Teoría de Cinemática:
Movimiento rectilíneo uniforme:
Principio: [ de Velocidad constante ]
[Ev][ d_{t}[x(t)] = v ]
Ley: [ de Posición ]
x(t) = vt+h
Deducción:
x(t) = int[ d_{t}[x] ]d[t] = int[v]d[t] = v·int[1]d[t]= vt+h
Movimiento rectilíneo uniformemente acelerado:
Principio: [ de Aceleración constante ]
[Ea][ d_{tt}^{2}[x(t)] = a ]
Ley: [ de Velocidad ]
d_{t}[x(t)] = at+v
Deducción:
d_{t}[x(t)] = int[ d_{tt}^{2}[x(t)] ]d[t] = int[a]d[t] = a·int[1]d[t]= at+v
Ley: [ de Posición ]
x(t) = a·(1/2)·t^{2}+vt+h
Deducción:
x(t) = int[ d_{t}[x(t)] ]d[t] = int[at+v]d[t] = int[at]d[t]+int[v]d[t]= ...
... a·int[t]d[t]+v·int[1]d[t] = a·(1/2)·t^{2}+vt+h
Ley:
( El Principio de Aceleración constante )
... implica ...
( El Principio de Velocidad constante ):
Si a = 0 ==> [Ev][ d_{t}[x(t)] = v ]
Deducción:
d_{t}[x(t)] = at+v = 0·t+v = v
Movimiento angular uniforme:
Principio: [ de Velocidad angular constante ]
[Eu][ d_{t}[w(t)] = (v/r) = u ]
Ley: [ de posición angular ]
w(t) = ut+s
Deducción:
w(t) = int[ d_{t}[w] ]d[t] = int[u]d[t] = u·int[1]d[t]= ut+s
Movimiento angular uniformemente acelerado:
Principio: [ de Aceleración angular constante ]
[Eb][ d_{tt}^{2}[w(t)] = (a/r) = b ]
Ley: [ de Velocidad angular ]
d_{t}[w(t)] = bt+u
Deducción:
d_{t}[w(t)] = int[ d_{tt}^{2}[w(t)] ]d[t] = int[b]d[t] = b·int[1]d[t]= bt+u
Ley: [ de posición angular ]
w(t) = b·(1/2)·t^{2}+ut+s
Deducción:
w(t) = int[ d_{t}[w(t)] ]d[t] = int[bt+u]d[t] = int[bt]d[t]+int[u]d[t]= ...
... b·int[t]d[t]+u·int[1]d[t] = b·(1/2)·t^{2}+ut+s
Ley:
( El Principio de Aceleración angular constante )
... implica ...
( El Principio de Velocidad angular constante ):
Si b = 0 ==> [Eu][ d_{t}[w(t)] = (v/r) = u ]
Deducción:
d_{t}[w(t)] = bt+u = 0·t+u = 0+u = u = (r/r)·u = (1/r)·(ru) = (v/r) = u
Ley:
( El Principio de Velocidad constante )
... sí, y solo sí ...
( El Principio de Velocidad angular constante ):
[Ev][ d_{t}[x(t)] = v ] <==> [Eu][ d_{t}[w(t)] = (v/r) = u ]
Deducción:
[==>] Si d_{t}[x(t)] = v ==>
(1/r)·d_{t}[x(t)] = (v/r)
d_{t}[w(t)] = d_{t}[(1/r)·x(t)] = (1/r)·d_{t}[x(t)] = (v/r) = u
[<==] Si d_{t}[w(t)] = (v/r) = u ==>
r·d_{t}[w(t)] = r·(v/r)
d_{t}[x(t)] = d_{t}[r·w(t)] = r·d_{t}[w(t)] = r·(v/r) = ( r·(1/r) )·v = 1·v = v
Ley:
( El Principio de Aceleración constante )
... sí, y solo sí ...
