When a garden hose with an output diameter of 20 mmmm is directed straight upward, the stream of water rises to a height of 0.19 mm . You th

Question

When a garden hose with an output diameter of 20 mmmm is directed straight upward, the stream of water rises to a height of 0.19 mm . You then use your thumb to partially cover the output opening so that its diameter is reduced to 10 mmmm. Part A How high does the water rise now? Ignore drag and assume that the smaller opening you create with your thumb is circular.

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Ben Gia 2 months 2021-08-30T15:33:51+00:00 1 Answers 0 views 0

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    2021-08-30T15:35:26+00:00

    Answer:

    3.04 m

    Explanation:

    20 mm = 0.02 m

    10 mm = 0.01 m

    When the water escape from the hose opening with a velocity v to rise up to a height of 0.19m, its potential energy is converted from kinetic energy:

    E_p = E_k

    mgh = mv^2/2

    where m is the mass and h is the vertical distance traveled, and g = 9.81m/s2 is the gravitational acceleration

    We can divide both sides by m and get:

    gh = v^2/2

    h = v^2/(2g) (1)

    When you cover part of the hose and reduce the cross sectional area, water velocity increases:

    A_1v_1 = A_2v_2

    v_2 = v_1\frac{A_1}{A_2}

    As the area of the circular opening is:

    A = \pi d^2/4

    where d is the diameter

    v_2 = v_1\frac{\pi d_1^2/4}{\pi d_2^2 /4} = v_1(d_1/d_2)^2 = v_1(0.02 / 0.01)^2 = 4 v_1

    v_2^2 = (4v_1)^2 = 16v_1^2

    v_2^2/v_1^2 = 16

    From equation (1) we have the following ratio

    \frac{h_2}{h_1} = \frac{v_2^2/2g}{v_1^2/2g} = v_2^2 / v_1^2 = 16

    h_2 = 16h_1 = 16*0.19 = 3.04 m

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