Question #50, Physical Sciences, Kaplan Full Length #4

Question #50, Physical Sciences, Kaplan Full Length #4

The question asks which factor would NOT increase the acceleration of a ball going down an inclined plane.

The correct answer is the mass of the ball. However, I didn't even read that answer because the first answer choice was the angle of the incline.

I thought that increasing the angle of incline would decrease the acceleration and decreasing the angle would increase the acceleration.

Can you please explain?

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Be careful with questions like this. Our intuition, as humans, is not particularly good when it comes to changing angles and thinking about the effects. Since it mentions acceleration in the question stem, think... force!

Three steps:

1. Make a free-body diagram.

2. Write Newton's Second Law.

3. Plug and chug (or, in this case, think about the variables they mention in the answers).

For step 1, let's think about the forces we have:

-Gravity. It will point straight down.

-The normal force, pointing 90 degrees from the incline.

-Friction, if it exists. It will point up the plane, if the object is going down the plane.

Since we don't have the forces acting in two directions only (like x/y axes), we'll have to break the gravity up into components. The force of gravity acting parallel to the plane (down the plane) will be equal to mg sinθ; the force of gravity acting perpendicular to the plane (opposite the normal force) will be equal to mg cosθ. If you don't remember why these components are what they are, make sure to check out the section on inclined planes in the Review Notes or Lesson Book.

So, let's move onto Newton's Second Law. The dimension we're concerned with here is everything parallel to the plane. This would be F = ma, where the net force is equal to mg sinθ - f (which is the force of gravity parallel to the plane, minus the friction. So mg sinθ - f = ma.

Remember that friction can also be expressed as the coefficient of friction, μ, times the normal force. We also pointed out that the normal force is equal and opposite to the component of gravity pointing perpendicular to the plane (mg cosθ). So, friction = μN = μmg cosθ.

Putting that into our equation, we have mg sinθ - μmg cosθ = ma. We can cancel out m, and have just g sinθ - μg cosθ = a. This is perfect for the question -- let's see how each variable will affect acceleration.

Starting with A (increasing the angle of inclination), what will that affect? It will affect θ throughout. As θ increases from 0 to 90 degrees, sinθ also increases (from 0 to 1); conversely, cosθ decreases (from 1 to 0). So acceleration must be increasing -- gsinθ is increasing, and we're subtracting a smaller and smaller number with μg cosθ. So increasing the angle increases the acceleration. Also, think intuitively -- the acceleration when θ=0 should be zero (it's flat on a table), and when θ=90, acceleration will be equal to g, since the ball will be in freefall then.

Now for B and C. Decreasing the friction will decrease μ, so that will increase acceleration (we'll be subtracting a smaller number). We didn't talk about air resistance, but removing that drag force should also logically increase the acceleration.

Finally, for D, note that mass actually cancelled out of our equation. Whether we increase the mass or decrease it, the acceleration shouldn't change. In reality, this scenario is very common. Mass often cancels out in kinematics equations, since most forces are dependent on the mass (gravity, for example) and we use the equation F = ma. We can often divide both sides by m once we've gotten it set up.

Question #44, Physical Sciences, Kaplan Full Length #4

Question #44, Physical Sciences, Kaplan Full Length #4

When looking at the passage and the 5 reactions given, it seems the ratio of moles of thiosulfate to oxygen would be 2:1, however that is not the case. And the explanation goes through a conversion.

I don't understand. Can you please explain?

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This question is probably easier to answer using the method you mention here -- looking at the 5 reactions they give you. The conversion they work with is tedious and long.

However, you have to be careful when you use these reactions. Reaction 2 is different from the rest -- because it has 2 Mn(OH)2 in it, and the rest only work with 1 Mn(OH)2. So, to make the reactions all equivalent, let's just multiply Reactions 1, 3, 4 and 5 by 2. This way, we can imagine that we start with 1 mol O2, and make 2 mol MnO(OH)2. Those 2 mol Mn(OH)2 make two mol Mn(SO4)2 (Reaction 3), which go on to make 2 mol I2 (Reaction 4), which finally are titrated with 4 mol S2O3(2-) (Reaction 5). Or, just jumping from Reaction 2 to Reaction 5, we have 1 mol O2 that we start with and, now that we've adjusted, we get out 4 mol thiosulfate. This gives us that 4:1 ratio they have as the answer.

