Either you read the whole textbook and find no new knowledge, or you find no purpose in using the textbook.

In astrophysics or orbital mechanics, the shape of the curve describing the negatively charged particle's path is known as the hyperbolic swing-by trajectory. Perhaps you can get a feel for the way the negatively charged particle performs the swing-by maneuver work by proposing a useful analogy, in the form a rather outlandish gravity-assist maneuver involving a spacecraft.

When the spacecraft is a long way away for the planet, before it even gets to point A, the gravity force of the planet is so feeble that the spacecraft effectively travels in a straight line. This line is called an asymptote of the hyperbola. However, as the spacecraft closes in on the planet, the gravity force steadily increases, and spacecraft's path describing the classic hyperbolic shape from point A through point B to point C. Beyond point C, the gravity force decreases rapidly, and the trajectory tends to the straight line given by the asymptote once again.

During this process, the planet has changed direction of the travel of the spacecraft, and this change is given by the deflection angle as shown. The amount by which the trajectory is deflected is dependent on three things: (1) how massive the planet is, (2) how close point B is to the planet, and (3) how fast the spacecraft is traveling on approach.



To answer your question in clarity without mathematical equations of the orbital motion, let's consider gravity-assist maneuvers using the planet Jupiter as an example and the velocities (v) of the objects are represented by vectors (arrows). The change in the heading direction and speed of the spacecraft can be determined using vector addition and vector calculus.



This post has been edited by Critical_Fallacy: Jul 28 2013, 01:37 AM

In fact, foxtrotalpha’s simple explanation of “attraction between electric charges” is sufficiently logical to deduce the correct path of the negatively charged particle. However, because you asked for the reason the particle behaves as shown in diagram C, it maybe necessary to to show you that the proton-packed nucleus is assumed to be way heavier than the negatively charged particle, by explaining from similarity in gravity-assist maneuvers involving a spacecraft.

Well, the fundamental laws of orbital mechanics are Newton’s law of universal gravitation and Newton’s laws of motion. And I wonder if you’ll be curious about the fact that Kepler's laws of planetary motion can be derived from Newton’s laws, and it is used to describe the hyperbolic trajectory when the orbiting body is subject only to the gravitational force of the central attractor.

Perhaps you can get a better picture by reading my Post #1129 again.

Nevertheless, I want to reiterate that the chosen example of gravity-assist is just analogy for the sake of a better imagination of the physical phenomena in question. You don’t use Newton’s law of universal gravitation to calculate the electrostatic interaction between electrically charged particles, but apply the Coulomb’s law instead, which is mathematically similar to Newton’s law of universal gravitation.

In fact, the theoretical physicists are struggling without success (none have been confirmed experimentally) to reconcile the Gravitational force (large-scale structure of space-time) and the Electronuclear force (small-scale structure of elementary particles) under a single physical framework called the “Theory of Everything” (ToE).



I have no idea how you synthesized Radioactivity in your mind, but it is related to Electricity. If you are interested, please study about the Holy Grail of Electrodynamics, the “Maxwell’s equations” because Gauss’s law for electrostatics is one of the four well-known Maxwell’s equations, which can be used to derive Coulomb’s law.

You probably won't find any higher res from Prof. Stanek's Webpage. However, I reassembled the memo maps for Prof. Stanek in his honor, and in the hope that it is beneficial to Saimun's Physics students.

Do you mean this?

Is this the only difference (in Science) between SPM and HKCEE (Hong Kong) / GCSE 'O' Level (British) / Grade 11 (US)?

Nevertheless, you will learn Chemical Bonding and Molecular Structure in Molecular Geometry. If you are interested in determining the bond angles, you can refer to the Valence Shell Electron Pair Repulsion (VSEPR) table.

The parallel lines and the two unknowns x & y are designed to trick you into believing that you need 2 equations to be solved simultaneously in order to find the values of x & y. However, only one equation involving x & y can be formulated.

Sum of Interior Angles of a Heptagon = (7−2) × 180° = 900°

(x + y) + y + y + 2x = 900° − 120° − 172° − 48° − 98°

3 * (x + y) = 462°

Need any shorter?

This is more of a IQ question than a math problem. In fact, your workings are necessary to determine x and y individually. However, always use the fundamental laws and principles. You will know they are not sufficient if the solution cannot be found. And don't be misled into believing most SPM math problems are direct. So, tune up your mentality.

For me? OK! According to k-combinations, C = 10; k = 2; n = ?

Yes, so that the formula for binomial coefficient can be reduced to

n × (n − 1) × ... × [n − (k − 1)] = k! × C

Knowing this little trade secret enables you to manipulate the numbers in your typical CASIO fx-991/570 calculators.

Say, in SPM exam, you encounter a mind-boggling question, in which you are required to count the number of combinations of 100 things taken 3 at a time without repetition. The question is 100% LOGIC and VALID.

