Num tubo em U, coloca-se mercúrio (p=13600 kg/m³) e água (p=1000 kg/m³). A altura da coluna do mercúrio acima da interface é 100 cm. Calcule a altura da coluna da água no verso da folha (1.0v).

The hydrostatic pressure ($P_{hydrostatic}$) is given by the formula $P_{hydrostatic} = \rho g h$, where $\rho$ is the density of the fluid, $g$ is the acceleration due to gravity, and $h$ is the height of the fluid column.
Given: $\rho_{oil} = 800 \text{ kg/m}^3$, $g = 10 \text{ m/s}^2$, $h = 8 \text{ m}$.

$$P_{hydrostatic} = (800 \text{ kg/m}^3) \times (10 \text{ m/s}^2) \times (8 \text{ m})$$ $$P_{hydrostatic} = 64000 \text{ Pa}$$

The absolute pressure ($P_{absolute}$) is the sum of the atmospheric pressure ($P_{atm}$) and the hydrostatic pressure.
Given: $P_{atm} = 101.325 \text{ Pa}$.

$$P_{absolute} = P_{atm} + P_{hydrostatic}$$ $$P_{absolute} = 101.325 \text{ Pa} + 64000 \text{ Pa}$$ $$P_{absolute} = 165325 \text{ Pa}$$

The hydrostatic pressure is $\boxed{\text{64000 Pa}}$ and the absolute pressure is $\boxed{\text{165325 Pa}}$.

Given: $h_{Hg} = 100 \text{ cm}$.

$$h_{Hg} = 100 \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}} = 1 \text{ m}$$

In a U-tube, at the same horizontal level within a continuous fluid, the pressure is equal. Therefore, the pressure exerted by the mercury column above the interface is equal to the pressure exerted by the water column above the interface.

$$\rho_{Hg} g h_{Hg} = \rho_{water} g h_{water}$$

We can cancel $g$ from both sides:

$$\rho_{Hg} h_{Hg} = \rho_{water} h_{water}$$

Given: $\rho_{Hg} = 13600 \text{ kg/m}^3$, $\rho_{water} = 1000 \text{ kg/m}^3$, $h_{Hg} = 1 \text{ m}$.

$$(13600 \text{ kg/m}^3) \times (1 \text{ m}) = (1000 \text{ kg/m}^3) \times h_{water}$$ $$13600 = 1000 \times h_{water}$$ $$h_{water} = \frac{13600}{1000} \text{ m}$$ $$h_{water} = 13.6 \text{ m}$$

The height of the water column is $\boxed{\text{13.6 m}}$.