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http://onsolver.com/

Original First Post

College Algebra

Calculus :: Early Transcendentals

Advanced Engineering Mathematics

Numerical Methods (a.k.a. Computational Methods for Applied Sciences)

Euler Angles

Operators & Symbols ::
(–x, y) ± + − × ÷ √ ² ³ ^ ∫ Σ Δ ∇ ∂ ∠ ° Ω “” → ← ↑ ↓ ∵ ∴ ½ ∞ ≈ ≠ ≪ ≤ ≥ ≫ • · ∝ † ⊗ ✔ ✘ 2⁄2 x≈-1.25992 ∧ ‖a‖ y≈-1.5874 …

Common Greek alphabets in Rotational dynamics:
ϕ, θ, ψ, ω

PID Controller ::
u = − (Ki*∫ x + Kp*x + Kd*ẋ)

15°	π/12
30°	π/6
45°	π/4
60°	π/3
75°	5π/12
90°	π/2

① ② ③ ④ ⑤ ⑥ ⑦ ⑧ ⑨ ⑩

b² – 4ac

Superscripts and subscripts ::
x⁰ x¹ x² x³ x⁴ x⁵ x⁶ x⁷ x⁸ x⁹ x⁺ x⁻ x⁼ x⁽ ⁾ xⁿ x* x˙ xˣ

x₀ x₁ x₂ x₃ x₄ x₅ x₆ x₇ x₈ x₉ x₊ x₋ x₌ x₍ ₎ xᵢ xᵣ xₑ xₙ

Greek alphabets (lowercase) ::
α, β, γ, δ, ε, ζ, η, θ, κ, λ, μ, ξ, π, ρ, σ, τ, υ, ϕ, φ, χ, ψ, ω

Greek alphabets (uppercase & lowercase) ::

Αα Alpha	Νν Nu
Ββ Beta	Ξξ Xi
Γγ Gamma	Οο Omicron
Δδ Delta	Ππ Pi
Εε Epsilon	Ρρ Rho
Ζζ Zeta	Σσς Sigma
Ηη Eta	Ττ Tau
Θθ Theta	Υυ Upsilon
Ιι Iota	Φϕφ Phi
Κκ Kappa	Χχ Chi
Λλ Lambda	Ψψ Psi
Μμ Mu	Ωω Omega

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This post has been edited by Critical_Fallacy: Apr 29 2023, 03:50 PM

Impressive! Thanks for sharing the link. Tell your Bronze Medallist friend that he obviously has a good understanding of mathematics that is far superior to mine. Through the composed power of his highly intelligent mind, he could probably solve some of the unsolved problems in mathematics in near future. Perhaps you don’t know; I’m particularly fascinated by the Navier–Stokes existence and smoothness problem.

If you want to claim $1 million Millennium Prize from the Clay Mathematics Institute, you’ll need to either prove or disprove this statement: “In three space dimensions and time, given an initial velocity field, there exists a vector velocity and a scalar pressure field, which are both smooth and globally defined, that solve the Navier–Stokes equations.”

In physics, the Navier–Stokes equations are a system of nonlinear partial differential equations for describing the motion of fluid substances in almost every real situation. In fact, these equations are derived from applying Newton's 2nd law to fluid motion. The animation below shows how a Kármán vortex street develops behind a cylinder moving through a fluid.


(courtesy of Cesareo de la Rosa Siqueira at the University of Sao Paulo, Brazil)

GreatFish posted a physics question about Newton’s Law of Universal Gravitation on this thread. Because there is slight inaccuracy about the direction of the gravitational force vector on the Wikipedia, so I decided to provide a salient answer on this matter. The negative-sign equation arises partly due to our interest in researching the kinodynamics behavior of the object being accelerated.



This post has been edited by Critical_Fallacy: Oct 2 2014, 04:02 PM

You should be able to find a lots of detailed explanations from many physics textbooks and websites.

The derivation of Einstein's most famous equation E = mc² is purely theoretical, which arises as a direct consequence of his Special Theory of Relativity that involves some algebraic manipulations using the Pythagorean theorem (SPM level). To understand how he derived the equation, you must understand a portion of his Special Relativity; Time Dilation (see the embedded Spaceship figure). The following derivation is highly simplified for the mathematically untrained. To further enhance your understanding, you should watch the YouTube video by minutephysics.

