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    __sarah__'s Avatar
    __sarah__ Posts: 20, Reputation: 1
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    #1

    Jan 16, 2005, 03:11 PM
    HELP Please!! Gr 11 physics
    A boy is sitting in his tree house and observes a falling walnut passing from the top to the bottom of his window. The window is 2.5 m tall and it takes 0.43 s for the walnut to pass the window. How far is tehtop of the window from the squirrel above it that dropped the window? Ignore air resistance.

    a) 0.29 m
    b) 0.58 m
    c) 0.70 m
    d) 2.5 m
    3) 3.2 m

    You throw your pen straight up in your physics classroom. Ignore air resistance. Which statement concerning the net force acting on the pen at the top of its path is true?

    a) the net force is instantaneously equal to zero.
    b) the direction for the net force changes from up to down.
    c) the net force is greater than the weight of the pen.
    d) the net force is equal to the weight of the pen.
    e) the net force is less than the weight of the pen, but greater than zero.

    A pail filled with sand has a total mass of 60 kg. A crane is lowering it such that it has an initial downward acceleration of 1.5 m/s^2. a hole in the pail allows sand to leak out. If the force exerted by the crane on the pail does not change, what mass of sand must leak out before the downward acceleration decreases to zero?

    a) 9.2 kg
    b) 20 kg
    c) 40 kg
    d) 51 kg
    e) 60 kg

    thanks
    drwls's Avatar
    drwls Posts: 14, Reputation: 2
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    #2

    Jan 16, 2005, 10:54 PM
    re: gr 11 physics
    The answers are: c, d, a

    The first and third problems require a bit of algebra.

    For the first problem, let H be the height above the top of the window from this the object is dropped. Let t be the time it takes to reach the top of the window. You have two equaitons in two unknowns:
    H = (g/2) t^2, and
    H + 2.5 = (g/2) (t + 0.43)^2
    Is it easiest to eliminate H first and solve for it once you know t.
    2.5 = (g/2) (0.86 t + 0.185).. which leads to
    t = 0.378 s
    H = 4.9 t^2 = 0.700 m

    For the third problem, let M equal the initial mass of the bucket and sand, and F be the (constant) force applied by the crane. Initially,
    (60 kg) * g - T = (60 kg) * a = 90 N
    When the deceleration rate becomes zero, the mass changes from 60 kg to to a value M such that
    M g - T = 0
    Eliminate the unknown T from the two equations, by subtracting, and you get
    (60 - M) g = 90 N
    60 - M = 9.2 kg (the amount lost)

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