91Ó°ÊÓ

We will use the pressure formula for fluids, which is: \[P = \rho gh\] where \(P\) is the pressure, \(\rho\) is the density, \(g\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\)), and \(h\) is the height of the fluid column. First, we will calculate the pressure exerted by the given column of mercury (height \(760\,\text{mm}\) and density \(13.6\,\text{g/mL}\)). We will use the height in meters and convert the density to \(\text{kg/m}^3\). Height of mercury column in meters: \(0.76 \, \text{m}\) Density of mercury in \(\text{kg/m}^3\): \(13.6 \,\text{g/mL} * 1000\,\text{kg/g} * \frac{1}{1000}\,\frac{\text{m}^3}{\text{L}} = 13{,}600\,\text{kg/m}^3\) Pressure exerted by mercury column: \[P_{\text{mercury}} = \rho_\text{mercury} g h_\text{mercury} = (13,600\,\text{kg/m}^3)(9.81\,\text{m/s}^2)(0.76\,\text{m}) = 101{,}104.96\,\text{Pa}\] Now, we will find the height of a water column that exerts the same pressure \(P_{\text{mercury}}\). We are given the density of water as \(1.0\,\text{g/mL}\). Let's convert it to \(\text{kg/m}^3\). Density of water in \(\text{kg/m}^3\): \(1.0\,\text{g/mL} * 1000\,\text{kg/g} * \frac{1}{1000}\,\frac{\text{m}^3}{\text{L}} = 1000\,\text{kg/m}^3\) Now, we have: \[101{,}104.96\,\text{Pa} = (1000\,\text{kg/m}^3)(9.81\,\text{m/s}^2)(h_\text{water})\] To find the height of the water column (\(h_\text{water}\)), we will rearrange the equation and solve for \(h_\text{water}\): \(h_\text{water} = \frac{101{,}104.96\,\text{Pa}}{(1000\,\text{kg/m}^3)(9.81\,\text{m/s}^2)} = 10.33\,\text{m}\) So, the height of a water column that exerts the same pressure as a \(760\,\text{mm}\) column of mercury is \(10.33 \, \text{m}\).

Given the depth of the diver is \(39\,\text{ft}\), we need to convert this depth to meters to work with SI units. We will use the conversion factor: \(1 \, \text{m} = 3.281 \, \text{ft}\). Depth of diver in meters: \(39\,\text{ft} * \frac{1}{3.281}\,\frac{\text{m}}{\text{ft}} = 11.89\,\text{m}\) Now, we can calculate the pressure on the diver due to the water column above them, using the formula and the given density of water: \[P_{\text{water}} = \rho_\text{water} g h_\text{diver} = (1000\,\text{kg/m}^3)(9.81\,\text{m/s}^2)(11.89\,\text{m}) = 116{,}767.9\,\text{Pa}\] Now, we need to add this pressure to the atmospheric pressure at the surface, which is given as \(0.97 \, \text{atm}\). We will first convert this pressure to Pascal using the conversion factor: \(1 \, \text{atm} = 101{,}325 \, \text{Pa}\). Atmospheric pressure at the surface in \(\text{Pa}\): \(0.97\,\text{atm} * 101{,}325 \, \frac{\text{Pa}}{\text{atm}} = 98{,}284.25\,\text{Pa}\) Total pressure on the diver: \[P_\text{diver} = P_{\text{water}} + P_{\text{surface}} = 116{,}767.9\,\text{Pa} + 98{,}284.25\,\text{Pa} = 215{,}052.15\,\text{Pa}\] Finally, we will convert this pressure back to atmospheres: Pressure on the diver in atmospheres: \[\frac{215{,}052.15\,\text{Pa}}{101{,}325\,\frac{\text{Pa}}{\text{atm}}} = 2.12\, \text{atm}\] So, the pressure on the diver is \(2.12 \, \text{atm}\).