Calculate the speed of water (in m/s) flowing from a pipe at the bottom of an open topped barrel which has a water level 138 cm high in it.

(1/2) rho v^2 = rho g h

v = sqrt (2 g h ) = sqrt (2*9.81*1.38)
which is exactly what would happen if you just dropped a rock that distance
or see
http://en.wikipedia.org/wiki/Bernoulli%27s_principle

To calculate the speed of water flowing from a pipe at the bottom of a barrel, we can use the principles of fluid dynamics and Bernoulli's equation.

Bernoulli's equation states that the sum of pressure energy, kinetic energy, and potential energy per unit volume of a flowing fluid is constant along a streamline. In this case, potential energy and pressure energy can be neglected since the fluid is open to the atmosphere.

Let's assume the water level in the barrel creates a pressure difference between the top and bottom of the pipe. The pressure difference can be calculated using the hydrostatic pressure equation:

P = ρgh

Where:
P is the pressure difference
ρ is the density of water (approximately 1000 kg/m³)
g is gravitational acceleration (approximately 9.8 m/s²)
h is the height difference (in meters)

In this case, the height difference is 138 cm, which is equal to 1.38 meters. Plugging these values into the equation:

P = 1000 kg/m³ * 9.8 m/s² * 1.38 m
P ≈ 13404 Pa

Now, using Bernoulli's equation:

P + 0.5ρv² = constant

At the bottom of the barrel (where the pipe is located), the pressure is atmospheric pressure (approximately 101325 Pa), and we can consider the velocity of water as the unknown variable, v.

101325 Pa + 0.5 * 1000 kg/m³ * v² = 13404 Pa

Rearranging the equation to solve for v:

0.5 * 1000 kg/m³ * v² = 101325 - 13404 Pa
0.5 * 1000 kg/m³ * v² = 87921 Pa

v² = (87921 Pa) / (0.5 * 1000 kg/m³)
v² ≈ 175.842

v ≈ √(175.842) ≈ 13.25 m/s

Therefore, the speed of water flowing from the pipe at the bottom of the barrel is approximately 13.25 m/s.