miércoles, 30 de septiembre de 2015

domingo, 20 de septiembre de 2015

Suma de fracciones de distinto denominador

Matemática, espíritu y arte: Suma y resta de fracciones de igual denominador

Matemática, espíritu y arte: Suma y resta de fracciones de igual denominador

Suma y resta de fracciones de igual denominador

Operaciones con fracciones

Matemática, espíritu y arte: Suma de fracciones: ejercicios resueltos

Matemática, espíritu y arte: Suma de fracciones: ejercicios resueltos

Suma de fracciones: ejercicios resueltos



Clasificación de triángulos

Consider each of the triangles below. Circle all that apply to the triangle.

1.         Acute              Obtuse             Right               Scalene            Isosceles          Equilateral

2.         Acute              Obtuse             Right               Scalene            Isosceles          Equilateral

3.         Acute              Obtuse             Right               Scalene            Isosceles          Equilateral

4.         Acute              Obtuse             Right               Scalene            Isosceles          Equilateral

5.         Acute              Obtuse             Right               Scalene            Isosceles          Equilateral

Triángulos congruentes

miércoles, 9 de septiembre de 2015

Trabajo I

Let's see an exercise.
A 40 kg box is dragged 30 m on a horizontal floor, applying a Fp = 100 N exerted by a person. Such force acts doing a 60º angle. The floor exerts a friction force Fr = 20 N. Calculate the work done for each one of these forces Fp, Fr, the weight mg and the normal. Calculate also the net work done on the box.
Solution: There are four forces acting on the box, Fp, Fr, the weight mg and the normal.
The work done by mg and the normal N is zero, because they are perpendicular to displacement(=90º for them).
The work done by Fp is: 
Wp = Fpxcos (using x in place of d) = (100 N)(30 m)cos60º = 1500 J.
The work done by the friction force Fr is: 
Wr = Frxcos180º = (20 N)(30 m)(-1) = -600 J.
The angle between Fr and displacement is 180º because they point in opposite directions.
The net work can be calculated in two equivalents ways:
  • As the algebraic sum of the work done by each force:
    WNET = 1500 J +(- 600 J) = 900 J.
  • Finding first the net force on the object along the displacement:
    F(NET)x= Fpcos - Fr
    and then doing
    WNET = F(NET)x = (Fpcos - Fr)x
    = (100 Ncos60º - 20 N)(30 m) = 900 J.

Ley de Ohm III

Ohm's Law,
Example Three

Given the resistor arrangement shown below, prove that the relation between R1 and R2 must be R2=1.618R1 in order that the resistance of the system R be equal to R2.
The condition is R=R2, so we do
0= R22 - R1R2 - R12.
This is a second degree equation in R2, whose solution is R2= R1(1/2 ± )
Disregarding the minus sign we get R2 = 1.618R1

Ley de Ohm II

Ohm's Law,
Example Two

Determine the total resistance and potential difference between points a and b in the circuit shown. Also determine the current in, and the potential difference across, each resistor. The input current is 60 A.
a) Total resistance between a and b.
First, calculate the 12, 6, and 4 ohm parallel combination R//1
1/R//1 = 1/12 + 1/6 + 1/4 = 1/2. Then R//1 = 2 ohm.
The upper branch is 3 ohms in series with R//1, 3 ohm + 2 ohm = 5 ohm.
This 5 ohm resistor is in parallel with the 20 ohm resistor, equivalent to a single resistor R//2 = 5 ohm//20 ohm:
1/R//2 = 1/5 + 1/20 = ¼. Then R//2 = 4.
The total resistance between a and b is 4 ohm + 5 ohm = 9 ohms. The voltage drop or potential difference between points a and b is 60Ax9ohm = 540 V.
b) To obtain the individual currents and voltages in each resistor we can proceed as follows (there is no a unique method):
The voltage across R//2 is the difference between the total 540 voltage and the voltage drop in the 5 ohm resistor. This drop is 60Ax5ohm = 300 V. So the potential difference across R//2 is 540 V – 300 V = 240 V.
Now we can calculate the currents and voltages in the R//2combination:
In the 20 ohm resistor: current= 240 V/20 ohm = 12 A, voltage = 240 V
The upper branch has then a current of 60 A - 12 A = 48 A, or alternatively 240 V/5 ohm = 48A.
The 3 ohm resistor has a voltage of 48 Ax3 ohm = 144 V.
The R//1 combination presents a voltage of 48 AxR//1 = 48 Ax2 ohm = 96 V, or alternatively 240 V - 144 V = 96 V.
The currents in the 12 ohm, 6 ohm and 4 ohm are respectively 96/12 A=8 A, 96/6=16 A and 96/4=24 A

Ley de Ohm I

Ohm's Law,
Example One

The circuit below is called a Wheatstone bridge. It is used for measuring resistance. Show that when the current through the galvanometer G is zero, then R1 = R2(R3/R4). Thus, if we know R2 and the ratio (R3/R4), we can obtain the resistance R1.
IG = 0 means points C and D are at the same potential and
that I1 = I2, I3 = I4.
By Ohm’s law VAC = I1R1, VAD = I3R3, VCB = I2R2, VDB = I4R4.
As VAC = VAD and VCB = VDB, then I1R1= I3R3,
I2R2= I4R4.
As I1 = I2 and I3 = I4, then doing I1R1/ I2R2=I3R3/I4R4 we get R1= R2(R3/R4)

física y creatividad: Fuerzas actuantes en plano inclinado y polea fija

física y creatividad: Fuerzas actuantes en plano inclinado y polea fija

domingo, 6 de septiembre de 2015



a) Dos ángulos adyacentes son consecutivos
b) Dos ángulos adyacentes son suplementarios
c) Los ángulos opuestos por el vértice son complementarios
d) El complemento de un ángulo agudo es un ángulo obtuso
e) Dos ángulos consecutivos son complementarios
f) El complemento de un ángulo recto es un ángulo nulo
g) Dos ángulos opuestos por el vértice son suplementarios

Álgebra - Verdadero-Falso



martes, 1 de septiembre de 2015

Satz des Pythagoras (Mathe-Song)

Brüche addieren - mit vedischer Mathematik (Mathe-Song)

Quadratische Funktionen (Mathe-Song)

Quadratische Funktionen Crashkurs =D part1

Lineare Funktionen Crashkurs =D

Mathe F06: Quadratische Funktionen (Teil 3 von 7) - Allgemeinform und Qu...

Mathe G21: Irrationale Zahlen + Reelle Zahlen

Mathe F05: Lineare Gleichungssysteme (Teil 1 von 6) - Die 3 Lösungsverfa...