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11 June, 06:39

You spray your sister with water from a garden hose. The water is supplied to the hose at a rate of 0.111 liters per second and the diameter of the nozzle you hold is 5.79 mm. At what speed does the water exit the nozzle

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Answers (2)
  1. 11 June, 06:51
    0
    29.5 m/s

    Explanation:

    Volumetric flowrate = (average velocity of flow) * (cross sectional area)

    Volumetric flowrate = 0.111 liters/s = 0.000111 m³/s

    Cross sectional Area of flow = πr²

    Diameter = 0.00579 m,

    Radius, r = d/2 = 0.002895 m

    A = π (0.002895) ² = 0.0000037629 m²

    Velocity of flow = (volumetric flow rate) / (cross sectional Area of flow)

    v = 0.000111/0.0000037629

    v = 29.5 m/s
  2. 11 June, 07:04
    0
    Given Information:

    diameter of the nozzle = d = 5.79 mm = 0.00579 m

    flow rate = 0.111 liters/sec

    Required Information:

    Velocity = v = ?

    Answer:

    Velocity = 4.21 m/s

    Explanation:

    As we know flow rate is given by

    Flow rate = Velocity*Area of nozzle

    Where

    Area of nozzle = πr²

    where

    r = d/2

    r = 0.00579/2

    r = 0.002895 m

    Area of nozzle = πr²

    Area of nozzle = π (0.002895) ²

    Area of nozzle = 2.6329x10⁻⁵ m²

    Velocity = Flow rate/area of nozzle

    Divide the litters/s by 1000 to convert into m³/s

    0.111/1000 = 1.11x10⁻⁴ m³/s

    Velocity = 1.11x10⁻⁴/2.6329x10⁻⁵

    Velocity = 4.21 m/s

    Therefore, the water exit the nozzle at a speed of 4.21 m/s
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