1. Identify two points in the fluid flow. 2. List knowns and unknowns (P, v, h) at each point. 3. Simplify the equation by canceling out negligible terms. 4. Plug in values and solve for the unknown.

1. Identify two points in the fluid flow where the cross-sectional area changes. 2. Use the equation $V = vA$ (volume flow rate = velocity * area) to relate the velocities at the two points. 3. If the volume flow rate is constant, then $v_1A_1 = v_2A_2$.

1. Recognize that the velocity at the top of the tank is negligible ($v_1 \approx 0$). 2. Recognize that both the top and the leak are at atmospheric pressure ($P_1 = P_2$). 3. Simplify Bernoulli's equation to $\rho g h = \frac{1}{2} \rho v^2$. 4. Solve for the velocity of the leak: $v = \sqrt{2gh}$

1. Calculate the potential energy change: $\Delta PE = mgh = V\rho g \Delta h$. 2. Recognize that the work done by the pump equals the change in potential energy of the water. 3. Use the power equation: $W = Pt$, so $t = W/P = \Delta PE/P$. 4. Calculate the time and then the work done: $W = P*t$.

A statement of energy conservation for fluids, relating pressure, velocity, and height at different points in a flowing fluid: $P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2$

Force per unit area. In Bernoulli's equation, it's measured in Pascals (Pa). Higher pressure can push the fluid faster.

Mass per unit volume of the fluid, typically measured in kg/m³.

The speed of the fluid at a given point, measured in m/s. Faster fluid velocity implies more kinetic energy.

The volume of fluid that passes a given point per unit time. It's related to velocity and cross-sectional area by the equation $V = vA$.

Energy of motion, related to fluid velocity. Faster fluid = more KE.

Energy of position, related to height. Higher fluid = more PE.

Work: Energy transferred (measured in Joules). Power: Rate of energy transfer (measured in Watts).