Hi
@br2pi5 -
Re: NS 2 C/P #54, yes, that's a Bernoulli's equation thing. The inverse relationship between pressure & velocity if you're holding height more less constant is one of the most common way that Bernoulli's equation is tested on the MCAT, so I'd suggest making that into an instinct. The continuity equation (A * v = constant) links cross-sectional area into this, so increased cross-sectional area = lower velocity = greater pressure, and vice versa. (Also worth noting is that Bernoulli's equation only applies to ideal fluids, which is a simplification for blood, because in reality blood has viscosity -- but you sort of have to roll with the context of any given question in this regard, and for this one, the only choice that deals with viscosity is A, which is extremely wrong [and I see that you correctly eliminated it]).
For the QPack physics question, there are two key ideas here: (1) that specific gravity is a relative measure of density compared to water, and (2) that buoyant force is proportional to the density of the fluid. (This is a consequence of the idea that buoyancy = weight of the displaced fluid, since the volume displaced by the object is not going to change here). Personally I like to avoid complex equations and to keep it conceptual whenever possible, so my line of reasoning would be this: The buoyant force goes up by a factor of 2.4 when the object goes from benzene to the unknown liquid, so the density of the unknown liquid must be 2.4 times that of benzene. Since the density of water is ~1.00 g/cm3, the fact that benzene has a specific gravity of 0.7 means that its density is 0.7 g/cm3. The density of the unknown liquid will be 2.4*0.7 g/cm3 = 1.68 g/cm3, which = a specific gravity of 1.68, which we can round up 7. Even w/out a calculator, we can easily estimate 2.4*0.7 to be about halfway between 1.4 (=2*0.7) and 2.1 (=3*0.7), so 1.7 is the best fit.
Hope this clarifies things
🙂.
