water flow at 0.65m/s through a 3-cm
diameter hose.at the end of the hose is a 0.3cm
diameter nozzle,at what speed does the water pass
through the nozzle?
To find the speed at which water passes through the nozzle, we can use the principle of conservation of mass, which states that the mass entering an area must equal the mass leaving that area.
Given:
- Water flow speed through the hose (v1) = 0.65 m/s
- Diameter of the hose (D1) = 3 cm = 0.03 m
- Diameter of the nozzle (D2) = 0.3 cm = 0.003 m
We can use the equation of continuity, which states that the flow rate (Q) remains constant at any cross-sectional area of the flow:
Q = A1 * v1 = A2 * v2
Where:
- A1 and A2 are the cross-sectional areas of the hose and nozzle, respectively
- v2 is the speed of water passing through the nozzle
First, let's calculate the cross-sectional areas:
- A1 = (1/4) * π * D1^2 (Area of a circle)
- A2 = (1/4) * π * D2^2 (Area of a circle)
Plugging in the given values:
A1 = (1/4) * π * (0.03)^2
A2 = (1/4) * π * (0.003)^2
Now, rearrange the equation of continuity to solve for v2:
v2 = (A1 * v1) / A2
Substitute the values and calculate:
v2 = ((1/4) * π * (0.03)^2 * 0.65) / ((1/4) * π * (0.003)^2)
Simplifying, we get:
v2 = (0.09 * 0.65) / 0.000009 = 5.85 / 0.000009
Calculating, we find that:
v2 = 650,000 m/s
Therefore, the water passes through the nozzle at a speed of 650,000 m/s.
volumewaterin=volumewaterout
areain*velocity=areaout*velocityout
velocityout=PI*1.5^2*.65/PI*.15^2
velocity out=100*.65 m/s