Physicists know some things never change and they call them the fundamental physical constants. Such frequently used constants as the speed of light in vacuum, c, magnetic constant, μ0, electric constant, ε0, Newtonian constant of gravitation, G, Planck constant, h, elementary charge, e, and the Proton-to-electron mass ratio, μ, are assumed to be the same at all places and times in the universe. They form the scaffolding around which the theories of physics are built on, and they define the fabric of the cosmos.

Despite that, one of the most fundamental properties of Newton’s Mechanics, Maxwell’s Macroscopic Electrodynamics, and the Laws of Thermodynamics, is the absence of any physical constants in their basic equations. In fact, all necessary fundamental physical constants appear only at the stage of applications of these theories to specific phenomena. Nevertheless, these constants play an important role in physics and metrology because scientists need reference values for measurements in experimental physics and making theoretical prediction on papers.

Other than Newton’s and Maxwell’s equations, centripetal acceleration (a = v²/r), density (Q = m/V), pressure (P = F/A), and electric power (P = V*I) seem to be self-evident counterexamples to your above question. You can also see other frequently used physics equations for yourself in this link.


Although physical constants can appear to be dimensional or dimensionless, the imaginary constant factor k in your F = kma does not exist. That’s because the equation F = ma is self-contained and Newton did NOT require adding a dimensionless constant factor k = 1 to his mechanics to reproduce that “observation of force”. In fact, as guided by VengenZ in Post #20, Newton’s 2nd Law originally states that the net force on an object is equal to the rate of change of its linear momentum p in an inertial reference frame, which can be manipulated algebraically to be stated in terms of an object's acceleration, i.e. F = ma.

--->

Since the dimensionless constant factor k does not exist in F = dp/dt in the first place, then it follows that k is not needed in F = ma either. If your friends or physics teacher insist on the existence of k = 1, be sure to ask them what they think of a = dv/dt, p = mv, v = dx/dt, and other counterexamples as shown, so that other similar fundamental physics equations cannot be ignored, and they will thus be psychologically compelled to consider their merit.

~ HAVE A NICE DAY! ~

This set of frequently used Fundamental Physical Constants is recommended for international use by CODATA (The Committee on Data for Science and Technology). The full 2010 CODATA set of constants may be found at its website (click the image) at the time of this writing.



For your convenience, a PDF document (Extensive Listing) is available which can be read online or printed.
 Fundamental_Physical_Constants_____Extensive_Listing__2010_.pdf ( 124.83k ) Number of downloads: 1

What you see in the following figure is an excerpt from The Principia: Mathematical Principles of Natural Philosophy (2010), the reprinted text which based on Motte’s 1729 translation of the 1726, 3rd edition of Philosophiæ Naturalis Principia Mathematica, that was the final version corrected by Sir Isaac Newton.




As you can see, Newton never explicitly stated the formula F = ma in his Principia. The “motion” and “impressed” in which Newton used his terminology, and how he understood the second law and intended it to be understood, have been extensively discussed by historians of science. It was until James Clerk Maxwell published a treatise in Matter and Motion in 1876 (see excerpt below) that the modern ideas of how Newton was using his terminology is understood.



According to Maxwell, Newton meant by motion “the quantity of matter moved as well as the rate at which it travels” and by impressed force he meant “the time during which the force acts as well as the intensity of the force”. And so in modern terms, motion is Newton’s name for momentum, which allows Newton’s 2nd Law to be formulated as follows:

→

Now, allow me to make a few additional points after my Post #29. Based on what the mathematicians and physicists understood from the experimental observations of Newton’s 2nd Law: the change of momentum of a body is proportional to the impulse impressed on the body, a proportionality constant k, shall be assigned to the formula F = kma. In order to standardize the unit force and to eliminate the proportionality constant k, Conférence Générale des Poids et Mesures (CGPM) in 1948, adopted the name "newton" with symbol N for the SI unit force to define the amount needed to accelerate 1 kg of mass at the rate of 1 m/s² by choosing a proportionality constant of 1.

