Sudhir asked in Science & MathematicsPhysics · 10 months ago

Physics Homework Help Needed ASAP?

A child's water pistol shoots water through a 1.0-mm-diameter hole. If the pistol is fired horizontally 75 cm above the ground, a squirt hits the ground 1.2 m away.

What is the volume flow rate during the squirt? Ignore air resistance.

QQ =

mL/s

2 Answers

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  • oubaas
    Lv 7
    10 months ago

    falling time t = √2h/g = √1.5/9.806 = 0.391 sec

    water speed Vw = x/t = 1.2/0.391 = 3.068 m/sec

    let's put dimension in cm

    Φ = 1.0 mm = 0.10 cm

    Vw = 3.068 m/sec = 306.8 cm/sec

    A = 0.7854*Φ^2 = 0.007854 cm^2

    Flow Q = A*Vw = 0.007854*306.8 = 2.410 cm^3/sec (ml/sec)

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  • 10 months ago

    The fall time is

    t = sqrt(2h/g)

    The initial velocity of the water therefore was

    v = D/t = D sqrt(g/(2h))

    where D is the horizontal distance.

    The flow rate is

    Q = v A

    = A D sqrt(g/(2h))

    where A is the area of the opening. If the diameter of the circular opening is d, then A = 1/4 pi d^2.

    Q = 1/4 pi d^2 D sqrt(g/(2h))

    If I computed correctly upon plugging in the quantities, it gives 2.4 ml/s, but do check it. You will need to convert from m^3/s to cm^3/s.

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