Four lawn sprinklerheads are fed by a 1.9 cm diameter pipe. The water comes out

Question

Four lawn sprinklerheads are fed by a 1.9 cm diameter pipe. The water comes out of theheads at an angle of 33° to thehorizontal and covers a radius of 7.8m.

(a) What is the velocity of the water coming out ofeach sprinkler head? (Assume zero air resistance.)
m/s
(b) If the output diameter of each head is 3.0 mm, how many litersof water do the four heads deliver per second?
                          L/s
(c) How fast is the water flowing inside the 1.9 cm diameterpipe?
                          m/s

Four lawn sprinklerheads are fed by a 1.9 cm diameter pipe. The water comes out of theheads at an angle of 33° to thehorizontal and covers a radius of 7.8m. Four lawn sprinklerheads are fed by a 1.9 cm diameter pipe. The water comes out of theheads at an angle of 33° to thehorizontal and covers a radius of 7.8m. Four lawn sprinklerheads are fed by a 1.9 cm diameter pipe. The water comes out of theheads at an angle of 33° to thehorizontal and covers a radius of 7.8m. Four lawn sprinklerheads are fed by a 1.9 cm diameter pipe. The water comes out of theheads at an angle of 33° to thehorizontal and covers a radius of 7.8m. Four lawn sprinklerheads are fed by a 1.9 cm diameter pipe. The water comes out of theheads at an angle of 33° to thehorizontal and covers a radius of 7.8m.

(a) What is the velocity of the water coming out ofeach sprinkler head? (Assume zero air resistance.)
m/s
(b) If the output diameter of each head is 3.0 mm, how many litersof water do the four heads deliver per second?
                          L/s
(c) How fast is the water flowing inside the 1.9 cm diameterpipe?
                          m/s

(a) What is the velocity of the water coming out ofeach sprinkler head? (Assume zero air resistance.)
m/s
(b) If the output diameter of each head is 3.0 mm, how many litersof water do the four heads deliver per second?
                          L/s
(c) How fast is the water flowing inside the 1.9 cm diameterpipe?
                          m/s

Explanation / Answer

. Problem statement: Four lawn sprinklerheads are fed by a 1.9 cm diameter pipe. The water comes out of theheads at an angle of 33° to thehorizontal and covers a radius of 7.8m. (a) What is the velocityof the water coming out of each sprinkler head? (Assume zero airresistance.)
m/s
(b) If the output diameter of each head is 3.0 mm, how many litersof water do the four heads deliver per second? _____L/s
(c) How fast is the water flowing inside the 1.9 cm diameterpipe? _____m/s . 01.   I willaccept your response to part_A and work the remainingsolutions. 02.  Calculate the volume-rate of the water from the sprinklerhead 03.      VSprinklerHead =(4)*{[()*(0.003 meters)2]/(4)}*(9.14 meters/second)= 258.43E-06 meters3/second 04.      1 cubic meter = 1000liters      ;   conversionof units 05.      VSprinklerHead =(258.43E-06 meters3/second)*[(1000 liters)/(1meter3)] = 258.43E-03 Liters/second 06.  Calculate the flow-rate through the pipe =(volume_rate)/(cross-sectional area) 07.      Fpipe = pipe flowrate = (258.43E-06 meters3/second)/{[()*(0.019meters)2]/(4)} = 911.47E-03 meters/second . Four lawn sprinklerheads are fed by a 1.9 cm diameter pipe. The water comes out of theheads at an angle of 33° to thehorizontal and covers a radius of 7.8m. (a) What is the velocityof the water coming out of each sprinkler head? (Assume zero airresistance.)
m/s
(b) If the output diameter of each head is 3.0 mm, how many litersof water do the four heads deliver per second? _____L/s
(c) How fast is the water flowing inside the 1.9 cm diameterpipe? _____m/s . 01.   I willaccept your response to part_A and work the remainingsolutions. 02.  Calculate the volume-rate of the water from the sprinklerhead 03.      VSprinklerHead =(4)*{[()*(0.003 meters)2]/(4)}*(9.14 meters/second)= 258.43E-06 meters3/second 04.      1 cubic meter = 1000liters      ;   conversionof units 05.      VSprinklerHead =(258.43E-06 meters3/second)*[(1000 liters)/(1meter3)] = 258.43E-03 Liters/second 06.  Calculate the flow-rate through the pipe =(volume_rate)/(cross-sectional area) 07.      Fpipe = pipe flowrate = (258.43E-06 meters3/second)/{[()*(0.019meters)2]/(4)} = 911.47E-03 meters/second .

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