In a sense, the proportionality constant k can be seen as a conversion factor if we apply F = kma in other system of measurement, such as the Imperial units. For example, 1 pound-force (lbf) is defined as the amount needed to accelerate 1 pound (lb) of mass at the standard gravity of 32.174 ft/s². Consequently, we can deduced from this observation that k = 1/32.174. Thus, to eliminate the proportionality constant k, the preferred unit of mass is the slug because a slug is defined to have a mass of 32.174 lb, so that 1 lbf = 1 slug × 1 ft/s².

To summarize this post and Post #29, naturally, the proportionality constant k = 1 doesn’t appear in momentum (p = m*v), centripetal acceleration (a = v²/r), density (Q = m/V), pressure (P = F/A), electric power (P = V*I), and a bundle of other primary equations of physics, because the physicists relate the basic physical quantities by defining the formulation of physical laws in a way to achieve universality and generality. All in all, to determine whether a significant proportionality constant is needed or not, at the first place, we should at least understand how strong the correlation of physical quantities is by performing the dimensional analysis, and ensure that the experimental data are consistent.

Oh! It is not as daunting as the quantum mechanics. Coincidentally, 12 August marks the birthday of Erwin Schrödinger, the Nobel prize-winning quantum physicist whose eponymous equation lies at the heart of quantum mechanics.

In Newtonian Kinetic Energy, as long as the object is not at microscopic level and the speed involved is very much lower than the speed of light, the mass is considered an absolute, or in a state of constant in an inertial frame of reference.



P.S. Thanks sharp-eyed Krevaki for pointing out the typo in Line 4 of the previous image. Correction has been made.

This post has been edited by Critical_Fallacy: Aug 14 2013, 07:28 PM

Though Newton’s Principia (1687) theorized the presence of the gravitational constant G, it was not until 1798 that the constant was determined through observing the density of the Earth in the Cavendish experiment. Because Newton’s calculations could not use the mathematical value of G, he could only calculate a force relative to another force as a ratio using Kepler’s Third Law. The following propositions for Newton’s Law of Universal Gravitation are the excerpts from The Book III of Principia. If you are interested, you probably find this link useful (Read Proposition VIII. Theorem VIII. Corollary 2).





As you probably already know, most modern physics textbooks states Newton’s law of universal gravitation according Maxwell’s Matter and Motion (1876); “Every portion of matter attracts every other portion of matter, and the stress between them is proportional to the product of their masses divided by the square of their distance.” Maxwell also acknowledged that although the idea of inverse square law had been the prior work of others, but in the hands of Newton, the doctrine of gravitation assumed its final form.



This post has been edited by Critical_Fallacy: Oct 24 2014, 12:05 AM

Erratum: Please take note that there was a typo in the last line of the previous image. Last line “√(2p)” should be “√2*p”. Thanks Krevaki for pointing it out. It has been corrected.

Basically, it is very similar to proving Trigonometric Identities in SPM Add Maths. When you have the knowledge of momentum p = m*v, you will think of ways to express KE in terms of momentum and at the same time, conserving the dimensions of the original KE equation.

By the way, care to tell me about the reception of your Science Fair project?

Under normal circumstances, your daily experience, common sense plus Newton’s 1st Law generally would tell you the “rational” answer seems to be B. However, if B were the intended answer, why would the person who set the question bother to add the floating helium-filled balloon in the boxcar? Is it a tricky question?

There are 2 clues provided:
(1) If the helium-filled balloon wasn't tied to the floor of the boxcar, it would float all the way up to the ceiling.
(2) In physics, Impulse of Force is used to refer to a fast-acting force that is applied briefly: I = F*δt = m*δv.

Lastly, logical thinking is not just using our “common sense.” After all, “common sense” tells us the earth is flat. But even when “common sense” is correct, it may tell us, for example, that gravity cannot be pulling a thing down and pushing up the other at the same time. Therefore, critical thinking goes well beyond just using such basic principles.

Perhaps Krevaki and work_tgr can offer some clues as well.

Perhaps watching the YouTube video below would give you new insights into the physics of helium balloon.

