Sure! Let's solve this problem step-by-step.
### Problem Details:
- We have a tank with dimensions 3 meters in length, 2 meters in width, and 1 meter in height.
- The tank is half-filled with water.
- We need to calculate the pressure exerted by the water at the bottom of the tank.
### Given Data:
1. Dimensions of the tank:
- Length (L) = 3 meters
- Width (W) = 2 meters
- Height (H) = 1 meter
2. Density of water:
- Density ([tex]\(\rho\)[/tex]) = 1000 kg/m³ (standard density of water)
3. Gravitational acceleration:
- Gravitational acceleration (g) = 9.8 m/s²
### Steps to Calculate the Pressure:
1. Calculate the height of the water when the tank is half-filled:
- The total height of the tank is 1 meter.
- When half-filled, the height of the water (h) = [tex]\( \frac{1}{2} \)[/tex] meters = 0.5 meters.
2. Use the formula to calculate the pressure exerted by a liquid at a certain depth:
- The formula for pressure is given by:
[tex]\[ P = \rho \cdot g \cdot h \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure.
- [tex]\( \rho \)[/tex] is the density of the liquid.
- [tex]\( g \)[/tex] is the gravitational acceleration.
- [tex]\( h \)[/tex] is the height (or depth) of the liquid column.
3. Substitute the known values into the formula:
- [tex]\( \rho = 1000 \)[/tex] kg/m³
- [tex]\( g = 9.8 \)[/tex] m/s²
- [tex]\( h = 0.5 \)[/tex] meters
[tex]\[
P = 1000 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2 \times 0.5 \, \text{meters}
\][/tex]
4. Calculate the product:
[tex]\[
P = 1000 \times 9.8 \times 0.5 = 4900 \, \text{Pascals}
\][/tex]
### Conclusion:
The pressure exerted by the water when the tank is half-filled is 4900 Pascals (Pa).