This is similar to a mistake many people make in biology, actually -- in the area of cell respiration. Remember how 1 glucose makes 2 pyruvate? And then the reactions from then are are written with just 1 pyruvate? It's the same idea. We have to be consistent with the Reactions here.

Question #42, Physical Sciences, Kaplan Full Length #4

Question #42, Physical Sciences, Kaplan Full Length #4

The question asks for which half reaction would the electrode potential increase as the pH of the solution increases.

The explanation states that as the pH is increasing, the H+ is decreasing and OH- is increasing. However, the correct answer has OH- in the reactants rather than the products. I would think it would be the other way around. Can you please further explain?

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Exactly as you stated, when the pH goes up, that means [H+] is decreasing, and [OH-] is increasing (it is getting more basic). The question itself says: "For which of the following half-reactions should the electrode potential be expected to increase as the pH of the solution increases?"

What is this "electrode potential"? Think of it as a measure of how much the reaction wants to occur, similar to Gibbs free energy. Indeed, this is confirmed mathematically by the equation ΔG = -nFE, where E (the emf of the cell) is based off of the standard reduction potentials of the cathode and anode (essentially the same thing as electrode potentials). Note that if ΔG is negative (spontaneous), then E will be positive because of the sign in the equation. Therefore, a positive E will represent a more favorable reaction.

By that logic, what this question is really asking is when having additional OH- around will be "good" for the reaction - more favorable.

The reduction potentials in the passage are not written as equilibrium reactions (with double-headed arrows), but these reactions must be reversible. A Galvanic cell and an electrolytic cell with the same electrodes will be running reverse reactions (that is, if Mn is being oxidized in a Galvanic cell, it's being reduced in the electrolytic cell). Therefore, these reactions are reversible.

That (reversibility) should make you think about Le Châtelier's Principle. If having OH- is favorable, that must mean that OH- is a reactant -- it will push the reaction forward, encouraging it to occur. The only answer that shows that is D.

For answer choice C, having additional OH- around would actually encourage this reaction to reverse, and thus makes it (as written) less favorable.

Question #29, Physical Sciences, Kaplan Full Length #4

Question #29, Physical Sciences, Kaplan Full Length #4

Per the explanation of this question and the correct answer, I understand that Pipe A opens to air at ground level where the air is moving slower, therefore exerting more pressure. I understand that Pipe B opens above ground where the air is moving quicker and exerting less pressure. I don't quite understand how those 2 situations affect the underground tunnel and the direction of the air flow.

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For this question, they give us Bernoulli's equation right in the question stem. Therefore, let's consider what variables we're concerned about in the equation.

Bernoulli's equation says P + 1/2ρv² + ρgy = constant. The constants 1/2 and g, and the variable ρ aren't really something we're considering here (the density of the air, it tells us, should be assumed to be constant). So what we're actually concerned about is P, v, and y. Let's consider what these represent here. P is the pressure of the fluid, v is the velocity of the fluid, and y is the height (altitude) of the fluid. In the question, we are told that both the velocity of air and the altitude of the air are higher at Pipe B. If this is true, the only way for the sum (P + v + y) to stay constant is if the pressure at Pipe B is much lower. This should also make some intuitive sense to you because pressure goes down with increasing altitude. So, P(A) > P(B).

How does this affect the air in the tunnel? Well, remember that air (and any fluid, for that matter) will follow a pressure gradient, going from the area of high pressure to the area of low pressure. We just identified Pipe A's opening as that point of high pressure and Pipe B's opening as the point of low pressure. Thus, the air will flow down Pipe A, across the tunnel (a left-to-right direction), and up and out of Pipe B.