( El Principio de Aceleración angular constante )
[Ea][ d_{tt}^{2}[x(t)] = a ] <==> [Eb][ d_{tt}^{2}[w(t)] = (a/r) = b ]
Deducción:
[==>] Si d_{tt}^{2}[x(t)] = a ==>
(1/r)·d_{tt}^{2}[x(t)] = (a/r)
d_{tt}^{2}[w(t)] = d_{tt}^{2}[(1/r)·x(t)] = (1/r)·d_{tt}^{2}[x(t)] = (a/r) = b
[<==] Si d_{tt}^{2}[w(t)] = (a/r) = b ==>
r·d_{tt}^{2}[w(t)] = r·(a/r)
d_{tt}^{2}[x(t)] = d_{tt}^{2}[r·w(t)] = r·d_{tt}^{2}[w(t)] = r·(a/r) = ( r·(1/r) )·a = 1·a = a
Problemas de Cinemática:
Ley:
Si ( x(t) = (-v)·t+h & y(t) = wt ) ==> ( [1] & [2] )
[1] y(t_{k}) = x(t_{k}) <==> t_{k} = ( h/(v+w) )
[2] [Ea][ d_{t}[y(t_{j})] = d_{t}[x(t_{j})]+at_{j} <==> t_{j} = ( (v+w)/a ) ]
Deducción:
[1]
[==>] Si y(t_{k}) = x(t_{k}) ==>
wt_{k} = (-v)·t_{k}+h
vt_{k}+wt_{k} = vt_{k}+( (-v)·t_{k}+h )
(v+w)·t_{k} = vt_{k}+wt_{k} = vt_{k}+( (-v)·t_{k}+h ) = ( vt_{k}+(-v)·t_{k} )+h = 0+h = h
(v+w)·t_{k} = h
( 1/(v+w) )·( (v+w)·t_{k} ) = ( h/(v+w) )
t_{k} = 1·t_{k} = ( ( 1/(v+w) )·(v+w) )·t_{k} = ( 1/(v+w) )·( (v+w)·t_{k} ) = ( h/(v+w) )
[<==] Si t_{k} = ( h/(v+w) ) ==>
(v+w)·t_{k} = (v+w)·( h/(v+w) ) = ( (v+w)·( 1/(v+w) ) )·h = 1·h = h
(v+w)·t_{k} = h
vt_{k}+wt_{k} = (v+w)·t_{k} = h = 0+h = ( vt_{k}+(-v)·t_{k} )+h = vt_{k}+( (-v)·t_{k}+h )
vt_{k}+wt_{k} = vt_{k}+( (-v)·t_{k}+h )
wt_{k} = ( (-v)·t_{k}+h )
y(t_{k}) = x(t_{k})
[2]
d_{t}[y(t)] = d_{t}[wt] = w·d_{t}[t] = w·1 = w
d_{t}[x(t)] = d_{t}[(-v)·t+h] = d_{t}[(-v)·t]+d_{t}[h] = (-v)·d_{t}[t]+d_{t}[h] = (-v)·1+0 = (-v)
[==>] Si d_{t}[y(t_{j})] = d_{t}[x(t_{j})]+at_{j} ==>
w = (-v)+at_{j}
v+w = v+( (-v)+at_{j} )
v+w = v+( (-v)+at_{j} ) = ( v+(-v) )+at_{j} = 0+at_{j} = at_{j}
v+w = at_{j}
at_{j} = v+w
(1/a)·( at_{j} ) = ( (v+w)/a )
t_{j} = 1·t_{j} = ( (1/a)·a )·t_{j} = (1/a)·( at_{j} ) = ( (v+w)/a )
[<==] Si t_{j} = ( (v+w)/a )
at_{j} = a·( (v+w)/a ) = ( a·(1/a) )·(v+w) = 1·(v+w) = v+w
at_{j} = v+w
(-v)+at_{j} = (-v)+(v+w)
(-v)+at_{j} = (-v)+(v+w) = ( (-v)+v )+w = 0+w = w
(-v)+at_{j} = w
w = (-v)+at_{j}
d_{t}[y(t_{j})] = d_{t}[x(t_{j})]+at_{j}
Ley:
Si ( x(t) = ( (-a)·(1/2) )·t^{2}+h & y(t) = ( b·(1/2) )·t^{2} ) ==> ( [1] & [2] )
[1] y(t_{k}) = x(t_{k}) <==> t_{k} = ( (2h)/(a+b) )^{(1/2)}
[2] [Ev][ d_{t}[y(t_{j})] = d_{t}[x(t_{j})]+v <==> t_{j} = ( v/(a+b) ) ]
Deducción:
[1]
[==>] Si y(t_{k}) = x(t_{k}) ==>
( b·(1/2) )·( t_{k} )^{2} = ( (-a)·(1/2) )·( t_{k} )^{2}+h
2·( ( b·(1/2) )·( t_{k} )^{2} ) = 2·( ( (-a)·(1/2) )·( t_{k} )^{2}+h )
2·( ( b·(1/2) )·( t_{k} )^{2} ) = 2·( ( (-a)·(1/2) )·( t_{k} )^{2}+h ) = 2·( ( (-a)·(1/2) )·( t_{k} )^{2} )+2h
b·( t_{k} )^{2} = (-a)·( t_{k} )^{2}+2h
a·( t_{k} )^{2}+b·( t_{k} )^{2} = a·( t_{k} )^{2}+( (-a)·( t_{k} )^{2}+2h )
(a+b)·( t_{k} )^{2} = a·( t_{k} )^{2}+b·( t_{k} )^{2} = a·( t_{k} )^{2}+( (-a)·( t_{k} )^{2}+2h ) = ...
