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MCAT physics QUESTIONS, please help!

  1. May 27, 2008 #1

    rexhi

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    I would really appreciate to know how some problems are solved!

    1)Two different objects are dropped from rest off of a 50-m-tall cliff. One lands going 30% faster than the other. The two ojects have the same mass. How much more kinetic energy does one object have at the landing than the other? The answer is 69% more why???

    2) A projectile is fired vertically at a speed of 30 m/s. It reaches a maximum height of 44.1 m. What fraction of its initial energy has been lost to air resistance at this point? the answer is 2%, can somebody explain me why?

    3) At what height above the earth's surface is the acceleration due to gravity 10% of that at sea level? 1.36 x 10^7 m

    4) The moon has 1/2 the acceleration of gravity of the earth. What would the mass of the earth have to be to have this acceleration at its surface? 1.03x10^3

    I wouuld really appreciate if someone can help me, physics is my weakness :(
     
    rexhi, May 27, 2008
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  3. May 28, 2008 #2

    SporadicSmile

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    1.
    [tex]E = 1/2 m v^2 [/tex]
    This depends on [tex] v^2 [/tex] and so, the ratio of energies will be the ratio of the velocities squared, so [tex] E_1/E_2 = (1.3v_1)^2/v_2 [/tex]
    which gives 1.69, ie 69% more kinetic energy

    2. with no air resitance [tex] v^2 = 2gs [/tex] where v is the initial velocity, g the acceleration due to gravity and s the maximum height. calculation this gives 45, now just go 44.1/45 to give 0.02, ie 2%

    3. [tex] g = GM/r [/tex], use this and the fact that 10% of g is 0.1g, divide g by 0.1 g to get the answer.

    4. Im not sure you have this answer right, because that is a migthy small mass for a planet to have. Solve this one by using the equation in 3, with 0.5g at the front.
     
    SporadicSmile, May 28, 2008
  4. May 28, 2008 #3

    rexhi

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    thank you so much for your help SporadicSmile :)

    by the way 1.3 was it achieved by (1/2)/(1/2) + 0.3?
     
    rexhi, May 28, 2008
  5. May 28, 2008 #4

    Phlogistonian

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    Actually,

    [tex]
    g = GM/r^2
    [/tex]
     
    Phlogistonian, May 28, 2008
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