Search 72,603 tutors
FIND TUTORS
Newest Most Active

A popular book of physics puzzles (Mad About Physics, Jargodski & Potter, 2001) states that if a car rolling down a slippery ramp locks its rear wheels only (while not braking the front wheels) it will turn 180° to slide down rear first. The reason given is that because the front wheels are in “static friction” with the surface (because they are rolling, not sliding), they have a greater force on them than exists on the rear wheels, which are in “sliding friction” mode. What is wrong with this argument?   Consider the parallel case of a car rolling on an icy level road. As you may know from driving (or being driven), as long as you don’t accelerate, brake, turn, get hit by wind, etc. your car will continue to run straight – Newton’s First Law of Motion. If you then braked only your rear wheels, what would happen? You would slow down, i.e. accelerate backwards, proportional directly to your rear tires’ load and coefficient of sliding friction, and inversely to... read more

This blog post will discuss some of the physics behind the things molecules do – as gases, liquids, or solids – and also get you thinking about the concepts momentum, kinetic energy, rate (of a physical molecular-scale process), and equilibrium constant (of a physical molecular-scale process). These may all sound like difficult, high-level ideas – but if molecules, which don’t have much in the line of brains, can act according to these ideas, you, with brains, can follow them too! What are molecules? First, at their simplest, they’re just clusters of atoms stuck together in such a way that they don’t come apart on their own. (If they did, they would do so immediately; there’s no such thing as delayed disintegration, the way that isotopes of some elements are radioactive.) Wait a minute, you say, what if I heat a pure substance and it breaks down? That happens when one molecule of the substance crashes into another molecule of the substance with enough energy to break at least... read more

Of course not! There's a whole x-y Cartesian plane out there to play on, so there's no reason they should be. They can be tilted and positioned any which way. Of course, when you start out learning about conic sections (including parabolas), you start with the simplest parabola, y = x^2, then graduate to (y-b) = (x-a)^2 and similar forms interchanging x and y. And when you use parabolas, to figure the trajectory of an object which was launched with a certain velocity, in a certain direction, and is subject to gravity, you use only untilted parabolas (well, parabolae, if you want to use the correct plural word). Most algebra classes don't have you figure out what a tilted parabola equation would look like, but it isn't that tough to figure out. First, though, it helps to reconsider just what it is that you do when you graph a point on the curve of an equation (and you thought you knew!). Let's say you're working with an ordinary function on ordinary x-y axes (it could... read more

The question: a scalene triangle determines exactly how many circles, where only the vertices of the triangle are uniquely used, or the body of the triangle in general (i.e. no arbitrarily specified points, such as the midpoint of any of the sides, are used to determine a circle)?   The answer is surprisingly high! Key thoughts: a circle can be uniquely determined by 3 non-collinear points (3 points), or by a diameter (requiring 2 points), or by a center and a radius (also 2 points)[seems a little like the ID's for getting a driver's license, doesn't it?]. Sounds simple -- until you realize that geometrical constraints can be used to yield additional points to use beyond just the three vertices [come to think of it, can it just be a coincidence that your driver's test included a *3-point* turn?].   So, there are inscribed and circumscribed circles (total, 2); circles centered at each vertex and using each side as a radius (total, 6); circles... read more

Proof of the Assertion that Any Three Non-Collinear Points Determine Exactly One Circle This is an interesting problem in geometry, for a couple of reasons. First, you can apply some earlier, basic geometry principles; and secondly, you can choose two different strategies for solving the problem. The basic geometry underlying: any three non-collinear points determine a plane, somewhere in 3D space. Once that has been done, imagine that the plane has been rotated into the x-y plane, which will make the problem much easier to solve! The two strategies for solution are: (Proof A) actually solve to find the circle. This is equivalent to finding the center of the circle (finding the equation of the circle is simple from there). But, you actually have to do some math to get this! If, while doing this, there is no possibility to obtain other values for the coordinates of the center of the circle, you have proved the assertion as well as obtained a method (and perhaps... read more

Students are often confused when starting to study acids and bases, by the multiple definitions and duplicate definitions of acidity (K(a)) and basicity (K(b)). Let me attempt to shed some light.   The element hydrogen is frequently found in substances, but never as an atom. It's always bonded, either to itself (as H2), or to some other element, either in a binary covalent (e.g. H2S), more complex molecule (e.g. C6H12O6).   Such molecules are often quite stable, but sometimes and under the right conditions the hydrogen can come off as a proton, immediately donated to whatever solvent molecule(s) surround the dissociating molecule. We call the original molecule an acid, and we say the molecule has dissociated. This process is rapidly reversible; that hydrogen ion or any other can go back onto the remaining molecule fragment, which we call a conjugate base.    The reaction for an acid dissociation is therefore:   Reactant1(initial... read more

Students are often confused when starting to study acids and bases, by the multiple definitions and duplicate definitions of acidity (K(a)) and basicity (K(b)). Let me attempt to shed some light.   The element hydrogen is frequently found in substances, but never as an atom. It's always bonded, either to itself (as H2), or to some other element, either in a binary covalent (e.g. H2S), more complex molecule (e.g. C6H12O6).   Such molecules are often quite stable, but sometimes and under the right conditions the hydrogen can come off as a proton, immediately donated to whatever solvent molecule(s) surround the dissociating molecule. We call the original molecule an acid, and we say the molecule has dissociated. This process is rapidly reversible; that hydrogen ion or any other can go back onto the remaining molecule fragment, which we call a conjugate base. The proportion of time that the proton in question spends on its original molecule, vs. the time... read more

It seems a bunch of Polynuclear Aromatic Hydrocarbon (PAH) chemists were all vying to head up their local ACS PAH subchapter -- but they decided to choose by a random draw rather than suffer the tumult of an election. So they all threw their rings into the hat....

We customarily teach atoms as “wanting” to fill their valence shells with electrons, thus setting up the whole of chemistry with covalent or ionic bonds. But what’s really in it for atoms to form covalent bonds? Turns out not to be so simple! Isolated atoms DON'T particularly tend to pick up, nor lose, free electrons; they are quite happy as neutral particles. Extra electrons would repel each other on an atom (ion), which is not energetically favorable unless the ion is stabilized somehow. And a loss of electrons would lead to a cation disposed to reacquire the electrons lost. That’s NOT evidence of “wanting to fill” the valence shell! So, consider instead what’s in it for each of the subatomic particle types in two atoms near each other that could share electrons. First, for the nuclei, there’s no advantage in bonding! If anything, if the shared electrons didn’t completely screen out the positive charges of the two nuclei from each other, they would repel each other... read more

RSS Stanton's Blog RSS feed

Woodbridge tutors