Technically speaking, the method employed by ystiang is called Complementary Cumulative Distribution Function. The solution is exactly the same as the one beautifully brought by ystiang, with a little extra information of how the principle of Cumulative Distribution Function works.

It takes the knowledge of Density of Steel, Density of Helium, Newton’s Law, Archimedes’ principle (Buoyancy), Impulse of Force, and a few assumptions to understand the tricky physics problem.

Statics ::

(1) The boxcar is at rest on a frictionless surface. Between the two objects inside the boxcar, a solid steel ball is much more dense than a helium-filled balloon. The steel pendulum and helium balloon are rigid bodies.

(2) Newton’s Law tells us the downward force of gravity pulls the steel pendulum and helium balloon, which results in their respective weights, W↓ = mg.

(3) Archimedes’ principle tells us the upward buoyant forces exerted on the steel pendulum and helium balloon “immersed fully” in the air, are equal to the respective weights of the air that the objects displace, F↑ = ρVg.

(4) For the heavier-than-air steel ball, because W↓ > F↑, the steel ball is pulled downward by the Net force.

(5) For the lighter-than-air helium balloon, because F↑ > W↓, the helium balloon is pushed upward by the Net force.

(6) Because the freedom of movements of the steel pendulum and helium balloon are constrained in the direction of the respective Net forces (vertical) axes, Newton’s Law tells us their respective normal Reaction forces are generated, so that they appear to be held in their places.



Dynamics ::

(7) For the impulse of force, in which we assume the force exerted on boxcar acts for a very short time δt but is much greater than any other force present. Imagine flicking your finger to apply the impulse of force on the outer left side of the boxcar! The duration of the collision is very short and bring the boxcar to an abrupt STOP.



(8) Now, because the steel pendulum and helium balloon are not constrained horizontally, any applied horizontal force will cause the objects to swing left and right.

(9) Since the boxcar comes to an abrupt STOP, when we reapply the Newton’s Laws and Archimedes’ principle from (1)−(5), the most probable positions of the steel pendulum and helium balloon caused by the impulse is depicted in Image C. What do you think?

Have you discovered this link?

Possible, if it is an “open” boxcar.

In the physics of general relativity (non SPM), we can certainly describe the motion of the steel pendulum and helium balloon by the equivalence principle.

True, but since the impulse acts for a very short time δt ~ 0 sec, the closest image that describes the positions (during δt) of the steel pendulum and helium balloon caused by the impulse that brings the boxcar to an abrupt STOP would probably be Image C. Just my 2 cents.

Refer to Tensile testing and Stress–strain curve.

The formula to calculate the “apparent” engineering stress, σ, is given by σ = F / Ao, assuming that the original cross-sectional area Ao is UNCHANGED (not reducing) through which the force is applied. In fact, as deformation continues, cross-sectional area decreases.

Similarly, the formula to calculate the engineering strain, ε, is given by ε = ΔL / Lo. Some questions give the elongation measurement L instead, and if that is the case, use ΔL = L − Lo.



Refer to Properties of Water, Continuity equation, Bernoulli's principle and Venturi effect.

Water is treated as an incompressible fluid because the changes in relative volume ΔV / V is negligible as the pressure increases.

In fluid mechanics, the continuity equation states that, in any steady state process, the rate of change of mass within the control volume is equal to the rate at which mass leaves the surface S of volume V.



In other words, mass inflow = mass outflow. However, mass is also given by the product of ρV (density × volume). Because water is essentially incompressible, therefore the density ρ of water is treated as constant throughout the streamline from the hose to the head. For that reason, we can rewrite the continuity equation as follows:

Volumetric inflow rate, Q1 = Volumetric outflow rate, Q2[/B][/COLOR]



Because Volumetric flow rate, Q is defined as Q = dV / dt,

and the volume of a mass, V is given by V = A*l (cross-sectional area × traversed length), therefore

A1 × (dl1 / dt) = A2 × (dl2 / dt)

Since dl / dt is defined as the velocity, υ, we can simplify the equation to

A1 × υ1 = A2 × υ2

If the gas is continuously supplied at a constant pressure, the new level of “A” would probably look like this: