Tutor profile: Shannen P.
Subject: Physics (Newtonian Mechanics)
I can solve a normal projectile motion problem just fine, but I just got one where an object is thrown from an initial height h with a velocity v at an angle of 60 degrees, and the initial height is really tripping me up when I try to plug into the kinematic equations. How do you solve one like this?
In my AP Mechanics class, our teacher didn't let us use the kinematic equations. In the beginning of the year, I was furious-- every other kid gets to use these equations on their AP exam, and we won't even know them. In the end, not using these equations was one of the most valuable things I learned in that class, and you're already starting to see why. Kinematics equations aren't the best to work with on a test-- there are a lot of places where you can plug in the wrong thing, misinterpret, or just plain get stuck if you're not familiar enough with them all and how they can interconnect. That's why the strategy our teacher had us use was graphs. Graph everything. Graph your simple projectile motion, your complex projectile motion, and everything in between. Not only does it help you understand graphs better, but you can see the motion of the object better as well (instead of leaving it as a bunch of variables). In this scenario, I'd say the best graph to start with is your Y-velocity vs. time. Your object is being thrown at some initial height h, and an initial velocity as well. With a theta, which we'll represent with p, of 60, you have x and y components to this v0. v0x will be v0 * cos(p), and v0y will be v0 * sin(p). On your Y-velocity graph, at t = 0 you start at v0y, and then then v(y) value decreases with a slope of -g (where g is 9.8 m/s^2) until reaching some final vfy value (final y velocity value when the object hits the ground at t = t2). We can already get a lot out of this graph. The area under the first triangle (between t = 0 and t = t1, where t1 is when the object reaches apogee/v=0), calling it y1, is going to be the distance your ball travels up from your initial height all the way to apogee. After that, it falls back down to the ground, traveling some distance y1 + h. Why do we know it travels y1 + h? That's because what up must come down (y1), but then on top of that, the cliff the object was shot off of isn't there to catch the ball, so it goes down an additional h value. Another graph we can make is X-velocity vs. time. This one is nice and simple, because an important concept of projectile motion is that your X-velocity doesn't change (unless wind is explicitly mentioned). Often times with kinematic equations, it can be easy to forget this fact and have v(x) vary with time in accordance to -g. This X-velocity graph, however, come with a built in checking mechanism. Start with a value of v0x at t = 0, and then draw a straight line until you get to t = t2. If you look at the area under the curve, you will get the total distance your object travels until it hits the ground-- let's call it x. Now, if you had varied v(x) with time on accident, your graph would look like v(y) vs. time, and the slope would indicate that your X-velocity decreases to 0 and then goes down to a negative number... that would mean your object starts to travel back toward you! Unless you have a mean physics teacher that gave you a boomerang problem, this can't be true, and thus you can easily catch your mistake using simple physics reasoning. With these two graphs, you have everything a kinematics equation could ever give you. The equations will come from performing simple geometry calculations. For example, the area of the rectangle for the v(x) graph will yield that x = v0x * t2. More useful though is the v(y) graph, where you can use the area of the triangles (1/2 * base * height) to get useful information, like how y1 = 1/2 * t1 * v0y. Want to get rid of an unknown? Use your other known information, which is that the slope is -g, and that the slope of a velocity graph (AKA the derivative) is acceleration, to solve even further. In this case, a = -g = rise / run = -v0y / t1. Flipping things around, t1 = v0y / g, meaning y1 = v0y^2 / (2 * g). No matter the exact combination of givens your problem provides, with this strategy, you will be able to solve any kinematics problem with ease. No memorization is really required, just some graph analysis. I find this is really great for high pressure tests where sometimes stress can make me forget things. Also, this method is a lot less error prone, can make a bit more sense in your head if you're a physical person rather than a mathematical one, and gives you a better understanding of the material.
Subject: Physics (Electricity and Magnetism)
How do you find the formula for the electric potential of a point above a disk?