... ( a·( t_{k} )^{2}+(-a)·( t_{k} )^{2} )+2h = 0+2h = 2h
(a+b)·( t_{k} )^{2} = 2h
(1/(a+b))·( (a+b)·( t_{k} )^{2} ) = ( (2h)/(a+b) )
( t_{k} )^{2} = 1·( t_{k} )^{2} = ( (1/(a+b))·(a+b) )·( t_{k} )^{2} = ...
... (1/(a+b))·( (a+b)·( t_{k} )^{2} ) = ( (2h)/(a+b) )
( t_{k} )^{2} = ( (2h)/(a+b) )
( t_{k} )^{(2/2)} = ( (2h)/(a+b) )^{(1/2)}
t_{k} = ( t_{k} )^{1}= ( t_{k} )^{(2/2)} = ( (2h)/(a+b) )^{(1/2)}
t_{k} = ( (2h)/(a+b) )^{(1/2)}
[<==] Si t_{k} = ( (2h)/(a+b) )^{(1/2)}
( t_{k} )^{2} = ( (2h)/(a+b) )^{(2/2)}
( t_{k} )^{2} = ( (2h)/(a+b) )^{(2/2)} = ( (2h)/(a+b) )^{1} = ( (2h)/(a+b) )
b·( t_{k} )^{2} = (-a)·( t_{k} )^{2}+2h
(1/2)·( b·( t_{k} )^{2} ) = (1/2)·( (-a)·( t_{k} )^{2}+2h )
( b·(1/2) )·( t_{k} )^{2} = ( (-a)·(1/2) )·( t_{k} )^{2}+h
y(t_{k}) = x(t_{k})
[2]
d_{t}[y(t)] = d_{t}[ ( (-a)·(1/2) )·t^{2}+h ] = d_{t}[ (-a)·( (1/2)·t^{2} )+h ] = ...
... d_{t}[ (-a)·( (1/2)·t^{2} ) ]+d_{t}[h] = (-a)·d_{t}[ ( (1/2)·t^{2} ) ]+d_{t}[h] = (-a)·t+0 = (-a)·t
d_{t}[x(t)] = d_{t}[ ( b·(1/2) )·t^{2} ] = d_{t}[ b·( (1/2)·t^{2} ) ] = b·d_{t}[ (1/2)·t^{2} ] = bt
Ley:
Si ( x(t) = (-v)·t+h & y(t) = ( a·(1/2) )·t^{2} ) ==> ( [1] & [2] )
[1] y(t_{k}) = x(t_{k}) <==> t_{k} = (1/a)·( (-v)+( v^{2}+2ah )^{(1/2)} )
[2] d_{t}[y(t_{k})] = d_{t}[x(t_{k})]+( v^{2}+2ah )^{(1/2)}
Deducción:
[1]
[==>] Si y(t_{k}) = x(t_{k}) ==>
( a·(1/2) )·( t_{k} )^{2} = (-v)·t_{k}+h
vt+( a·(1/2) )·( t_{k} )^{2} = vt_{k}+( (-v)·t_{k}+h )
( a·(1/2) )·( t_{k} )^{2}+vt_{k} = vt_{k}+( a·(1/2) )·( t_{k} )^{2} = ...
... vt_{k}+( (-v)·t_{k}+h ) = ( vt_{k}+(-v)·t_{k} )+h = 0+h = h
( a·(1/2) )·( t_{k} )^{2}+vt_{k} = h
( a·(1/2) )·( t_{k} )^{2}+vt_{k}+(-h) = h+(-h)
( a·(1/2) )·( t_{k} )^{2}+vt_{k}+(-h) = h+(-h) = 0
( a·(1/2) )·( t_{k} )^{2}+vt_{k}+(-h) = 0
t_{k} = (2/2a)·( (-v)+( v^{2}+4·(1/2)·ah )^{(1/2)} )
t_{k} = (1/a)·( (-v)+( v^{2}+2ah )^{(1/2)} )
[<==] Si t_{k} = (1/a)·( (-v)+( v^{2}+2ah )^{(1/2)} ) ==>
( a·(1/2) )·( t_{k} )^{2}+vt_{k}+(-h) = 0
( a·(1/2) )·( t_{k} )^{2}+vt_{k}+(-h)+h = 0+h
vt_{k}+( a·(1/2) )·( t_{k} )^{2} = ( a·(1/2) )·( t_{k} )^{2}+vt_{k} = ...
... ( a·(1/2) )·( t_{k} )^{2}+vt_{k}+0 = ( a·(1/2) )·( t_{k} )^{2}+vt_{k}+(-h)+h = 0+h = h
vt_{k}+( a·(1/2) )·( t_{k} )^{2} = h
(-v)·t_{k}+( vt_{k}+( a·(1/2) )·( t_{k} )^{2} ) = (-v)·t_{k}+h
( a·(1/2) )·( t_{k} )^{2} = 0+( a·(1/2) )·( t_{k} )^{2} = ...
... ( (-v)·t_{k}+vt_{k} )+( a·(1/2) )·( t_{k} )^{2} = (-v)·t_{k}+( vt_{k}+( a·(1/2) )·( t_{k} )^{2} ) = ...
... (-v)·t_{k}+h
( a·(1/2) )·( t_{k} )^{2} = (-v)·t_{k}+h
y(t_{k}) = x(t_{k})
Ley:
Si ( d_{t}[x(t)]+d_{t}[y(t)] = at & d_{t}[x(t)]+(-1)·d_{t}[y(t)] = bt ) ==>
[1:a] d_{t}[x(t)] = (1/2)·(a+b)·t
[1:b] d_{t}[y(t)] = (1/2)·(a+(-b))·t
[2:a] x(t) = (1/4)·(a+b)·t^{2}
[2:b] y(t) = (1/4)·(a+(-b))·t^{2}
[3:a] d_{tt}^{2}[x(t)] = (1/2)·(a+b)
[3:b] d_{tt}^{2}[y(t)] = (1/2)·(a+(-b))
Deducción:
[1:a] Si ( d_{t}[x(t)]+d_{t}[y(t)] = at & d_{t}[x(t)]+(-1)·d_{t}[y(t)] = bt ) ==>
2·d_{t}[x(t)]+( d_{t}[y(t)]+(-1)·d_{t}[y(t)] ) = at+bt
2·d_{t}[x(t)]+0 = 2·d_{t}[x(t)]+( d_{t}[y(t)]+(-1)·d_{t}[y(t)] ) = at+bt = (a+b)·t
2·d_{t}[x(t)] = (a+b)·t
(1/2)·( 2·d_{t}[x(t)] ) = (1/2)·(a+b)·t
d_{t}[x(t)] = 1·d_{t}[x(t)] = (2/2)·d_{t}[x(t)] = (1/2)·( 2·d_{t}[x(t)] ) = (1/2)·(a+b)·t
d_{t}[x(t)] = (1/2)·(a+b)·t
Ley:
Si x(t) = ( d^{n}+(vt)^{n} )^{(1/n)} ==> ( [1] & [2] )
[1] d_{t}[x] = x^{1+(-n)}·(vt)^{n+(-1)}·v
[2] d_{tt}^{2}[x] = ...
... ( (1+(-n))·x^{1+(-1)·2n}·(vt)^{2n+(-2)}+x^{1+(-n)}·(n+(-1))·(vt)^{n+(-2)} )·v^{2}
Deducción:
[1] d_{t}[x] = d_{t}[ ( d^{n}+(vt)^{n} )^{(1/n)} ] = ...