This is the same phenomenon as what is occurring when you have two windows open in different rooms of your apartment, and may sometimes experience an "air tunnel". The air is rushing because it's following a pressure gradient in the building.

Question #16, Physical Sciences, Kaplan Full Length #4

Question #16, Physical Sciences, Kaplan Full Length #4

I'm having trouble understanding the question as well as the explanation.

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Just to copy the question, it says: "If a helium-neon laser is used as the light source (λ = 6328 Å), how far should the movable mirror be displaced to move from one bright fringe to the next one?"

First of all, as a strategic point, don't let the fact that there are Greek letters and strange units/Swedish letters throw you off. Think logically about the experiment they're doing.

First of all, why do we use the helium-neon laser? Apparently because it gives one wavelength, i.e., is monochromatic. Also, what are these "bright fringes"? According to the passage, a bright fringe is a "maxima" (paragraph 4). But again, "maxima" of what? In this case, the "maxima" are distances that give us constructive interference. And then we ask again -- constructive interference? Of what?

Constructive interference (and any type of interference, for that matter) requires two waves to interact. For the interference to be judged as "constructive," the two waves need to add together when they are in phase. This occurs when the two waves' electric and magnetic fields are oscillating with the same wavelength, and when they are hitting their peaks and troughs at the same time.

The two waves in this passage are created by the beamsplitter. The beamsplitter allows half of the waves to pass through and travel to mirror M2; another half is reflected and goes to mirror M1. Both waves reflect off of M2 and M1, respectively, and then travel back to the beamsplitter. They are all directed to the detector at that point.

When the distances d1 and d2 are equal, you can imagine that the two waves are traveling the exact same distance -- however far it is from the radiation source to the beam splitter plus two times the distance to the mirror M1 or M2 (remember, it has to travel there and also come back), and then the distance from the splitter to the detector. The only part varying here is that distance to the mirror M1 or M2, so let's just focus on that. In this first case, if d1 = d2, then each wave has traveled 2d for its distance. How do we get two waves to constructively interfere? Again, they have to be in phase. If we start pushing out d2, when will the two waves be in sync again? Well, when one wave is a full wavelength "ahead" of the other -- their peaks and troughs will still line up, even if one is one wavelength "behind". But remember that the wave actually has to travel double the distance we push out the mirror, since it will travel this new distance to the mirror as well as away from the mirror and back towards the beamsplitter.

In this example, the wavelength is 6328 Å. To add that distance to one of the waves, we'd have to push its mirror back half that distance (remember the doubling), or 3164 Å. Then the two waves will be in sync again, and constructively interfere. This creates the next bright fringe.

Question #7, Physical Sciences, Kaplan Full Length #4

Question #7, Physical Sciences, Kaplan Full Length #4

Can you please explain why the solid phase is the thermodynamically stable one here? I'm confused by the explanation. I understand the concept: for a given temperature and pressure the phase that exists is the most thermodynamically stable. However, the "given temperature and pressure" for the freeze-dried product in experiment 2 are not provided in the passage. So how do we know that the frozen sample that is placed in the chamber is the thermodynamically stable phase?

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As you state, to know the most thermodynamically stable state, you'd have to be provided a temperature and a pressure to be able to use the phase diagram they handily provide in the passage. We are actually given this information -- the pressure is given in the question stem ("at an initial pressure of 1 atm"), and the temperature is given in the passage in Experiment 2 ("placed in an identical chamber at -20 degrees Celsius"). Using the diagram, we're at 253 K and 1 atm. This is pretty close to the solid/liquid line shown there. So what do we do?

Well, use what you know. If the water is a solid before they even put it in the -20 C container ("The sample is frozen and placed in..."), it's definitely going to be solid at -20 C and 1 atm.

Make sure when you read passages, to highlight the numbers, especially in experimental passages. You can easily find it that way, without copying down all the variables on your scratch paper.

