Final answer:
The gage pressure at a depth of 9m in the same liquid, assuming constant density, is triple that at 3m, resulting in a pressure of 126kPa.
Step-by-step explanation:
The gage pressure in a liquid at a certain depth can be calculated using the formula: P = hρg, where P is the pressure, h is the depth, ρ is the density of the liquid, and g is the acceleration due to gravity. Given that the gage pressure in a liquid at a depth of 3m is 42kPa, we can determine the pressure at a depth of 9m, assuming the density of the liquid remains constant. Since pressure is directly proportional to depth, tripling the depth from 3m to 9m will triple the pressure, resulting in a gage pressure of 126kPa at a 9m depth.
The gauge pressure in a liquid at a depth can be calculated using the equation P = hρg, where P is the pressure, h is the depth, ρ is the density of the liquid, and g is the acceleration due to gravity. In this case, the depth is 3m and the gauge pressure is 42kPa. To determine the gauge pressure at a depth of 9m, we can use the same equation and substitute the new depth and the same liquid's density.
Let's assume the density of the liquid is constant. Plugging in the values, we get:
P2 = h2ρg = (9m)(ρ)(g) = 3P1 = 3(42kPa) = 126kPa
Therefore, the gauge pressure in the same liquid at a depth of 9m is 126kPa.