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Look at the information about different electric motors - OCR Gateway - GCSE Physics - Question 18 - 2018 - Paper 1

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Look at the information about different electric motors. Electric motor Energy input per hour (J) Useful energy output per hour (J) Energy 'wasted' per hour (J) A 7... show full transcript

Worked Solution & Example Answer:Look at the information about different electric motors - OCR Gateway - GCSE Physics - Question 18 - 2018 - Paper 1

Step 1

Calculate the energy input per hour in J for electric motor D.

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Answer

The energy input for electric motor D is already provided in the table as 60,000 J.

Step 2

Which electric motor has the lowest 'wasted' energy in one hour?

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Answer

Electric motor E has the lowest 'wasted' energy, which is 6,000 J, calculated as follows:

  • Motor A: Wasted = 72,000 J - 60,000 J = 12,000 J
  • Motor B: Wasted = 54,000 J - 36,000 J = 18,000 J
  • Motor C: Wasted = 18,000 J - 3,000 J = 15,000 J
  • Motor D: Wasted = 48,000 J - 12,000 J = 36,000 J
  • Motor E: Wasted = 54,000 J - 48,000 J = 6,000 J

Step 3

Which electric motor has the highest 'wasted' energy in one hour?

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Answer

Electric motor D has the highest 'wasted' energy, which is 36,000 J.

Step 4

Describe how energy is 'wasted' in an electric motor.

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Answer

Energy is 'wasted' in electric motors primarily as heat due to resistance in the windings, friction in the moving parts, and sound energy produced during operation. This wasted energy does not contribute to the useful work output of the motor.

Step 5

Suggest how this 'wasted' energy can be reduced in an electric motor.

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Answer

To reduce 'wasted' energy in an electric motor, measures can include:

  • Lubrication: Reducing friction between moving parts through proper lubrication.
  • Improving Design: Utilizing efficient motor designs that minimize resistance in electrical components and aerodynamics.
  • Maintenance: Regular maintenance to ensure that all components are functioning optimally.

Step 6

Calculate the % efficiency of electric motor E.

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Answer

To calculate the efficiency of electric motor E, use the formula:

extEfficiency=(Useful output energy transferInput energy transfer)×100 ext{Efficiency} = \left( \frac{\text{Useful output energy transfer}}{\text{Input energy transfer}} \right) \times 100

Substituting the values:

  • Useful output = 48,000 J
  • Input = 54,000 J

Thus,

Efficiency=(48,00054,000)×100=88.89%89%ext(to2significantfigures)\text{Efficiency} = \left( \frac{48,000}{54,000} \right) \times 100 = 88.89\% \approx 89\% ext{ (to 2 significant figures)}

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