Question #5, Physical Sciences, Kaplan Full Length #4

Question #5, Physical Sciences, Kaplan Full Length #4

The answer to this question states: Light has to travel the distance between the two mirrors twice. So 2d + 2 = 58, d = 28

Can you please explain this?

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The question here, by the way is: "If incoming parallel rays, when reflected by the mirrors, converge 2 m to the left of the reflective side of the primary mirror, what is the approximate distance between the two mirrors?"

For this question, keep in mind what they mean by "incoming parallel rays...converg[ing]". This is simply the definition of the focal point -- the point at which parallel rays are all converged to. When we look at the diagram, that means the light has to travel a distance d between the "Primary Mirror" and the "Secondary Mirror" (an arbitrarily defined variable). Then, it has to go back a distance d, still converging, and travel another two feet past the Primary Mirror to converge at the "FOC/FOS/HRS" in the picture. You can see the light rays coming in -- just follow their path, and you'll see they go d + d + 2 while they converge, or 2d + 2 where d again is the distance between the Primary and Secondary Mirror.

If 2d + 2 is that distance, defined as the focal length, then it is equal to 58 m, as stated in the third paragraph. So d is 28.

GPA

Hello,

Alex, I had a question on the GPA selection. How much emphasis do you think the admissions committee gives on GPA for each term or do they always see the overall GPA and do not really care about how much your GPA was for each term.

Thanks

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In general, medical schools will focus on your overall GPA, although you should know that AAMC will list both an overall GPA and a "BCPM" GPM (Biology, Chemistry, Physics, Mathematics). Medical schools will definitely look intensely at both of these.

However, medical schools are aware of your grades in all classes. This is significant only if you have gotten a failing grade in a class, or if your academic portfolio has changed over time. For instance, the admissions committee will look more favorably at someone who has a 3.0 GPA made up of C's in the freshman year that have improved to A's by senior year than someone with the opposite circumstances - dropping grades from A's to C's over their college career. The first case shows a growth of maturity and taking ownership of one's academic progress; the latter shows someone who has "given up," hit the hardest material they can master, or who has become lazy.

So, my answer is that -- unless you have someone particularly egregious in one particular term -- medical schools will really be looking at your overall and science GPAs.

Med School Admissions - Balanced or Unbalanced Scores

How much emphasis personally do you think medical school admissions committees value a balanced score over an unbalanced one? My strengths lie VERY heavily on the physical sciences and biological sciences part and my verbal score absolutely sucks (as I said above, 6-10 is my range) whereas my physical science and biological science scores go anywhere from 11-15...
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In general, a balanced score tends to be better than an unbalanced score, but there's a better way to approach it - try to balance your percentiles. For example, a 34 put together as 14, 7, 13 (PS, VR, BS) is not as powerful as 12, 11, 11 or 12, 10, 12. The difference here is that it gives the person pretty even (and high) percentiles.

In general, do know that VR is the most important section to medical schools. This is because the section relies solely on logic and reasoning, and not on content knowledge. I encourage you to talk with me if you have more questions on verbal tips and admissions.

Verbal Reasoning Tips and Timing

Any suggestions you have for verbal reasoning? My problem isn't answering the questions because I find with ample time I can answer even the most difficult questions correctly... but my problem is I spend way too much time on the first few passages, which I usually answer all correctly, and then run out of time and rush through the last few where I get the questions largely wrong.. This trend is obvious because the red Xs increase with frequency as question # increases... I've tried pacing myself but my score doesn't really improve. Instead I just end up spreading my wrong answers across the entire test instead of concentrating them all at the end. Any suggestions here? I map but only write maybe one or two words as I feel if I map anything else I'd be wasting my time...
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I'll discuss this in a lot more detail in Verbal III and MSCT III, as well as my re-cap emails. I'm also going to send out my emails for the first two Verbal sessions and MSCT II, since they have more information. Allow me to genuinely make the argument for passage mapping to you in MSCT III - there's a part at the beginning of the lesson where we discuss each of the strategies and why we're doing them.