The main ingredient to solving problems like this is an integral. How I used to try and get through my AP E&M course was memorizing the end result integral for each geometry scenario-- definitely do not recommend that strategy. I think the best way is to go from your basic integral set up and break down the components logically, then doing a simple integral to find the final answer. The "basic integral set up" for this type of problem has two steps. First, take the equation you usually use for a point charge-- in this case with electric potential, we have V = (k* q)/s, where V is electric potential, k is the electrostatic constant, q is the point charge, and s is the distance from the point you're analyzing to the point charge itself. Second, change our equation so that we're looking at the effect of a chunk of total charge-- here, replace q with dq and V with dV. Now our equation is set up for an integral on both sides, where we are looking at how each chunk of charge (dq) in our shape (disk) contributes to a total value (V). When we integrate on the side of dV, we'll end up just getting V. That's because all of our chunks of potential (dV) that are adding up due to our chunks of charge (dq) will total to V. Cool! What about the other side? Well, here's where a bit of abstract thinking is involved. What we want to integrate over is the total charge of the disk, Q. If the problem tells you the disk is charge to Q, then your integral can stay as dq, because that right there would you be your integral bounds: 0 to Q! Not all problems will be so nice. Another common given for a problem is the "surface charge density" of the disk. This value represents the total charge Q of the disk over the total area of the disk, A. If R is the total radius of the disk, A is 4*pi*R^2. One of the most important things to remember about charge density is that (in the case of surface area), not only is it equal to Q/A, but it's also equal to dq / dA. If you think about it, they should be in proportion! When we talk about dq being a "chunk" of the total charge, this "chunk" is infinitesimally small, and thus why we can integrate it. Same idea goes for dA-- it's an infinitesimally small chunk of the total area. Even though "infinitesimally" is a bit abstract, the point is that both dq and dA are this same "very small chunk size", making them in proportion to Q/A as well. If we present the surface charge density as p, then we can say that p = Q/A = dq / dA. Since we're looking for dq, and again assuming this version of the problem doesn't give you Q, let's simplify down to dq = p * dA. We rearranged to rid ourselves of one unknown, but now another is in our way-- what's dA? It's time to think in "infinitesimally small chunk" sizes again. As mentioned before, dA is a piece of the total ring area A. The problem has given you radius R of the disk. In the initial scenario, we had a total charge Q, which we could integrate over for dq to get a final answer. Why not go about R the same way? Let's get our integration variable to become dr, and we can integrate over R! What does dr really mean? This is where things get weird again. Imagine a ring in your head of radius r, where r is less than total disk radius R. One might imagine that dr would be equivalent to that r value, but that's not the case. dr actually represents the infinitesimally small thickness of the ring of radius r, allowing each ring to be an actual chunk of the disk instead of integrating a circle that grows to radius R. The next step requires a bit of geometry and more abstract thinking-- what is the area of this ring chunk if it's infinitesimally small? Well, for one thing, usually a ring in our minds would have an outer diameter and a smaller inner diameter. Here, the inner diameter is the same as the outer one, because there is an infinitely small distance between them. And then while in the real world we might consider an infinitely small distance to be 0, for integrate purposes the thickness is still represented by dr. If you were to unwrap this ring from around the disk axis, you would get a rectangle with a length of the ring circumference (2 * pi * r) and a thickness dr. And bam. There you have it. You have your dA. It's really weird to imagine, but the trick is to just keep them as variables. Forget that dr is infinitely small-- it's still a thickness, and that's why we can make a rectangle with an area instead of a simple line. Alright, so we have dA, so let's plug-in to get a final dq value: dq = p * (2 * pi * r) * dr. What's left? We need to integrate from 0 to R, plugging in dq to your previous V = (k * dq) / s equation. The s in your equation will have to be changed before you integrate, because if your point of interest is a distance z above the center of the disk, then you can imagine that each ring will be a distance away from the point. Luckily, simple Pythagorean theorem helps us out here. If you draw it out, each ring will leave s equal to sqrt(z^2 + r^2). Now that your equation deals only with known variables, you're good to go. Sometimes a problem might give you a more complicated surface integral value, but typically a physics E&M exam will give you the basic integral formula to use, making your only concern the setting up. Once you get down how to figure out what these different d"X" values are, you can solve any of these types of problems, no matter if you're dealing with a ring, a sphere, a plane, etc. Getting over the "infinitely small" stuff can be challenging, but once you learn the strategies and continue to practice them, it'll only get easier.
My teacher told me that adenine and thymine go together, and cytosine and guanine go together... but why does this actually happen?
It has to do with the H bonds! CATG-- they're all nucleobases that make up the rungs of the DNA ladder. Like you said, when looking at the DNA of an organism, you'll find C and G in proportion to each other, and A proportional to T, but this didn't just happen for no reason. Adenine and thymine form 2 hydrogen bonds between one another, whereas cytosine and guanine form 3. Think of it like trying to plug a 3-pronged plug into a 2-pronged socket... not going to work. Same sort of idea here-- if C has 3 spots where it can form hydrogen bonds with another nucleobase, do you think it'll want to only have 2 filled? Nope, it'll want them all! Now, why is it that a base doesn't just pair with itself? That'd be a lot easier, right? Well, that ties back to a set of terms you might have heard before: purines and pyrimidines. Adenine and guanine are both purines (double rings), where as cytosine and thymine are pyrimidines (single ring). In these base pairs, purines only want to pair with pyrimidines, and vice versa, and that's how we eliminate that self-pairing possibility. Understanding the why of base pairing instead of just memorizing A is proportional to T will definitely help out further down the road of biology content. For example, if your teacher later tells you that A/T rich regions of DNA are weaker than C/G regions, you don't have to memorize the fact. Instead, by simply knowing A and T only share 2 H bonds, you can infer they'll be inherently less strong than the C and G pairs that share 3 H bonds!
needs and Shannen will reply soon.