Calculating the Speed of Water Exiting from a Sprinkler

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To calculate the speed of water exiting the sprinkler, the principle of conservation of mass applies, where the volume flow rate must remain constant. The area of the hose and the total area of the sprinkler holes must be used to determine the exit speed. The internal diameter of the hose is 1.5 cm, and the sprinkler has 14 holes, each 0.20 cm in diameter. Given the water speed in the hose is 3.0 m/s, the speed of water exiting the sprinkler will be higher due to the smaller exit area. This calculation confirms that water exits the sprinkler at a faster speed due to being forced through multiple smaller openings.
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1. A garden hose has an internal diameter of 1.5cm. It's connected to a sprinkler that consists merely of an enclosure with 14 holes, each .20 cm in diameter. The water in the hose moves with a speed of 3.0m/s. At what speed does the water leave the sprinkler?


OK, so some relevant equations I think would be area- 1/2*pi*r^2. Where i go from there I'm not sure. I also know that the water should leave at a faster speed since it is being forced through many smaller holes.

Your thoughts?
 
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The volume of water passing a point in the thick section in a certain time must equal the volume passing through the sprinkler in the same time. That makes sense since if this was not true some water would have gone missing somewhere. What will be the volume of water passing a point in the hose for a length of time \Delta T?
 

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