². Plugging in: m = 1500 kg, v = 20 m/s.
动能公式：KE = ½ m v²。代入数值：m = 1500 kg, v = 20 m/s。
KE = ½ × 1500 × (20)² = 0.5 × 1500 × 400 = 300 000 J (300 kJ)
The motor converts electrical energy to mechanical work with 90% efficiency, so the electrical energy needed from the battery is: Eelec = KE / efficiency.
电机将电能转换为机械能的效率为90%，因此电池需要提供的电能为：Eelec = KE / 效率。
Eelec = 300 000 J / 0.90 = 333 333 J (≈ 333 kJ)
an be found from Q = E / V. This shows how energy and voltage relate in electric vehicles.
若电池工作电压为400 V，则可利用Q = E / V求出所用电荷量。这表明了电动汽车中能量与电压的关系。
3. Bridge Forces and Moments | 桥梁受力与力矩平衡
A uniform bridge of length 10 m and weight 5000 N is supported at both ends. A truck of weight 8000 N is stopped 3 m from one end. Find the support forces using the principle of moments.
一座长10 m、重5000 N的均匀桥梁两端支撑。一辆重8000 N的卡车停在距一端3 m处。利用力矩原理求两端的支持力。
Take moments about support A. The bridge' other support B is 10 m from A. Clockwise moments = anticlockwise moments for equilibrium.
对支撑点A取矩。桥的重力作用在中心，距A点5 m。卡车距A点3 m。另一支撑B距A点10 m。平衡时顺时针力矩等于逆时针力矩。
FB × 10 = (5000 N × 5 m) + (8000 N × 3 m)
FB × 10 = 25 000 + 24 000 = 49 000 N·m
FB = 4900 N
The total upward forces equal total downward forces: FA + FB = 5000 N + 8000 N, so FA = 13000 N − 4900 N = 8100 N.
向上的合力等于向下的合力：FA + FB = 5000 N + 8000 N，因此FA =
Published by TutorHao | Year 11 Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图