Water (p=1000kg/m^3) enters a pipe on the ground with an inside diameter of 2cm at a pressure of 4x10^5 Pa. Then a 1-cm diameter pipe leads the water to a height of 5m above the ground. The flow speed at the inlet is v1=1.5m/s.

A) Find the flow speed at the outlet.
B) Find the pressure at the outlet.

To solve this problem, we can use the principle of conservation of mass and Bernoulli's equation.

A) Flow speed at the outlet (v2):
According to the principle of conservation of mass, the mass flow rate at the inlet (m1) is equal to the mass flow rate at the outlet (m2). The mass flow rate (m) can be calculated using the equation:

m = ρ * A * v

where ρ is the density of water (1000 kg/m^3), A is the cross-sectional area of the pipe, and v is the flow speed.

At the inlet (A1 = π*(0.02/2)^2), the velocity (v1 = 1.5 m/s), and at the outlet (A2 = π*(0.01/2)^2), we need to find the velocity (v2).

Using the conservation of mass:
m1 = m2

ρ * A1 * v1 = ρ * A2 * v2

Simplifying the equation:

A1 * v1 = A2 * v2

π*(0.02/2)^2 * 1.5 = π*(0.01/2)^2 * v2

Solving for v2:

v2 = (π*(0.02/2)^2 * 1.5) / (π*(0.01/2)^2)
= (0.02/2)^2 * 1.5 / (0.01/2)^2
= (0.01/2)^2 * 1.5 / (0.01/2)^2 (Cancelling the units and π)

v2 = (0.01/2)^2 * 1.5
v2 = (0.01^2)/4 * 1.5
v2 = 0.0001 * 1.5
v2 = 0.00015 m/s

Therefore, the flow speed at the outlet is 0.00015 m/s.

B) Pressure at the outlet (P2):
We can use Bernoulli's equation, which states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume is constant along a streamline. The equation can be written as:

P + 0.5 * ρ * v^2 + ρ * g * h = constant

At the inlet (P1 = 4x10^5 Pa, v1 = 1.5 m/s) and at the outlet (P2 = ?, v2 = 0.00015 m/s), we need to find the pressure (P2).

Since the problem doesn't mention any change in height or potential energy between the inlet and outlet, the potential energy term can be ignored.

Applying Bernoulli's equation at the inlet and outlet:

P1 + 0.5 * ρ * v1^2 = P2 + 0.5 * ρ * v2^2

Replacing the known values:

4x10^5 + 0.5 * 1000 * (1.5)^2 = P2 + 0.5 * 1000 * (0.00015)^2

Simplifying:

4x10^5 + 0.5 * 1000 * 2.25 = P2 + 0.5 * 1000 * 0.0000225

4x10^5 + 2250 = P2 + 0.045

P2 = 4x10^5 + 2250 - 0.045
P2 = 402,000.045 Pa

Therefore, the pressure at the outlet is approximately 402,000.045 Pa.