If you input 100!, you will only cry in your futile attempts. That's because most calculators use scientific notation with 2-digit decimal exponents, and the largest factorial that fits is then 69!, due to max range of 69! < 10^100 < 70!. However, the alternative computation is applied correctly, you'll get the answer unremittingly Faster, Cheaper, & Better. I will leave this as an exercise for you to practice using your calculator effectively.

This post has been edited by Critical_Fallacy: Aug 5 2013, 01:44 PM

Technically speaking, in mathematics, it doesn't really matter whether plotting a graph A against B or vice versa. For example, in the Velocity-Time graph, we normally plot the velocity (v) against time (t). Even if it is plotted inversely, we still can extract some useful information from the graph. Therefore, sometimes plotting a graph A against B out of the norm is NOT technically considered “wrong” as one would imagine.

I'd probably raise up my left hand in the middle of the exam and request for clarification politely. If no satisfactory answer is provided, then I'll do my way which I deem intuitively correct, and of course, with a little remark written to justify my workings. In a professional setting, the exam moderator shall minute all queries and clarifications. Better ask mrsaimun regarding this matter!

Would you be so kind in return as to snap a picture the explanation and post it here, please?

Unfortunately, what is lacking in the Teaching of Mathematics in Malaysian secondary education is the Art of Problem Solving. This happened since Polya's 4-step approach to problem solving was expelled from the Maths syllabus a few years ago, according to maximR. Not every student is naturally born with high mathematical intuition like you and Carl Friedrich Gauss. In fact, maximR developed it through countless hours of reading extra materials and practicing math skills beyond SPM level.

The coordinated of R(4,2), P(0, y_P) and Q(x_Q, 0) are known, where y_P = 4 and x_Q = 8

The equation for Line SRT is given by 2y = kx + p.

The gradient of Line PRQ can be determined with m_PRQ = -y_P / x_Q.

The gradient of Line SRT can be calculated using m_SRT = -1 / m_PRQ.

Using the standard equation for Line SRT of y = (k / 2)*x + (p / 2), k can be found.

Substituting the coordinate of R(4,2) into Line SRT, you can work out the value of p.

Similarly, x-intercept of the Line SRT can be determined exactly the same way you worked on the gradient of Line PRQ.

Generally speaking, the CRT can be viewed as a part of the CRO. If one really wants to find out, perform the experiment cautiously in the lab. However, it seemed that the CRT in this question is highly simplified and designed to function principally like a vacuum tube Diode (forget about the X-Y deflection plates and the Phosphorous screen). If that is the case and it isn't a trick question, then we can analyze such electrical circuit in theory:

As you probably already know that diodes are used in AC circuits where it is desired to convert AC voltages to DC voltages, because diode has low resistance-to-current flow in the forward-biased direction. The simplest of all the rectifier circuits is a half-wave rectifier (Click it), when the diode is connected to a load resistance R as shown in the following figure. During the positive half-cycle from 0 to π of the voltage supply Vs, the diode is forward-biased so current flows through the diode and resistor where it develops an output voltage drop, Vout = Vs − Vd, and the built-in potential of a silicon diode Vd is approximately 0.7 V. The diode is reverse-biased during the negative half-cycle from π to 2π, so no current flows, and thus Vout = 0 V. And this is the well-known half-wave ripple in the signal across the load resistor where the name of the rectifier circuit comes from.



To get rid of the undesirable large ripple, we add a “smoothing” capacitor across the resistor, and the voltage across the resistor-capacitor (RC) combination will be as shown in the next figure. When the instantaneous voltage of the AC source Vs, is higher than the instantaneous value of capacitor voltage Vc, the diode conducts, and the capacitor C charges up to the positive peak of the output voltage Vout. When the instantaneous voltage of the AC source Vs, falls below the instantaneous value of capacitor voltage Vc, the diode is reverse biased and the capacitor C is discharged through load resistance R. The resulting capacitor voltage Vc varies between a maximum value of Vout and a minimum value of Vout − Vr(pp). Theoretically, the peak-to-peak ripple voltage Vr(pp), can be made sufficiently small by choosing the time constant τ = RC much larger that the period T = 1/f of the input sinusoidal AC waveform. However, it is not very practical to use larger capacitors due to the constraints of both cost and size. Therefore, the full-wave rectifier with capacitor-input DC filter (Click it) is a much more practical approach.



Come back to the original question, I'd probably choose “A” because it is the closest image to the predicted waveform as indicated by the solid green “slight” ripple below.

As expected, and the best of luck to you!

Always leave a backdoor open for others when the situation allows. Sometimes I'm dissatisfied with the reality, and I become a high-minded idealist, feeling that it is up to me to improve everything. But at your best, you can become extraordinarily wise and discerning. By accepting what is, you become transcendentally realistic, knowing the best action to take in each moment. If there is any question about that intention or about any implicit part of the statement, the originator should be given the benefit of any doubt in the reformulation or, when possible, given the opportunity to amend it.