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SYSTEM VECTOR CONVERSION OF UNITS and Physics

SYSTEM UNITS

A unit system is a consistent set of units of measurement. Define a core set of measurement units from which we derive the rest. There are several systems of units:

International System of Units or SI : is the most used. Its basic units are the meter, kilogram, second, the ampere, the kelvin, the candela and the mole. Other units are derived from the International System.
Metric: first unified action.
or System CGS cegesimal : So named because its basic units are the centimeter, gram and second.
Natural System, in which units are chosen so that certain physical constants are worth exactly 1.
Technical system of units: metric derived from the previous units. This system is deprecated. Anglo System
units, still used in some common law countries. Many of them are replaced by the International System of Units.

CONVERSION OF UNITS

The Conversion is transformation from one unit to another.

This is done using the conversion factors and useful conversion tables. Suffice

multiply by a fraction (conversion factor) and the result is another equivalent, in which units have changed.

When units change involves the transformation of several units can use multiple conversion factors one after another so that the end result will be the equivalent measure in the units we want, for example if we spend 8 meters yards, all you have to do is multiply 8 x (0914) = 7312 yards.

VECTORS

In physics, a vector is a geometric tool used to represent a physical quantity which depends only on a module (or length) and direction (or orientation) for be defined.

Notation

The vector quantities are represented in the texts of letters printed in bold to distinguish them from scalar quantities are represented in italics . In the manuscripts, the vector magnitudes are represented placing an arrow on the letter that identifies your module (which is a scalar).

Vector Addition

Free To add two vectors (vector and vector) are chosen as representatives of two vectors such that the tail end of a match with the source end of another vector.

parallelogram method

This method allows only adding vectors in pairs. Is to have two vectors graphically so that the origins of both agree on one point, drawing lines parallel to each of the vectors, in the end of the other and of equal length, forming a parallelogram (see chart at right). The result of the sum is the diagonal of the parallelogram of the common origin of both vectors. Method

triangle consists

have a vector graphic after another, ie the origin of each vector is carried over the end of another. The resulting vector is one that is born in the origin of the first vector and ends at the end of the latter.

Analytical method for the sum and difference of vectors

Given two free vectors,

$\mathbf{a} = (a_x \mathbf{i} + a_y \mathbf{j} + a_z \mathbf{k})$

$\mathbf{b} = (b_x \mathbf{i} + b_y \mathbf{j} + b_z \mathbf{k})$

The result of the sum or the difference is expressed in the form

$\mathbf{a} \pm \mathbf{b} =(a_x \mathbf{i} +a_y \mathbf{j} +a_z \mathbf{k}) \pm(b_x \mathbf{i} +b_y \mathbf{j} +b_z \mathbf{k})$

and ordering the components,

$\mathbf{a} \pm \mathbf{b} = (a_x \pm b_x) \mathbf{i} + (a_y \pm b_y) \mathbf{j} + (a_z \pm b_z)\mathbf{k}$

With the matrix notation would

$\mathbf{a} \pm \mathbf{b}= \begin{bmatrix} a_x\ a_y\ a_z\\end{bmatrix}\pm\begin{bmatrix} b_x\ b_y\ b_z\\end{bmatrix}= \begin{bmatrix} a_x\pm b_x\ a_y\pm b_y\ a_z\pm bz\\end{bmatrix}$

Known modules given two vectors, and $\mathbf{a}$ $\mathbf{b}$ and the angle θ

that form each other, the module is $\mathbf{a} \pm \mathbf{b}$ :

$|\mathbf{a} \pm \mathbf{b}| = \sqrt{a^2 + b^2 - 2ab \cos \theta}$

The derivation of this expression can be found in module deduct the sum.

product of a vector by a scalar

The product of a vector by a scalar is another vector whose magnitude is the scalar product of the magnitude of the vector, whose direction is equal to the vector, or against it if the scale is negative.

Based on the graphical representation of the vector, on the same line of address took the form as many times as indicated by the scale vector.

Sean $p \,$ $\mathbf{a}$ a scalar and a vector, the product is represented $p \,$ by $\mathbf{a}$ $p \, \mathbf{a}$ and is done by multiplying each of the components of the vector by the scalar, ie

$p \, \mathbf{a} = pa_x \mathbf{i} + pa_y \mathbf{j} + pa_z \mathbf{k}$

With the matrix notation would

$p \, \mathbf{a} = p \, \begin{bmatrix} a_x\ a_y\ a_z\\end{bmatrix} = \begin{bmatrix} p\,a_x\ p\,a_y\ p\,a_z\\end{bmatrix}$

Scalar Product

The scalar product is an operation defined on two vectors of a Euclidean space that results in a number or scale.

A scalar product can be expressed as an application where $\langle \cdot,\cdot \rangle: V \times V \longrightarrow \mathbb{K}$ V is a vector space and $\mathbb{K}$ is the body that is defined on V . $\langle \cdot,\cdot \rangle$ must meet the following conditions:

Linearity on the left and right: $\langle ax+by,z \rangle = a \langle x,z \rangle + b \langle y,z \rangle$ , and similarly $\langle x, ay+bz \rangle = a \langle x, y \rangle + b \langle x,z \rangle$
Hermitian: $\langle x,y \rangle = \overline{\langle y,x \rangle}$ ,
positive definite: $\langle x,x \rangle \geq 0 \,$ and $\langle x,x \rangle = 0 \,$ if and only if x = 0,

which are vectors of $x, y, z \in V$ V , $a, b \in \mathbb{K}$ represent scalars and $\mathbb{K}$ $\overline{c}$ body is complex conjugate c .

If the body has no imaginary part (eg, $\mathbb{R}$ ), the property of being sesquilinear comes into being and being Hermitian bilinear becomes be symmetrical.

also often represented by $(\cdot|\cdot)$ or $\bullet$ .

A vector space over the body or $\mathbb{R}$ equipped $\mathbb{C}$ from an inner product space is called prehilbertiano prehilbert or space. If addition is complete, is said to be a Hilbert space, and if the dimension is finite, we can say that is a Euclidean space.

All scalar product induces a norm on the space that is defined as follows:

$\|x\|:= \sqrt{\langle x,x \rangle}$ .

The vector product is a binary operation between two vectors of a three-dimensional Euclidean space that results in a vector orthogonal to the two original vectors. Sean

$\mathbf a$ and $\mathbf b$ two vectors in the vector space $\mathbb{R}^3$ . The vector product between and $\mathbf a\,$ $\mathbf b\,$ results a new vector, $\mathbf c\,$ . To define this new vector is necessary to specify its magnitude and direction:

The of $\mathbf c\,$ module is given by

$c = a \, b \, \sin\theta$

where θ is the angle determined by the vectors to and b .

The address c vector that is orthogonal to to and orthogonal to b is given by the right-hand rule.

The vector product between to and b is denoted by a × b therefore is also called cross product . In the manuscripts, to avoid confusion with the letter x (X) is often used to denote the vector product by a b ∧ .

The vector product can be defined in a more compact way as follows:

${\mathbf a \times \mathbf b = {a} \, {b} \, {\sin}{\theta}\ \hat\mathbf n}$

$\hat\mathbf n$ where is the unitary vector orthogonal to the vectors to and b and its direction is given by the right hand rule and θ is, as before, the angle between a and b . On the right hand rule is also often called the corkscrew rule.

Derivative of a vector

Given an array is a function of independent variable

$\mathbf{a}(t)= a_x(t) \mathbf{i} +a_y(t) \mathbf{j} +a_z(t) \mathbf{k}$

estimate the derivative of the vector with respect to variable t , calculating the derivative of each one of its scalar components as if it were:

$\frac{d}{dt}\mathbf{a}(t)= \frac{d}{dt}a_x(t) \mathbf{i} + \frac{d}{dt}a_y(t) \mathbf{j} + \frac{d}{dt}a_z(t) \mathbf{k}$

given that the unit vectors are constant in magnitude and direction.

With matrix notation would

$\frac{d}{dt}\mathbf{a}(t)= \frac{d}{dt} \, \begin{bmatrix} a_x\ a_y\ a_z\\end{bmatrix} = \begin{bmatrix} \frac{d}{dt}a_x\ \frac{d}{dt}a_y\ \frac{d}{dt}a_z\\end{bmatrix}$

$\mathbf{r}(t)=\sin(t) \mathbf{i}+\cos(t)\mathbf{j}+ 5t\mathbf{k}$

An example of derivation of a vector, based on a vector function:

$\mathbf{r}(t) = \sin(t) \mathbf{i} + \cos(t) \mathbf{j} + 5t \mathbf{k}$

This function represents a helical curve around the axis z , radio unit, as illustrated in Fig. We can imagine that this curve is the trajectory of a particle and $\mathbf r (t)\,$ function represents the position vector function of time t . Differentiating we have:

$\frac{d\mathbf{r}(t)}{dt} = \frac{d}{dt}\sin(t) \mathbf{i} + \frac{d}{dt}\cos(t) \mathbf{j} + \frac{d}{dt}5t \mathbf{k}$

Performing the derivative:

$\frac{d\mathbf{r}(t)}{dt} = \cos(t) \mathbf{i} - \sin (t) \mathbf{j} + 5 \mathbf{k}$

The derivative of the position vector with respect to time is the velocity, so this second function determines the particle velocity vector versus time, we can write:

$\mathbf{v}(t) = \cos(t) \mathbf{i} - \sin(t) \mathbf{j} + 5 \mathbf{k}$

This velocity vector is a vector tangent to the trajectory at the point occupied by the particle in each moment. If derivásemos again would get the acceleration vector.

angle between two vectors

The angle determined by the directions of two vectors $\mathbf{a}$ $\mathbf{b}$ and is given by:

$\cos \theta = \frac {\mathbf a \cdot \mathbf b}{|\mathbf{a}| \, |\mathbf{b}|}$

vector base change

rotations are linear transformations that preserve norms in vector spaces in which you have defined an inner product operation. The transformation matrix has the property of being a unitary matrix, ie is orthogonal and its determinant is 1. $\scriptstyle \mathbf A$ a vector expressed in a Cartesian coordinate system ( x, y, z ) with a vector base defined by the associated $\mathcal{B}$ $\left( \mathbf i, \mathbf j,\mathbf k\, \right)$ vestors, ie

$\mathbf A=\begin{bmatrix} A_x \ A_y \ A_z \end{bmatrix}_{\mathcal{B}}$

Now suppose we turn the system coordinate axes, holding the source of the problem, so we get a new trihedral orthogonal axes ( x ', y', z ') with a vector base defined by the associated $\mathcal{B}'$ $\left( \mathbf i', \mathbf j',\mathbf k'\, \right)$ vestors. $\mathbf A \,$ vector components in this new basis vector will be:

$\mathbf A=\begin{bmatrix} A'_x \ A'_y \ A'_z \end{bmatrix}_{\mathcal{B}'}$

rotation operation based vector can always be expressed as the action of a linear operator (represented by a matrix) acting on the vector (multiplying the vector):

$\mathbb R \, \mathbf A_{\mathcal{B}} = \mathbf A_{\mathcal{B}'}$

is the transformation matrix for the vector base change.

Examinations. Free High School Graduation Tests (ESO). Madrid

Sunday, March 6, 2011