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 decrease in pressure.

The fluid velocity increases.

The potential energy increases.

The exit velocity increases (specifically, $v = \sqrt{2gh}$).

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