Verbal Reasoning Comparison - AAMC, Kaptest and the Real MCAT

I've been struggling with the verbal section on my practice exams... on the AAMC exams I've scored in the range of 8-10 whereas on Kaplan I've struggled to get past a 9 and range anywhere from 6-9... is Kaplan considerably more difficult than the actual exam and more importantly is the verbal section of the AAMC exams we get more representative of the real exam? On that note... how do the physical sciences and biological sciences sections on Kaplan and the AAMC we're given compare to the actual exams? On my most recent Kaplan exam I scored a -- which makes me wonder even more.
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Ah, Verbal Reasoning. A few notes on it:
-It tends to be the lowest score for people taking the MCAT. If you look at average MCAT scores overall and in med school admissions, it is consistently a point or two behind physical and biological sciences on average.
-It is the slowest section to improve. This is because it's not the way the vast majority of pre-med students are used to being tested. The verbal section is almost LSAT-like in the way you have to think -- interpret data, understand assumptions, find faults in people's arguments, understand how a new piece of evidence may fit into or dissuade the reader from agreeing with the author's argument. Practice verbal! A lot! I recommend students do at least two verbal passages every single day. Some days, do whole sections. And take serious time to really look back over your work. Review your answers. Use post-phrasing (I'll send out an email explaining this in case you haven't heard about it yet) to check your answers before reading Kaplan's explanations. Make "Why-I-Missed-It" tables (to be described again in Verbal III on Monday). But whatever you do, don't get trapped into thinking that just because you know how to study for the sciences (content and practice, and you tend to know how to improve) and may not know what to do for verbal (practice, review, make sure you really understand what to do with each question type -- again, to be discussed in Verbal III) that you shouldn't bother studying for verbal. If anything, it's bigger motivation to do so.

Now for the comparison.

The questions and passages are really not made more difficult in Kaplan or AAMC than the actual MCAT. The distribution is roughly similar -- 2 "easier" passages per section, 2-3 "medium" and 2-3 "hard"; the hard ones tend to be very detailed natural sciences passages, public policy, or philosophy -- we'll talk about why at MSCT III. There also tend to be 2-3 passages each in natural sciences, social sciences and humanities. Each type of passage tends to have the same types of questions (lots of inference and assumption in humanities, lots of evaluation and incorporation in social sciences, lots of detail and application in natural sciences). One key difference though, as mentioned in the previous blog post, is that the variety of passages in Kaplan may be wider, and AAMC may have multiple passages on a full-length in the same topic. On Test Day, it will tend to be more like Kaplan.

As for scoring, remember that one question can correspond to an increase or decrease in a point. So, it's not really significant yet if your AAMC scores tend to be a point or two higher than Kaplan. Our goal, though, is to get you at least into double-digits consistently on both types of tests, and hopefully into the 12+ range (optimal for "Top 20" schools, but not an absolute requirement). How to do this? We'll talk about it at Verbal III, MSCT III, and in my re-cap emails for both of those lessons.

I hope this helps!

AAMC Exams

April told me that the AAMC exams we get are in fact "harder" than the actual AAMC exams that we'd purchase online in that the questions are ordered differently and Kaplan picks and chooses the harder AAMC questions... from what I've seen in comparison with friends who actually purchased the AAMC exams online they're pretty much the same thing. Could you maybe clarify what the difference is for me?
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I'm not exactly sure where this information came from. Here's the actual low-down on the AAMC tests (by the way, I was part of the committee involved in developing the full-lengths, so I'm speaking from experience here):

The AAMC exams you get on the Kaplan website are exactly the ones you can buy directly from AAMC. You can opt to only work with harder questions (available when you "make your own test" on the site), but all the questions available from AAMC to anyone in the world are available to you when you access AAMC from your Kaptest.com syllabus. It's the same for people buying it a la carte.

