Mechanics och 3.08 i Ergo 1.

Publicerat den september 18, 2013 av mattelararen

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The discipline of physics dealing with the concepts of motion and the forces that cause them are called mechanics.
Motion is a fascinating phenomenon as expressed in Zeno’s paradoxes.
He states that it is impossible to move from A to B since first you have to cover half the disance and after that half the remaining distance and so on. The result is that you have to travel an infinite number of half-distances.
This ought to take an infinite amount of time.
Velocity is defined as translation divided by time.
The unit is m/s.
v=δs/δt
Acceleration is defined as velocity change divided by time.
a=δv/δt
unit: m/s2.

The distance can be found by determining the area under the st-graph.

Den del av mekaniken som beskriver rörelse utan hänsyn tagen till de krafter som är anledningen till rörelsernas uppkomst kallas kinematik.

Om analysen av rörelsen även omfattar dessa krafter handlar är utövar man om dynamik.

Några centrala kinematiska lagar:

s(t) = s0 + vt

v(t) = v0 + at

s(t) = s0 + vt + at2/2

v2 – v02 = 2as


Då hastigheten är ändringshastigheten är hastgheten, v(t) = s'(t) (derivatan av sträckan med avseende på tiden) och accelerationen utgör ändringshastigheten av hastigheten är a(t) = v'(t). Således blir s”(t) = a(t).
Integrerar man dessa derivatasamband får man ovanstående kinematiska formler.
∫v(t)dt = s

Begreppet tid är i sig själv ett gåtfullt begrepp som inte är lätt att beskriva. Effekterna av den kan man dock se t.ex. i följande kompilation av rembrandt’s självporträtt.


 

Uppgift 3.08 i Ergo 1:

Olof kör med konstant fart 105 km/h på en motorväg utom under 12 mi då han tar sig en paus.

Hur lång tid tar resan om medelhastigheten är 92 km/h för hela resan?

Lösning:

Antag att den totala restiden är x h.

DEn totala resträckan kan då tecknas på två sätt:

s = 92x

s = 105(x-12/60)

Dessa kan sedan sättas lika med varandra:

92x = 105(x-12/60)

x=1,62h = 97min

 

 

 

 

Publicerat i Ergo 1, Fysik 1, Gymnasiefysik(high school physics) | Märkt , , | Lämna en kommentar

HTML coding

Publicerat den september 6, 2013 av mattelararen

A good instructionfilm about HTML-coding.

 

Publicerat i Uncategorized | Lämna en kommentar

The standard model

Publicerat den augusti 21, 2013 av mattelararen


The standard model is the broadly accepted theory for the building blocks of the universe. The fundamental forces and the elementary particles.

According to this theory the particles can be divided into hadrons and leptons. They are the building blocks of matter. The hadrons forms the nucleus of the atom and the electrons orbiting the nucleus is a lepton.

Leptons means light particle.

Because of their colour charge the quarks always appear in triplets or in quark-anti-quark pairs.

The proton is composed of 2 up-quarks and one down quark (uud) and the neutron is (ddu).

In the standard model every force-interaction must be accompanied by the exchange of particles. According to the standard model there are four fundamental forces in nature: gravity, electro-magnetic forces, strong nuclear force and weak nuclear force.

Gravity is the exchange of gravitons, electromagnetic forces are the exchange of virtual photons, the strong interaction is mediated by gluons whereas the weak force is mediated by Z-bosons.

In addition there are top-quarks, down quarks and a lepton called muon and another called neutrino.

Particles can also be categorized as particles with half-numbered spins and bosons -particles with integer spin. Fermions are the constituents of matter whereas bosons are responsible for the transfer of forces.

The constituents of matter must have half-integer spin since they must follow Pauli’s exclusion principle. This requires the wavefunctions to be asymmetric.

The question why particles have masses can be explained with the Higgs particle. 

The laws of physics are time-invariant so even the dinosaurs obeyes the same natural laws as we do.

exempel på uppgift:

hur stor blir fotonenergin då ett elektron-positronpar förintas?

Enligt Einsteins formel E=mc2 fordras energin 2˙ 0,511 MeV= 1,022 MeV eftersom massenergin för en elektron är 0,511 MeV.

Publicerat i Fysik 2, Gymnasiefysik(high school physics), Mathematical physics | Märkt , | 1 kommentar

Lagrangian mechanics

Publicerat den augusti 5, 2013 av mattelararen

In functional analysis the variable itself is a function.

This is used e.g. in the Lagrangian formulation of mechanics where one derives the Lagrangian i.e.
L = kinetic energy – potential energy.

This transforms classical Newtonian mechanics into differentialcalculus.
The variables, or degrees of freedom, can be selected to make the problem as easy as possible. They can be cartesian coordinates, velocities or momentums for example.

By solving Lagrange’sM differential equation the Lagrangian can be found.

A similar system was devsed by William Rowan Hammilton. He studied the hamiltonian for the system. This is the sum of the kinetic and potential energy of the system.

It is used for example in the Schrödinger equation.

https://www.google.se/search?q=lagrange&client=safari&hl=sv-se&prmd=mivn&source=lnms&tbm=isch&sa=X&ved=2ahUKEwj7nNudz47gAhUrhaYKHfTmAMwQ_AUoAnoECA0QAg#imgrc=OJ1wq2zLr0WKuMedbeca77-6993-41aa-a628-95f37e159063

Publicerat i Advanced, Calculus, Fysik 2, Mathematical physics | Märkt , , | Lämna en kommentar

Hyperbolic functions

Publicerat den juni 8, 2013 av mattelararen

The hyperbolic functions have similar names to the trigonometric functions, but they are defined
in terms of the exponential function. In this unit I define the three main hyperbolic functions,
and sketch their graphs. I also discuss some identities relating these functions, and mention
their inverse functions and reciprocal functions.

The hyperbolic functions cosh x and sinh x are defined using the exponential function ex. We
shall start with cosh x. This is defined by the formula
cosh x = (ex + e−x)/2.
We can use our knowledge of the graphs of ex and e−x to sketch the graph of cosh x. First, let
us calculate the value of cosh 0. When x = 0, ex = 1 and e−x = 1.

So
cosh 0 =(ex + e−x)/2 = (1 + 1)/2 = 1 .
Next, let us see what happens as x gets large. We shall rewrite cosh x as
cosh x =e x/2 + e−x/2.

To see how this behaves as x gets large, recall the graphs of the two exponential functions.
y = ex/2

y= e−x
As x gets larger, ex increases quickly, but −x decreases quickly. So the second part of the sum
e , ex/2 + e−x /2 gets very small as x gets large. Therefore, as x gets larger, cosh x gets closer and
closer to ex/2. We write this as
cosh x ≈
(ex+e-x)/2
for large x.
But the graph of cosh x will always stay above the graph of ex/2. This is because, even though
(e−x)/2 (the second part of the sum) gets very small, it is always greater than zero. As x gets
larger and larger the difference between the two graphs gets smaller and smaller.

As x becomes more negative, ex increases quickly, but ex decreases
quickly, so the first part of the sum ex/2 + e−x/2 gets very small. As x gets more and more
negative, cosh x gets closer and closer to e−x/2. We write this as
cosh x ≈
(e−x)
/2
for large negative x.
Again, the graph of cosh x will always stay above the graph of e−x/2 when x is negative. This is
because, even though ex/2 (the first part of the sum) gets very small, it is always greater than
zero. But as x gets more and more negative the difference between the two graphs gets smaller
We can now sketch the graph of cosh x. Notice the graph is symmetric about the y-axis, because
cosh x = cosh(−x).

Key Point
The hyperbolic function f(x) = cosh x is defined by the formula
cosh x = (ex)/2 + e−x/2
The function satisfies the conditions cosh 0 = 1 and cosh x = cosh(−x). The graph of cosh x is always above the graphs of ex/2 and e−x/2.

Publicerat i Geometri, Uncategorized | Märkt | Lämna en kommentar

Matrices

Publicerat den maj 19, 2013 av mattelararen

A matrix is a square or retangular array of numbers or function of numbers that obeys certain laws. The numbers are distinguished by two subscripts ij. The first indicating the row and the second indicating the column (vertical). in which the number appears. a13 is ther number in the first row snd third column.
It is worth mentioning that it is not a number as the determinant but an array of numbers.

Aij =Bij if and only if all the corresponding elements are equal in the two matrices.

Addition is performed by adding the respective numbers in the corresponding places in each matrix wuth each other.
A+B = B´+A so matrix addition it is commutative.

Division is performed by multipying yjr elements in row i by the elemetns in column j.
Therfore matrixmultiplication is anti-commutatative and AB not equal to BA.
A diagonal matrix is a matrix with zeroes in every position except in the diagonal. The trace of such a matrix is the sum of the diagonal elements.

A diagonal matrix with ones in all position of the diagonal is called an identity matrix. 

The identity matrix is defined by the operation AI = A.
The inverse of a matrix is defined by the relation A A-1 =I.

A tensor can be thought of as a three dimensional matrix.

Publicerat i Algebra, matematik 5 | Märkt , | Lämna en kommentar

Spherical harmonics

Publicerat den maj 12, 2013 av mattelararen

The spherical harmonics are functions describing the angular dependence of many physical problems e.g solutions to the Schrödinger equation for the hydrogen atom.

If the latitiude is denoted by v and x= cosv then  the equation is

(d/dx){(1 – x2)dPndx} + n(n+1)Pn = 0.

Pn is the spherical harmonics or the Legendre polynomials of degree n.

a more detailde description is found below:

Spherical harmonics

Publicerat i Calculus | Märkt , , | Lämna en kommentar

The Meton Cycle

Publicerat den april 18, 2013 av mattelararen

The ancient greek astronomer Meton observed that the lunar phases are repeated on the same weekday once every 19th year. therefore this period is referred to as the

Meton Cycle.  Theses days were carved in a stone at Forum Romanum in ancinet rome and they also play an importatnt part in the Baha’i calender as well as other calendars based on the lunar-phases.

He also observed that one tropical year consists of 365 days 6 hours and 19 minutes.

Thus one Meton cycle equals 6 940 days.

x2

 

link

Publicerat i Astronomy | Lämna en kommentar

The KAngaroocompetition 2013

Publicerat den april 3, 2013 av mattelararen

On March 21th I participated in the Kängurutävlingen arranged by the Royal Swedish academy of Science. My result was 45p.

See examples from last years competition here

Publicerat i Uncategorized | Lämna en kommentar

Eulers polyederformula

Publicerat den mars 27, 2013 av mattelararen

Definition: A graph is called planar if can be drawn in one plane without any arcs crossing each other.

Definintion: The graph G = (V,E) is called bipartite if the nodes can be divided into two disjunct parts V = V1∨ V2. where V1dosen’t have any elemenets in common with V2.

Eulers polyhedronformula: Let G = (V, E) be a planar, connected graph and let v denote the number of nodes, e the number of arcs and r be the number of surfaces. Then

v – e + r = 2.

Ex. For the dodecaedron, the number of surfaces is 12 similar pentagons. v = 20, e = 30 and r = 12.

Publicerat i Algebra, Gymnasiematematik(high school math), matematik 5 | Märkt , | 5 kommentarer