Your town is installing a fountain in the main square. If the water is to rise 16 m (52.5 feet) above the fountain, how much pressure must the water have as it moves slowly toward the nozzle that sprays it up into the air? Assume atmospheric pressure equal to 100,000 Pa.

assuming no energy loss through the nozzle, which is a silly assumption...

energy to raise water to 16m...

Force*distance=energy= mgh
netpressure*area*distance=mgh
(P-100kN/m^2)*areastream*16m=areastream*16m*densitywater*g*16m
(P-100kN/m^2)=16*densitywater*9.8N/kg

P=16m*1E6kg/m^3*9.8N/kg +100kPa

495

To determine the pressure required by the water to rise 16 m above the fountain, we can use the concept of hydrostatic pressure.

Hydrostatic pressure is given by the formula: P = ρ * g * h, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height or depth of the fluid.

In this case, we need to find the pressure exerted by the water at a height of 16 m above the fountain. Since the water is moving slowly, we can assume it has reached an equilibrium state with no kinetic energy.

First, we need to determine the density of water. The density of water is approximately 1000 kg/m³.

Next, we need to determine the acceleration due to gravity, which is approximately 9.8 m/s².

Finally, we can calculate the pressure:

P = ρ * g * h
P = 1000 kg/m³ * 9.8 m/s² * 16 m

Now, let's convert the answer into the desired unit, pascals (Pa).

1 Pa = 1 N/m², and 1 N = 1 kg * m/s²

P = 1000 kg/m³ * 9.8 m/s² * 16 m
P = 156800 N/m² = 156800 Pa

Adding the atmospheric pressure of 100,000 Pa (given in the question) to the pressure exerted by the water, we find:

Total Pressure = 156800 Pa + 100000 Pa
Total Pressure = 256800 Pa

Therefore, the water must have a pressure of 256,800 Pa as it moves slowly toward the nozzle that sprays it up into the air.