Where do I start?
Calculate the water pressure at the bottom of the 50-m {\rm m}-high water tower shown in the photo.
Express your answer to two significant figures and include the appropriate units.
The question says to neglect the pressure due to atmosphere when doing my calculations. But I don't see how I can figure this out with only one number given ?
If it is 50 meters high, the pressure is the weight of a one meter square column 50 meters high.
Water density is about 1000 Kilograms/ cubic meter
So the mass of this column is 1000 Kg/m^3 * 50 m * g
or
1000 * 50 * 9.81 = 490,500 Newtons
that weight is spread out over the one square meter of base so the pressure is
490,500 Newtons per square meter
A newton per square meter is called a "Pascal"
so we have
4.9*10^5 Pascals
by the way atmospheric pressure is about 1 * 10^5 Pascals so indeed we might as well ignore it. An atmosphere is about 33 feet or 10 meters of water so we have about 5 atmospheres here. If we did account for the atmophere we would say about 6 atm
To calculate the water pressure at the bottom of the water tower, you need to use the concept of hydrostatic pressure, which depends on the height of the column of fluid above a given point.
The formula to calculate hydrostatic pressure is:
P = ρgh
Where:
P is the pressure
ρ (rho) is the density of the fluid (in this case, water)
g is the acceleration due to gravity
h is the height of the fluid column
In this case, the only given information is the height of the water tower, which is 50 m. Although the density of water is not given, we can consider it as a standard value of approximately 1000 kg/m³. The acceleration due to gravity can be approximated as 9.8 m/s².
Now we can use the formula to calculate the pressure at the bottom of the water tower:
P = 1000 kg/m³ * 9.8 m/s² * 50 m
P ≈ 490,000 Pa
So, the water pressure at the bottom of the 50 m-high water tower is approximately 490,000 Pascal (Pa). Remember to round your answer to two significant figures, as requested.