A dam is used to block the passage of a river and to generate electricity. Approximately `5.73xx10^(4)` kg of water fall each second through a height of 19.6 m. If one half of the gravitational potential energy of the water were converted to electrical energy, how much power would be generated ?

Options

A. `5.50xx10^(6)W`
B. `1.10xx10^(7)W`
C. `2.70xx10^(9)W`
D. `1.35xx10^(9)W`

Answer

To solve the problem, we need to calculate the gravitational potential energy (GPE) of the water falling through the dam and then find out how much power is generated when half of that energy is converted to electrical energy.

### Step-by-Step Solution:

**Step 1: Calculate the gravitational potential energy (GPE) of the water.**

The formula for gravitational potential energy is given by:

\[ \text{GPE} = mgh \]

where:
- \( m \) = mass of water (in kg)
- \( g \) = acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \))
- \( h \) = height (in meters)

Given:
- \( m = 5.73 \times 10^4 \, \text{kg} \)
- \( g = 9.8 \, \text{m/s}^2 \)
- \( h = 19.6 \, \text{m} \)

Substituting the values into the formula:

\[ \text{GPE} = 5.73 \times 10^4 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 19.6 \, \text{m} \]

**Step 2: Perform the calculation.**

Calculating the GPE:

\[ \text{GPE} = 5.73 \times 10^4 \times 9.8 \times 19.6 \]

Calculating step by step:

1. Calculate \( 9.8 \times 19.6 = 192.08 \)
2. Now calculate \( 5.73 \times 10^4 \times 192.08 \)

Using a calculator:

\[ \text{GPE} \approx 5.73 \times 10^4 \times 192.08 \approx 1.101 \times 10^7 \, \text{J} \]

**Step 3: Calculate the energy converted to electrical energy.**

Since only half of the gravitational potential energy is converted to electrical energy:

\[ \text{Electrical Energy} = \frac{1}{2} \times \text{GPE} = \frac{1}{2} \times 1.101 \times 10^7 \, \text{J} \]

Calculating this gives:

\[ \text{Electrical Energy} \approx 5.505 \times 10^6 \, \text{J} \]

**Step 4: Calculate the power generated.**

Power is defined as energy per unit time. Since the water falls each second, the power generated is:

\[ \text{Power} = \text{Electrical Energy} \, \text{(in Joules)} \, \text{per second} \]

Thus:

\[ \text{Power} = 5.505 \times 10^6 \, \text{W} \]

### Final Answer:

The power generated is approximately:

\[ \text{Power} \approx 5.5 \times 10^6 \, \text{W} \]

---