iency/questions","formulaUrl":null,"initialIndex":8,"initialQuestionWindow":[{"id":"4c43df98-60d4-4438-9d58-ae2e7774f8dd","revisionId":"e279f7d5-e726-48e9-8d2f-dc5c36dfc55d","questionNumber":null,"questionType":"LA","specification":"A prototype tidal energy harvester converts kinetic energy from ocean currents into electrical energy.\n\nThe moving water delivers a total of $22500\\text{ kJ}$ of kinetic energy to the harvester's turbine blades every $1.25\\text{ minutes}$.\n\nThe overall efficiency of the harvester in converting this kinetic energy into electrical power is $0.32$ ($32\\%$).","difficulty":5,"difficultyBand":"medium","totalMarks":8,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"09d7b7bb-40ea-4d03-816d-9babfbc85ce1","label":null,"content":"Calculate the useful electrical power output of the harvester.\n\nGive your answer in kW.","markscheme":"- Converts the time interval into seconds: $1.25\\text{ minutes} = 1.25 \\times 60 = 75\\text{ seconds}$ B1\n- Calculates the total power input by dividing energy by time: $\\text{Power input} = \\frac{22500\\text{ kJ}}{75\\text{ s}} = 300\\text{ kW}$ M1\n- Recalls and applies the efficiency formula: $\\text{Useful power output} = \\text{Efficiency} \\times \\text{Power input} = 0.32 \\times 300$ M1\n- Obtains a final useful power output of $96\\text{ kW}$ A1\n\n*(Alternative Method)*\n- Calculates the useful energy output over the time period: $\\text{Useful energy} = 22500\\text{ kJ} \\times 0.32 = 7200\\text{ kJ}$ M1\n- Converts the time interval into seconds: $1.25\\text{ minutes} = 1.25 \\times 60 = 75\\text{ seconds}$ B1\n- Calculates the useful power output: $\\text{Useful power} = \\frac{7200\\text{ kJ}}{75\\text{ s}}$ M1\n- Obtains a final useful power output of $96\\text{ kW}$ A1","marks":8,"order":0,"markingRubric":null,"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"5451de71-2d09-4044-9947-4b255654965a","revisionId":"d03d71ce-d97d-4680-bdff-2167dc81bc07","questionNumber":null,"questionType":"LA","specification":"","difficulty":5,"difficultyBand":"medium","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"f85c87a3-26c6-46c6-add3-a030be4da61c","label":null,"content":"An automated deep-sea winch hoists a scientific core sample weighing $14\\,000\\text{ N}$ from the ocean floor up to a research vessel.\n\nThe sample is lifted vertically through a height of $450\\text{ m}$.\n\nThe useful power developed by the winch motor is $3.5\\text{ kW}$.\n\nCalculate the time, in seconds, needed to hoist the sample to the vessel.","markscheme":"- Calculates the work done in lifting the core sample: $\\text{Work done} = 14\\,000 \\times 450 = 6\\,300\\,000\\text{ J}$ M1\n- Converts power to watts ($3500\\text{ W}$) and uses $\\text{time} = \\frac{\\text{Work done}}{\\text{Power}}$ M1\n- Evaluates the time to get $1800\\text{ s}$ A1","marks":3,"order":0,"markingRubric":null,"explanation":{"methods":[{"id":"work_and_power","label":"Work done and power","steps":[{"type":"text","order":1,"title":"Choose the equation","content":"The sample moves vertically upwards, against its weight. So the force is $14\\,000\\text{ N}$ and the distance moved in the force's direction is $450\\text{ m}$.\n\nUse:\n\n$$W=Fd$$\n\nHere, $W$ is the work done in joules, $F$ is the force in newtons, and $d$ is the distance in metres.","highlights":[{"text":"weighing $14\\,000\\text{ N}$","location":"specification"},{"text":"lifted vertically through a height of $450\\text{ m}$","location":"specification"}]},{"type":"text","order":2,"title":"Calculate the work done","content":"Use force multiplied by the vertical distance lifted. Calculating 14 000 × 450 = 6 300 000 J earns **M1**.","highlights":[{"text":"weighing $14\\,000\\text{ N}$","location":"specification"},{"text":"height of $450\\text{ m}$","location":"specification"}]},{"type":"text","order":3,"title"

Power and efficiency

Question 1

Power and efficiency Questions

Power and efficiency

Question 1

Power and efficiency Questions