To find the gauge pressure at the base of the dam, we use the formula for hydrostatic pressure:

$P_{\text{gauge}} = \rho \cdot g \cdot h$

where:

• $\rho$ is the density of water (assuming $\rho = 1000 \, \text{kg/m}^3$),
• $g$ is the acceleration due to gravity ($9.8 \, \text{m/s}^2$),
• $h$ is the depth of the water ($32 \, \text{m}$).

Step 1: Calculate the Gauge Pressure

1. Plugging in the values: $P_{\text{gauge}} = 1000 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2 \times 32 \, \text{m}$

2. Calculate: $P_{\text{gauge}} = 313,600 \, \text{Pa}$

Step 2: Convert to kPa

Since $1 \, \text{kPa} = 1000 \, \text{Pa}$: $P_{\text{gauge}} = \frac{313,600 \, \text{Pa}}{1000} = 313.6 \, \text{kPa}$

Answer

The gauge pressure at the base of the dam is 313.6 kPa.

Would you like a deeper explanation on any part of this calculation?

Here are 5 related questions to deepen your understanding:

1. How would the gauge pressure change if the depth of the reservoir were doubled?
2. What would the absolute pressure at the base be if we included atmospheric pressure?
3. How does the density of the liquid affect the gauge pressure in this scenario?
4. Why is gauge pressure different from absolute pressure in fluid mechanics?
5. How would the gauge pressure at the base change on a planet with higher gravitational acceleration?

Tip: Gauge pressure only considers the pressure due to the fluid's weight, not atmospheric pressure.