To also mention the difference between Kaplan and AAMC exams (I get a lot of questions on this, so I'm going to answer it here):

The Kaplan exams are developed to be a bit more broad than they are deep. The question difficulty is similar to what you'll see on Test Day, although the topics and scopes on Test Day are, admittedly, a bit more narrow than what you see on Kaplan full-lengths. Some advantages of this are that you will see a wider variety of science topics on any given test, as well as a wider variety of questions (both plug-and-chug physics questions and concept-oriented questions, and all question types in verbal). As a disadvantage, the MCAT tends not to be quite as even keel as that. You'll have some questions on Test Day that are easier, and some that are harder. Some topics you'll feel were covered a bit extensively, and others may not show up at all.

The AAMC exams tend to be more narrowly-focused. Significantly, they are NOT old exams. They contain old passages and questions, but are scrambled. In other words, an AAMC exam may have an August 2003 passage and an April 1998 passage in it. As part of that, the grading scales do NOT reflect actual grading scales used in the past. They are algorithmically calculated and, while somewhat accurate, tend to be a bit "difficult" compared to usual Test Day curves.

What comes out of this is that Kaplan exams tend to give you more consistent scores (seeing all the topics and question types, you'll tend to do about the same... but are hopefully improving!). AAMC exams tend to be more jumpy (if you're strong at the question types and topics they're testing on that practice exam, you'll do really well. If not, you may not fare as well). But in bulk, the two average out to be the same. In other words, there isn't a "good" and a "bad" set of tests, but you have to know how to interpret the scores and styles of each test.

Myopia, Hyperopia and Converging/Diverging Lenses

I've been doing some AAMC exams and Kaplan exams and have noticed some stuff on these exams about myopia and stuff... I know that nearsightedness is image converging in front of the retina, and thus a concave lens needs to diverge light onto the retina. And oppositely, a convex lens would converge light onto the retina in hyperobia... My question is where does the whole 1/o + 1/i = 1/f equation fit into all this? And where does the whole negative and positive sign convention fit in?
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In this case, 1/o + 1/i = 1/f isn't really the best equation to use to explain the answer to the question. To use that equation, we'd have to consider a two-lens system (the corrective lens and the person's own eye lens) and combine the two expressions before it really made sense. The one thing that does fit in is to remember that a converging species has a positive f, and a diverging species has a negative f. For lenses, a converging lens is a convex lens; a diverging lens is a concave lens.

To getter get a handle on what kind of lens we'd want, think about the very definition of a converging lens or diverging lens. A converging lens is so-called because it will refract parallel light rays to a given focal point. A diverging lens is so-called because it will refract parallel light rays as though they were all emanating from a given focal point. Converging lenses "pull" the light together, while diverging lenses "spread it apart".

If someone has hyperopia (farsightedness), it means that objects nearby tend to focus too far back in the eye, behind the retina (hence the prefix hyper- meaning "too much"). Adding a converging (convex) lens will cause the light to pull together closer to the lens -- it's converging it. For this person, that means that the image will begin to focus right on their retina.

This is also why contact lenses and glasses have those numbers. They represent the powers of the lenses, expressed in diopters. Remember that the power of a lens is calculated as 1/f. Thus, a converging (convex) lens will have a positive value for f, and a positive power. Farsighted people's lenses have positive values. Nearsighted people's lenses have negative values. The stronger the lens, the stronger an effect it has because it means that the focal length is getting closer and closer the the lens.

Bernoulli's Principle and Vasodilation

So from what I understand vasoconstriction causes an increased blood pressure and vasodilation decreases blood pressure. This makes sense to me in the intuitive sense. However, from what I also understand of Bernouli's equation, this doesn't make sense. I'll try and explain with the vasodilation example. I feel as if one of these statements might be incorrect which is why I might be confused...

1. Vasodilation --> increased cross-sectional area

2. Increased cross-sectional area --> decreased velocity (A1V1=A2V2)

3. Decreased velocity --> Increased pressure (Bernouli's equation)

4. Therefore vasodilation --> increased blood pressure

Yet vasodilation is supposed to decrease blood pressure...

