Certainly! Let's go step-by-step to calculate the acceleration due to gravity given the depth of water, the pressure at the bottom of the pond, and the density of water.
### Given Data:
- Depth of water (\(\(h\)\)): 5 meters
- Pressure at the bottom of the pond (\(\(P\)\)): 49,000 Pascals (Pa)
- Density of water (\(\(\rho\)\)): 1000 kilograms per cubic meter (kg/m\(\(^3\)\))
### Formula to use:
The pressure exerted by a fluid column of height \(\(h\)\) and density \(\(\rho\)\) is given by:
\(\[ P = \rho g h \]\)
where:
- \(\(P\)\) is the pressure,
- \(\(\rho\)\) is the density,
- \(\(g\)\) is the acceleration due to gravity,
- \(\(h\)\) is the height (depth in this case).
We need to solve for \(\(g\)\):
\(\[ g = \frac{P}{\rho h} \]\)
### Steps to Solve:
1. Substitute the given values into the formula:
\(\[ P = 49,000 \, \text{Pa} \]\)
\(\[ \rho = 1000 \, \text{kg/m}^3 \]\)
\(\[ h = 5 \, \text{meters} \]\)
2. Calculate \(\(g\)\):
\(\[ g = \frac{49,000}{1000 \times 5} \]\)
3. Perform the multiplication in the denominator:
\(\[ \rho h = 1000 \times 5 = 5000 \]\)
4. Now divide the pressure by this product:
\(\[ g = \frac{49,000}{5000} \]\)
5. Simplify the division:
\(\[ g = 9.8 \, \text{m/s}^2 \]\)
### Conclusion:
The acceleration due to gravity at the given location, based on the provided data, is approximately:
\(\[ g = 9.8 \, \text{m/s}^2 \]\)
Thus, we have calculated the acceleration due to gravity as 9.8 meters per second squared.
