Final answer:
The gage pressure in a liquid at 9m depth is three times the pressure at 3m depth; therefore, the pressure at 9m depth is 126 kPa, assuming the density of the liquid and gravity remain constant.
Step-by-step explanation:
When we wish to find the gage pressure at a specific depth in a liquid, we rely on the relationship P = hpg, which tells us that pressure (P) changes linearly with depth (h) given a constant fluid density (ρ) and acceleration due to gravity (g). To determine the gage pressure at a new depth, we can use the fact that at a depth of 3m, the pressure is already given to be 42 kPa. Using the proportional relationship of pressure with depth, we can calculate the pressure at a depth of 9m.
Since the initial conditions provide us 42 kPa at 3m, tripling the depth to 9m means tripling the pressure associated with the depth, since gravity (g) and the fluid's density (ρ) remain constant. Thus, the gage pressure at a depth of 9m would be 3 times 42 kPa, equating to 126 kPa.
To verify this using the equation P = hρg, if we assume that the density (ρ) of the liquid remains constant and the value of gravity (g) is not changing with depth, we can write the equation for both depths and form a ratio.
P1 / P2 = h1 / h2
Where P1 is the initial pressure, P2 is the pressure at the new depth, h1 is the initial depth, and h2 is the new depth. Plugging in the known values we get:
42 kPa / P2 = 3m / 9m
P2 = 42 kPa * (9m / 3m)
P2 = 42 kPa * 3
P2 = 126 kPa
This confirms that the gage pressure at a depth of 9m is 126 kPa.