... d_{(vt)^{n}}[ ( d^{n}+(vt)^{n} )^{(1/n)} ]·d_{vt}[ (vt)^{n} ]·d_{t}[vt] = ...
... (1/n)·( d^{n}+(vt)^{n} )^{(1/n)+(-1)}·n·(vt)^{n+(-1)}·v = ...
... ( d^{n}+(vt)^{n} )^{( (1+(-n))/n )}·(vt)^{n+(-1)}·v = x^{1+(-n)}·(vt)^{n+(-1)}·v
[2] d_{tt}^{2}[x] = d_{t}[ d_{t}[x] ] = d_{t}[ x^{1+(-n)}·(vt)^{n+(-1)}·v ] = ...
... d_{t}[ x^{1+(-n)}·(vt)^{n+(-1)} ]·v = ...
... ( d_{t}[ x^{1+(-n)} ]·(vt)^{n+(-1)}+x^{1+(-n)}·d_{t}[ (vt)^{n+(-1)} ] )·v = ...
... ( d_{x}[ x^{1+(-n)} ]·d_{t}[x]·(vt)^{n+(-1)}+x^{1+(-n)}·d_{vt}[ (vt)^{n+(-1)} ]·d_{t}[vt] )·v
Ley:
Si x(t) = ( d^{2}+(-1)·(vt)^{2} )^{(1/2)}+vt ==>
x(t) = d <==> ( t = 0 || t = (d/v) )
x(t_{k}) es máxima en t_{k} = (1/2)^{(1/2)}·(d/v)
x(t_{k}) = 2^{(1/2)}·d
Ley:
Sea ( d_{t}[y] = v & d_{t}[x] = uy ) ==>
x(y) = (u/v)·(1/2)·y^{2}
y(t) = vt
d_{tt}^{2}[x] = uv
d_{t}[x] = uvt
x(t) = uv·(1/2)·t^{2}
Deducción:
d[y] = v·d[t]
d[x] = uy·d[t]
d[x] = (u/v)·y·d[y]
Irodov-Garriga problems de cinemática de rotación uniforme:
Ley:
Sea d_{t}[w(t)] = d_{t}[a(t)]+d_{t}[b(t)] ==>
Si ( d_{t}[a(t)] = (v/h) & d_{t}[b(t)] = (v/r)·a(t) ) ==>
a(t) = (v/h)·t
b(t) = (v/r)·(v/h)·(1/2)·t^{2}
d_{t}[w(t)] = (v/h)·( 1+(v/r)·t )
d_{tt}^{2}[w(t)] = (v/r)·(v/h)
Ley:
Sea d_{t}[w(t)] = d_{t}[a(t)]+d_{t}[b(t)]+d_{t}[c(t)] ==>
Si ( d_{t}[a(t)] = (v/h) & d_{t}[b(t)] = (u/h) & d_{t}[c(t)] = (v/r)·a(t)+(u/r)·b(t) ) ) ==>
a(t) = (v/h)·t
b(t) = (u/h)·t
c(t) = (1/r)·(1/h)·( v^{2}+u^{2} )·(1/2)·t^{2}
d_{t}[w(t)] = (1/h)·( (v+u)+(1/r)·( v^{2}+u^{2} )·t )
d_{tt}^{2}[w(t)] = (1/r)·(1/h)·( v^{2}+u^{2} )
Ley:
Sea d_{t}[w(t)] = d_{t}[a(t)]+d_{t}[b(t)] ==>
Si ( d_{t}[a(t)] = (v/h) & d_{t}[b(t)] = (u/r)·( s(t) )^{n+1} & d_{t}[s(t)] = (u/r)·( 1/s(t) )^{n} ) ==>
a(t) = (v/h)·t
b(t) = (n+1)·(u/r)^{2}·(1/2)·t^{2}
d_{t}[w(t)] = (v/h)+(n+1)·(u/r)^{2}·t
d_{tt}^{2}[w(t)] = (n+1)·(u/r)^{2}
Deducción:
( s(t) )^{n+1} = (n+1)·(u/r)·t
Ley:
Sea d_{t}[w(t)] = d_{t}[a(t)]+d_{t}[b(t)] ==>
Si ( d_{t}[a(t)] = (v/h) & d_{t}[b(t)] = (u/r)·e^{n·s(t)} & d_{t}[s(t)] = (u/r)·e^{(-n)·s(t)} ) ==>
a(t) = (v/h)·t
b(t) = n·(u/r)^{2}·(1/2)·t^{2}
d_{t}[w(t)] = (v/h)+n·(u/r)^{2}·t
d_{tt}^{2}[w(t)] = n·(u/r)^{2}
Ley:
Naces en la Tierra que es el Malo,
porque no has estudiado,
y no condenas,
en no tener motivo de energía,
y no puedes estar en los Buenos.
