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EK Physics #103 and #107

Discussion in 'MCAT Study Question Q&A' started by aries12r40, Dec 29, 2008.

  1. aries12r40

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    I have trouble figuring out the answers for EK Physics 30min In Class Exam Lecture 5 #103 and #107. The question for #103 asks

    A 2kg object submerged in the unknown fluid has an apparent loss of mass of 0.5kg. What is the specific gravity of the object?

    The answer should be 20. The book's explanation is very confusing to me. Maybe someone can offer a better explanation? Also, how does the weight loss of the object has to do with the specific gravity of the unknown fluid?

    Also, for #107, why is that the speed equals frequency times circumference of the golf ball? The question asks

    The air very close to the surface of the ball is dragged along by a golf ball's spinning motion so that it moves at the same speed as the surface of the ball. If a golf ball in flight spins with a frequency of 60Hz, what is the approximate speed w of the air at its surface?


    Many thanks!
     
  2. futuredoctor10

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    by w, i think you mean the angular velocity omega. If it asks for approximate speed "w" (meaning angular velocity?) then:

    w = 2pi x f

    Not sure why it says 2pi r x f though... not sure why "r" is accounted for.
     
  3. aries12r40

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    Hello!
    The w for #107 is not omega, it is the speed moving on the surface of the golf ball (please see passage for more detail; p. 189, passage III in EK ).

    But your suggestion that w=2pi*f gives me an idea about where the answer comes from. Since w is asking for the speed on the surface of the ball, and speed (not velocity) is distance over time, then the speed will be circumference times frequency. Circumference is the distance the air travels on the surface of the ball and frequency is (1/sec). So altogether the speed would be circumference times the frequency.

    But I am still clueless for #103, any idea?

    Thank you! :)
     
  4. futuredoctor10

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    yeah that makes sense for 107 v = d/t = d x f.... I am not sure about 103 though.
     
  5. delphikijb

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    The speed of a point on the surface of a rotating object should be frequency of rotation times circumference, but circumference is 2 pi r, not pi r which would give you:

    w = (2)(3.14)(0.043m)(60Hz) = ~16m/s, right?

    why do they drop the two in the equation above?
     
  6. loveoforganic

    loveoforganic -Account Deactivated-
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    specific gravity is essentially another term for density (in g/mL). That should now make sense using archimedes' principle.
     
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    #6 loveoforganic, Mar 6, 2009
    Last edited: Mar 6, 2009
  7. amikhchi

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    I was thinking about #103, and I can't seem to figure out the specific gravity of the object, don't you need to know either the density or sp.grav of the fluid in order to calculate this?
     

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