91Ó°ÊÓ

Lets first calculate the absolute pressure at the bottom of the lake. This can be calculated by adding pressure due to the water column and the atmospheric pressure. The pressure due to a fluid column is given by the equation \(P = \rho g h\), where \(P\) is the fluid pressure, \(\rho\) is the fluid density, \(g\) is acceleration due to gravity and \(h\) is the height of the fluid column. Here, \(\rho = 1.00 \times 10^{3} \mathrm{~kg/m^3}\), \(g = 9.8 \mathrm{~m/s^2}\) (approximated value of gravity on Earth), and \(h = 27.5 \mathrm{~m}\). So, \(P_{Fluid} = \rho g h = 1.00 \times 10^{3} \mathrm{~kg/m^3} \times 9.8 \mathrm{~m/s^2} \times 27.5 \mathrm{~m} = 269.5 \mathrm{kPa}\). The atmospheric pressure \(P_{Atmospheric} = 101.3 \mathrm{kPa}\). Hence the absolute pressure \(P_{Absolute}\) at a depth of 27.5 m can be calculated by adding the atmospheric pressure to the fluid pressure. So, \(P_{Absolute} = P_{Fluid} + P_{Atmospheric} = 269.5 \mathrm{kPa} + 101.3 \mathrm{kPa} = 370.8 \mathrm{kPa}\)

Before determining the force exerted by the water on the window, let's first calculate the area of the window. Given the window is in circular shape, we use the area formula for a circle: \(\mathrm{Area} = \pi(\mathrm{Diameter}/2)^2\). Here, the diameter of the window is \(35.0 \mathrm{cm}\), so the radius is \(Diameter / 2 = 35.0 \mathrm{cm} / 2 = 17.5 \mathrm{cm} = 0.175 \mathrm{m}\), (converted from cm to m). So, \(\mathrm{Area} = \pi * (0.175 \mathrm{m})^2 = 0.096 \mathrm{m^2}\).

We can now calculate the force exerted by the water on the window. The force exerted by fluid pressure is given by the equation \(Force = Pressure \times Area\). Here, Pressure is the absolute pressure we calculated in Step 1 and Area is the area of the window we calculated in Step 2. Hence, \(Force = P_{Absolute} \times \mathrm{Area} = 370.8 \mathrm{kPa} \times 0.096 \mathrm{m^2} = 35.6 \mathrm{kN}\). As the pressure is high, Force is given in kilo Newtons (kN).