5 (a) Which of these is a non-renewable source of energy? (1) geotharml A natural gas B tidal C solar (b) Explain why renewable sources provide an increasing fraction of the electricity supply for many countries - Edexcel - GCSE Physics Combined Science - Question 5 - 2018 - Paper 1

Renewable energy sources, such as solar, wind, and hydro, are becoming more prevalent due to their environmental benefits, lower operational costs, and advancements in technology. As countries aim to reduce greenhouse gas emissions and combat climate change, the use of renewable energy helps decrease reliance on fossil fuels. Additionally, government incentives and policies promote the adoption of renewable technologies, thus increasing their share in national electricity supplies.

To find the minimum height, we can use the formula for gravitational potential energy, which is given by: $PE = mgh$ Where:

• $PE$ is the potential energy (1300 J),
• $m$ is the mass (7.0 kg),
• $g$ is the acceleration due to gravity (approximately 9.81 m/s²),
• $h$ is the height in meters.

Rearranging this formula to solve for height gives: $h = \frac{PE}{mg} = \frac{1300}{7.0 \times 9.81} \approx 18.24 m$

Thus, the minimum height is approximately 18.24 meters.

For this calculation, we use the kinetic energy formula: $KE = \frac{1}{2} mv^2$ Where:

• $KE$ is the kinetic energy (1100 J),
• $m$ is the mass of the water (8.0 kg),
• $v$ is the speed.

Rearranging this formula to solve for speed gives: $v = \sqrt{\frac{2 \times KE}{m}} = \sqrt{\frac{2 \times 1100}{8.0}} \approx 17.0 m/s$

Thus, the speed of the moving water as it enters the turbine is approximately 17.0 m/s.

Using the provided graph (Figure 6), read the kinetic energy values associated with wind speeds of 15 m/s and 13 m/s. For a wind speed of 15 m/s, the kinetic energy is approximately 5.2 kJ, and for a wind speed of 13 m/s, it is approximately 3.9 kJ. The difference in energy transferred to the turbine is:

$\Delta KE = 5.2 - 3.9 = 1.3 kJ$

To find the percentage transferred, we calculate: $\text{Percentage} = \left(\frac{\Delta KE}{KE_{initial}}\right) \times 100 = \left(\frac{1.3}{5.2}\right) \times 100 \approx 25\%$

Therefore, the percentage of kinetic energy transferred from the air to the turbine is approximately 25%.