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A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet.
Compared to the acceleration due to gravity near Earth’s surface, the acceleration due to gravity near the
surface of the planet is approximately
(1) the same (3) one-half as great
(2) twice as great (4) four times as great




    Hmmmm. You have to be a highschool boy. You use the immature name PIMP, and you expect serious responses from adults. Maybe you are just joking about you need help, how are we to know?


    On earth, the distance an object would fall in 3.0 seconds is
    (g/2)*t^2 = 4.9*9 = 44.1 m

    On planet X, it falls 22 meters.

    The distance it falls in a given time t is proportional to the acceleration of gravity.

    Reach your own conclusion from that.


    7 On the surface of Earth, a spacecraft has a mass
    of 2.00 × 104 kilograms. What is the mass of the
    spacecraft at a distance of one Earth radius
    above Earth’s surface?
    (1) 5.00 × 103 kg (3) 4.90 × 104 kg
    (2) 2.00 × 104 kg (4) 1.96 × 105 kg


    The mass does not change with position.


    bobpursley, instead of wasting time on speculating about the poster, just actually help out. Clearly this person needs help because he deliberately came on the site, asking for help. Immature name or not.

    Anyway, the answer is one half as great. Here's how you find out:
    First the givens are:
    vi=0 (b/c it starts at rest)
    m= 5kg
    d= 22m
    t= 3s
    a on Earth always equals 9.81 m/s*s

    Given these variable, I surmised that the best formula to use is d=vi(t)+(1/2)at^2.

    When you plug in the variables given, you are left with:
    22=(1/2)a(3)^(2) ,which gives you:

    Now its basic algebra. Divide both sides by 4.5 and you end up getting a=4.8 repeating. When you divide 9.81 by 2, you get 4.905, which is close enough, showing you that it one half as great.

    Hope that helps! And good luck on the regents.


    >Uses the name PIMP
    >Expects a serious homework question

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