Final answer:
The pressure at the deep end of a bath with a depth of 0.3 m is 104,268 Pa, which includes contributions from both the water's weight and atmospheric pressure, adhering to Pascal's Principle.
Step-by-step explanation:
The question at hand is determining the pressure exerted by water at the deep end of a bath, with a depth of 0.3 m. To calculate this, we use the principle that fluid pressure increases with depth. The formula to determine the pressure at a given depth in a fluid is P = ρgh + P_{atm}, where ρ is the density of the fluid (for water, this is approximately 1000 kg/m³), g is the acceleration due to gravity (9.81 m/s²), h is the depth of the fluid, and P_{atm} is the atmospheric pressure at the surface (approximately 101,325 Pa). Therefore, the total pressure at the deep end of the bath can be calculated.
To find the pressure due to the water alone, we multiply the density of water by the acceleration due to gravity and the depth:
P_water = 1000 kg/m³ × 9.81 m/s² × 0.3 m = 2943 Pa
The total pressure is then the pressure due to the water plus the atmospheric pressure:
P_total = P_water + P_{atm} = 2943 Pa + 101,325 Pa = 104,268 Pa
Therefore, the pressure at a depth of 0.3 m at the deep end is 104,268 Pa, which includes contributions from both the water's weight and atmospheric pressure, as explained by Pascal's Principle.
It is worth noting that just as 10.3 m of water creates the same pressure as the weight of the atmosphere, the pressure increase per meter depth of water is significant compared to the uniform atmospheric pressure that exists at the surface.