Water flows through a 4 cm diameter garden hose at a velocity of 2 m/s. What is the velocity of the water if a
nozzle is used with a diameter of 0.5 cm?
since the volume is constant,
4^2 * 2 = .5^2 * v
To find the velocity of the water when a nozzle with a different diameter is used, we can use the principle of conservation of mass. According to this principle, the mass flow rate of water must remain constant, regardless of the change in diameter.
The formula for mass flow rate is given by:
m_dot = ρ * A * V
where m_dot is the mass flow rate, ρ is the density of water, A is the cross-sectional area, and V is the velocity of the water.
Since the density of water remains constant, we can equate the mass flow rates before and after the change in diameter:
m_dot1 = m_dot2
We can express the mass flow rate in terms of the cross-sectional area and velocity:
ρ * A1 * V1 = ρ * A2 * V2
The cross-sectional areas A1 and A2 can be related using the formula for the area of a circle:
A = π * r^2
where r is the radius. So, the formula becomes:
ρ * π * r1^2 * V1 = ρ * π * r2^2 * V2
We can cancel out the density ρ and π from both sides of the equation:
r1^2 * V1 = r2^2 * V2
Now, we plug in the given values:
r1 = 4 cm = 0.04 m (diameter of the garden hose)
V1 = 2 m/s (velocity of water in the garden hose)
r2 = 0.5 cm = 0.005 m (diameter of the nozzle)
Substituting these values into the equation, we can solve for V2:
(0.04^2) * 2 = (0.005^2) * V2
0.0016 * 2 = 0.000025 * V2
0.0032 = 0.000025 * V2
V2 = 0.0032 / 0.000025
V2 ≈ 128 m/s
Therefore, the velocity of the water when a nozzle with a diameter of 0.5 cm is used is approximately 128 m/s.