AAMC CBT 11 PS #20 CALCULATION

[Oh_Gee]

when i rearranged i got
F= E * (L/DELTA L) * A

Is the .01 in their answer the delta L? wouldn't you use the reciprocal of this when you calculate. how do you even do this without L?

In E=(F/A)/(delta L/L)
a 1% change could be substituted as (1/100) right? so wouldn't you get

F=(2 x 10^8) (100) (5 x 10^-4)

also for #30 i thought force= (air speed^2) (area of person) like the passage says. but i guess you can't use this cause you don't know the person's area.

but if the person terminal velocity, that means her velocity is constant and therefore acceleration due to gravity is 0 so how can you use F=mg for this?


32. why would [NiF2] be useful?

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[K1NG]

Not sure if you figured these out already, but I was looking at unanswered topics so I thought I'd chime in..

(1) Delta L should be on top, I think you made a mistake in your rearrangement. Rearranged formula should be F = mg = A*E*(deltaL/L)

(2) There are basically two forces acting on a person falling. Try drawing a FBD to help you visualize it. When acceleration is constant, then F (upward force of air resistance) simply equals the weight (downward force due to gravity) of the person. Therefore F = mg = (75)(10). The force doesn't equal (air speed^2)(area of person), but it's proportional to it. So if the area of the person or speed increases, they will experience greater upward force.

(3) The molar solubility of the products would be useful, and you saw that, so A and C were eliminated. The concentration of NiF2 is useful in this case, so D is also eliminated. The concentration will show how many moles of NiF2 are present, and therefore will indicate how they interact with MgSO4 through stoichiometry. This post explains it better than I can: http://forums.studentdoctor.net/threads/aamc-11-ps-32-solubility.915034/#post-12535702