Home
/
Physics
/
A remote village in the UK uses a hydroelectric generator to provide electricity. In one day, 2500 000 kg of water passes through the hydroelectric generator. The change in gravitational potential energy of the water is 367.5 MJ. gravitational field strength=9.8N/kg Calculate the mean change in vertical height of the water as it moves through the hydroelectric generator. Use the Physics Equations Sheet. [4 marks __

Question

A remote village in the UK uses a hydroelectric generator to provide electricity.
In one day, 2500 000 kg of water passes through the hydroelectric generator.
The change in gravitational potential energy of the water is 367.5 MJ.
gravitational field strength=9.8N/kg
Calculate the mean change in vertical height of the water as it moves through the
hydroelectric generator.
Use the Physics Equations Sheet.
[4 marks
__

A remote village in the UK uses a hydroelectric generator to provide electricity. In one day, 2500 000 kg of water passes through the hydroelectric generator. The change in gravitational potential energy of the water is 367.5 MJ. gravitational field strength=9.8N/kg Calculate the mean change in vertical height of the water as it moves through the hydroelectric generator. Use the Physics Equations Sheet. [4 marks __

expert verifiedVerification of experts

Answer

4.4219 Voting
avatar
UlyssesElite · Tutor for 8 years

Answer

The mean change in vertical height of the water as it moves through the hydroelectric generator is 15 m.

Explain

<br /><br />1. The formula to calculate the change in gravitational potential energy (ΔU) is \( ΔU = m \times g \times Δh \), where: <br /> - \( m \) is the mass of the object.<br /> - \( g \) is the gravitational field strength.<br /> - \( Δh \) is the change in height.<br /><br />2. From the question, we know that: <br /> - The mass that moves through the hydroelectric generator \( m = 2500000 \, \text{kg} \),<br /> - The change in gravitational potential energy \( ΔU = 367.5 \mathrm{~MJ} = 367.5 \times 10^6 \mathrm{~J} \) (converted from MJ to J), because we need to maintain units consistency in the formula. <br /> - Gravitational field strength \( g = 9.8 \mathrm{~N} / \mathrm{kg} \).<br /><br />3. We can rearrange the formula to find the change in height \( Δh \): \( Δh = ΔU / (m \times g) \).<br /><br />4. Substituting the known values in the rearranged formula gives \( Δh = 367.5 \times 10^6 \mathrm{~J} / (2500000 \mathrm{~kg} \times 9.8 \mathrm{~N} / \mathrm{kg}) \).<br /><br />5. The calculation gives the mean change in vertical height of the water as it moved through the hydroelectric generator by about \( Δh = 15 \mathrm{~m} \).
Click to rate:

Hot Questions

More x