IGCSE AQA Science: Calculation Practice Special Training | IGCSE AQA 科学：计算题专项训练

📚 IGCSE AQA Science: Calculation Practice Special Training | IGCSE AQA 科学：计算题专项训练

Mastering calculations is essential for success in IGCSE AQA Science. Whether tackling Physics equations or Chemistry mole problems, consistent practice and understanding of formula manipulation make all the difference. This article provides targeted training on key calculation topics, with step-by-step examples and bilingual explanations.

掌握计算题对于在 IGCSE AQA 科学考试中取得成功至关重要。无论是处理物理方程还是化学摩尔问题，持续的练习和对公式变形的理解都至关重要。本文针对关键计算主题提供专项训练，配有分步示例和双语解释。

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1. Speed, Distance, Time | 速度、距离和时间

The average speed of an object is given by speed = distance / time (s = d/t). When distance is in metres (m) and time in seconds (s), speed is in metres per second (m/s). You can rearrange to find d = s × t or t = d/s. Always check units and convert if needed.

物体的平均速度由速度 = 距离／时间 (s = d/t) 给出。当距离单位为米 (m)、时间单位为秒 (s) 时，速度单位为米每秒 (m/s)。你可以通过变形得到 d = s × t 或 t = d/s。务必检查单位，必要时进行换算。

Example: A cyclist travels 450 m in 30 s. Calculate the average speed. Solution: s = d/t = 450/30 = 15 m/s.

例题：一名自行车手在 30 秒内行驶 450 米。计算平均速度。解：s = d/t = 450/30 = 15 m/s。

If speed is given in km/h, convert to m/s by dividing by 3.6. For instance, 72 km/h = 72 ÷ 3.6 = 20 m/s.

如果速度以千米每小时 (km/h) 给出，转换为 m/s 需除以 3.6。例如，72 km/h = 72 ÷ 3.6 = 20 m/s。

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2. Acceleration | 加速度

Acceleration is the rate of change of velocity: a = (v – u) / t, where v is final velocity (m/s), u is initial velocity, and t is time taken (s). The unit is m/s². If the acceleration is uniform, you can also use the equation v = u + a t.

加速度是速度变化的速率：a = (v – u) / t，其中 v 是末速度 (m/s)，u 是初速度，t 是所用时间 (s)。单位是 m/s²。如果加速度均匀，你还可以使用方程 v = u + a t。

Example: A car accelerates from rest (u=0) to 20 m/s in 5 s. a = (20-0)/5 = 4 m/s².

例题：一辆汽车从静止 (u=0) 加速到 20 m/s，用时 5 秒。a = (20-0)/5 = 4 m/s²。

When displacement (s) is known, use v² = u² + 2 a s. For example, to stop a car with deceleration, this equation helps find braking distance.

若已知位移 (s)，可使用 v² = u² + 2 a s。例如，计算汽车减速至停止所需的制动距离时，这个公式很实用。

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3. Force, Mass and Acceleration | 力、质量与加速度

Newton’s second law: resultant force F (N) = mass m (kg) × acceleration a (m/s²), F = m a. Rearrange to find m = F/a or a = F/m. This law links directly to free-body diagrams and unbalanced forces.

牛顿第二定律：合力 F (N) = 质量 m (kg) × 加速度 a (m/s²)，F = m a。变形可得 m = F/a 或 a = F/m。这一定律直接与受力分析和非平衡力相关。

Example: A 5 kg mass experiences a net force of 20 N. Find its acceleration. a = F/m = 20/5 = 4 m/s².

例题：一个 5 千克的物体受到 20 牛的合力。求其加速度。a = F/m = 20/5 = 4 m/s²。

In a free-fall situation where air resistance is negligible, the only force is weight, giving a = g = 9.8 m/s².

在空气阻力可忽略的自由下落中，唯一的作用力是重力，因此加速度 a = g = 9.8 m/s²。

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4. Weight and Gravitational Field Strength | 重量与重力场强度

Weight W (N) = mass m (kg) × gravitational field strength g (N/kg). On Earth, g = 9.8 N/kg (often rounded to 10 N/kg in IGCSE). Weight varies with location; mass remains constant.

重量 W (N) = 质量 m (kg) × 重力场强度 g (N/kg)。在地球上，g = 9.8 N/kg（IGCSE 中常近似为 10 N/kg）。重量随地点变化，质量保持不变。

Example: An astronaut has a mass of 70 kg. Calculate her weight on Earth (g = 9.8 N/kg). W = 70 × 9.8 = 686 N.

例题：一名航天员质量为 70 kg。计算她在地球上的重量 (g = 9.8 N/kg)。W = 70 × 9.8 = 686 N。

On the Moon, g ≈ 1.6 N/kg. Her weight there would be 70 × 1.6 = 112 N. Mass remains 70 kg everywhere.

在月球上，g ≈ 1.6 N/kg。她在月球上的重量为 70 × 1.6 = 112 N。质量始终为 70 kg。

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5. Work Done and Energy Transfers | 做功与能量转移

Work done W (J) = force F (N) × distance moved in the direction of force d (m). When a force moves an object, energy is transferred. Gravitational potential energy (GPE) = m g h, and kinetic energy (KE) = ½ m v².

做功 W (J) = 力 F (N) × 沿力的方向移动的距离 d (m)。当力使物体移动时，能量发生了转移。重力势能 (GPE) = m g h，动能 (KE) = ½ m v²。

Example: A 50 N force pushes a box 3 m. Work done = 50 × 3 = 150 J.

例题：用 50 N 的力将一个箱子推动 3 m。做功 = 50 × 3 = 150 J。

A 2 kg ball dropped from 10 m (g=10 N/kg) has GPE = 2 × 10 × 10 = 200 J. Just before hitting the ground, all GPE converts to KE, so 200 = ½ × 2 × v². Solve: v² = 200, v = √200 ≈ 14.14 m/s.

一个 2 kg 的球从 10 m 高处落下 (g=10 N/kg)，其重力势能为 2 × 10 × 10 = 200 J。在触地前，所有重力势能转化为动能，因此 200 = ½ × 2 × v²。解得：v² = 200，v = √200 ≈ 14.14 m/s。

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6. Power and Efficiency | 功率与效率

Power P (W) = work done (or energy transferred) / time taken, P = W/t. For constant speed, P = F × v. Efficiency (%) = (useful energy or power output / total energy or power input) × 100.

功率 P (W) = 做功（或转移的能量）/ 所用时间，P = W/t。对于恒定速度，P = F × v。效率 (%) = (有用能量或功率输出 / 总能量或功率输入) × 100。

Example: A motor lifts a load, doing 600 J of work in 4 s. Power = 600/4 = 150 W.

例题：一台电动机提升重物，在 4 秒内做 600 J 的功。功率 = 600/4 = 150 W。

An electric motor consumes 500 J of electrical energy but produces only 400 J of useful mechanical work. Efficiency = (400/500)×100 = 80%.

一台电动机消耗 500 J 电能，但只产出 400 J 有用机械功。效率 = (400/500)×100 = 80%。

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7. Electrical Power and Energy | 电功率与电能

Electrical power P = potential difference V × current I, P = V I. Using Ohm’s law, equivalents: P = I² R and P = V²/R. Energy transferred E (J) = power × time, E = P t, or E = V I t.

电功率 P = 电势差 V × 电流 I，P = V I。由欧姆定律可得等效公式：P = I² R 和 P = V²/R。转移的电能 E (J) = 功率 × 时间，E = P t，或 E = V I t。

Example: A 12 V device draws 2 A. Find power and energy

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