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An electric car is travelling at a speed of 16.0 m/s - Edexcel - GCSE Physics - Question 8 - 2023 - Paper 2

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An electric car is travelling at a speed of 16.0 m/s. The total mass of the car is 1200 kg. (i) Calculate the kinetic energy, in kJ, of the car. The kinetic energ... show full transcript

Worked Solution & Example Answer:An electric car is travelling at a speed of 16.0 m/s - Edexcel - GCSE Physics - Question 8 - 2023 - Paper 2

Step 1

Calculate the kinetic energy, in kJ, of the car.

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Answer

To calculate the kinetic energy, we use the formula: KE=12mv2KE = \frac{1}{2}mv^{2} Substituting the values: KE=12×1200×(16.0)2KE = \frac{1}{2} \times 1200 \times (16.0)^{2} Calculating: KE=12×1200×256=153600 J=153.6 kJKE = \frac{1}{2} \times 1200 \times 256 = 153600 \text{ J} = 153.6 \text{ kJ} Thus, the kinetic energy is 153.6 kJ.

Step 2

Calculate the time taken for the battery to become discharged on this journey.

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Answer

Given the average power transfer rate is 17.5 kW (or 17500 W) and the energy transferred is 126 MJ (or 126000000 J), we can use the formula: t=EPt = \frac{E}{P} Substituting: t=12600000017500t = \frac{126000000}{17500} Calculating: t7200 seconds=2exthourst \approx 7200 \text{ seconds} = 2 ext{ hours} So, the time taken for the battery to become discharged is 2 hours.

Step 3

Explain how using the device can help to increase the time that the car can be driven before the battery becomes discharged.

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Answer

The electrical device functions both as a motor when accelerating and as a dynamo when decelerating. When the car decelerates, the energy that would otherwise be wasted during braking is converted back into electrical energy and stored in the battery. This process recharges the battery, thereby increasing the overall efficiency of energy usage and allowing the car to drive longer before the battery gets discharged. Hence, the energy management improves the range of the vehicle.

Step 4

Comment on this claim.

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Answer

To evaluate the claim that 126 MJ can be transferred in less than 6 hours, we first calculate the time using the formula: t=EI×Vt = \frac{E}{I \times V} Where:

  • E=126×106 JE = 126 \times 10^{6} \text{ J}
  • I=15.0 AI = 15.0 \text{ A}
  • V=400 VV = 400 \text{ V} Substituting values: t=126×10615×400t = \frac{126 \times 10^{6}}{15 \times 400} Calculating: t=126×1066000=21000exts=5.83exthourst = \frac{126 \times 10^{6}}{6000} = 21000 ext{ s} = 5.83 ext{ hours} Since 5.83 hours is less than 6 hours, the claim is justified.

Step 5

Calculate the total charge that moves into the battery while it is being recharged.

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Answer

Using the equation: E=Q×VE = Q \times V Rearranging gives: Q=EVQ = \frac{E}{V} Substituting the values: Q=126×106400Q = \frac{126 \times 10^{6}}{400} Calculating: Q=315000 CQ = 315000 \text{ C} Therefore, the total charge that moves into the battery is 315,000 C.

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