So I feel like I might be mistaken at either step 1 or step 3, but I really don't know where.. could you perhaps help explain this concept to me?
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First of all, all of the statements you provided there are correct when we're talking about physics. Vasodilation by definition means that we have an increased radius of small blood vessels due to one of many triggers. From the continuity equation, as you wrote, an increase in A, the cross-sectional area, will correspond to a decreased v, velocity. All else being equal (density of fluid, height, etc.), a decreased velocity corresponds to increased pressure.

And yet vasodilation does decrease blood pressure.

First, there are a number of assumptions that are true when we're talking about MCAT physics that don't apply in the human body:
1. The continuity equation assumes that the flow rate is constant; indeed, the very point of the continuity equation stems from the idea that if the flow is constant, then decreasing the cross-sectional area means the fluid has to come out faster. In biology, this is not necessarily true. In many cases, for example, vasoconstriction is caused by an outpouring of epinephrine (the idea being that in a fight-or-flight response, you don't want a bleeding risk and so the bloodflow, especially to injured areas, decreases by vasoconstriction). At the same time, however, epinephrine increases heart rate and stroke volume. This leads to a much larger cardiac output, and therefore the continuity equation doesn't really apply. We're putting through a faster flow rate, so that partially increases the pressure.
2. MCAT physics assumes that the tubes are made of an inelastic material. While this may seem insignificant at first, it's actually a critical difference between what makes the pressure in a concrete tube, and what makes pressure in blood vessels. Pressure for a rigid surface is really just based on the force of the fluid over a cross-sectional area. We assume the tube to be rigid so that recoil doesn't really play a role. In the human body, however, this is a key part to how blood vessels work. In the intimal layer of arteries and arterioles (less so venules and veins), there are proteins that are designed to stretch when the artery is filled with blood, and recoil during the other half of the cardiac cycle. These proteins, called elastins (hence the name), act much like a balloon. As the vessel is dilated, they aren't able to provide as much recoil (think about how a balloon gets much easier to inflate once you've gotten it started). This is why aneurysms can occur - as the vessel stretches, it becomes easier to continue stretching and can blow out. In looking at vasoconstriction, the wall will have much more recoil. This causes a lot more pressure within that vessel.
3. One last assumption is that the volume of the fluid stays the same. When vessels vasodilate, they tend to lose some of their fluid into the interstitium. This "extravasation" of fluid explains why people can get very swollen when they're sunburnt - the vasodilated vessels, trying to give off heat and bring white blood cells to the affected area, allow fluid leakage. This causes a decrease in pressure in the remaining vessels as blood volume goes down.

Hopefully these two points help explain this connection. Vasodilation causes a decrease in blood pressure, and vasoconstriction causes an increase in blood pressure. The specifics are outside the scope of the MCAT, but hopefully it helped put together some points from physics and biology! There may be even more to answer this question, and if I hear anything about it when we start cardiology next week, I'll make sure to share.

-Alex

Question about Physics p. 186, question #3

Hey Alex,

I had a question regarding one of the problems in the Physics section. On page 186, question #3 (Chpt 6 Electrostatics), it asked "If an e- were placed midway between R and S above, the resultant electric force on the e- would be..."

I was wondering why the answer is B "towards S". Am I missing something because how would you tell which one is positive and which one is negative? I know that since the force is towards each other, that one is positive and one is negative but is there any other information that tells us S is positive?
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You are correct in feeling that something is missing here. This must be a piece of errata that has not yet been posted -- I am submitting it on your behalf.

In general, the only way to know that a point charge is positive or negative is in context. Either we would have to be told the charges directly, or they could tell us that S repels a known +q charge. In this case, we aren't given that info; thus, we can determine the magnitude of the force, but not the direction (sign).

Hope this helps!

Welcome to Club MCAT!

Please post your questions here so we can keep moving forward towards "45T"s together. Best of luck as you continue your prep for the August and September test dates!