Resucitas en el Cielo de Jûanat-Hád o en el Paraíso de Júpiter que son los Buenos,
porque has estudiado,
y condenas,
en tener motivo de energía,
y no puedes estar en el Malo.
Anexo:
Primero viene el Malo y después viene el Bueno.
Momento de Inercia de un anillo:
Ley:
I_{z} = m·(1/2)·r
Deducción:
I_{z} = int[x = 0]-[r][ ( m/(2pi·r) )·2pi·x ]d[x]
Momento de Inercia de un disco:
Ley:
I_{z} = m·(1/3)·r
Deducción:
I_{z} = int[x = 0]-[r][ ( m/(pi·r^{2}) )·pi·x^{2} ]d[x]
Momento de Inercia de una bola esférica:
Ley:
I_{z} = m·(1/4)·r
Deducción:
I_{z} = int[x = 0]-[r][ ( m/( (4/3)·pi·r^{3} ) )·(4/3)·pi·x^{3} ]d[x]
Momento de Inercia de una barra:
Ley:
r = radio de la anillo circunscrito = longitud de la barra.
I_{z} = m·(1/2)·r
Deducción:
I_{z} = int[x = 0]-[r][ (m/z)·x ]d[x]
Momento de Inercia de un rectángulo cuadrado:
Ley:
r = radio del disco inscrito.
I_{z} = m·(1/4)·r
Deducción:
I_{z} = int[y = 0]-[r]-int[x = 0]-[r][ (m/r^{3})·xy ]d[x]d[y]
Momento de Inercia de un rectángulo cúbico:
Ley:
r = radio de la esfera inscrita.
I_{z} = m·(1/8)·r
Deducción:
I_{z} = int[z = 0]-[r]-int[y = 0]-[r]-int[x = 0]-[r][ (m/r^{5})·xyz ]d[x]d[y]d[z]
Ley:
m·d_{t}[x(t)] = I_{z}·d_{t}[w(t)]
Ley:
Sea I_{z} = m·(1/n)·r ==>
Si d_{t}[w(t)] = (v/h)·w(t) ==>
w(t) = e^{(v/h)·t}
d_{t}[x(t)] = (1/n)·r·(v/h)·e^{(v/h)·t}
x(t) = (1/n)·re^{(v/h)·t}
d_{tt}^{2}[x(t)] = (1/n)·r·(v/h)^{2}·e^{(v/h)·t}
Ley:
Sea I_{z} = m·(1/n)·(vt) ==>
Si d_{t}[w(t)] = (v/h)·w(t) ==>
w(t) = e^{(v/h)·t}
d_{t}[x(t)] = (1/n)·(vt)·(v/h)·e^{(v/h)·t}
x(t) = (1/n)·h·( (v/h)·t )^{2}·er-h[2]( (v/h)·t )
d_{tt}^{2}[x(t)] = (1/n)·v·(v/h)·( 1+(v/h)·t )·e^{(v/h)·t}
Decret-Lley:
Parlament de Càteldor:
70 Pujjaló-Putxaló-Puixaló = Esquerra de Càteldor
65 Junts-Txunts-Ixunts = Dreta de Càteldor
Anex:
Impostos Socialistes.
Anex:
Es un delegat del govern de Càteldor de Putxaló,
el que gestiona els impostos a l'Aragó.
Es un delegat del govern de Càteldor de Puixaló,
el que gestiona els impostos en el País Valencià.
Decreto-Ley:
Congreso de los Diputados:
26 Bloque unionista.
180 escaños.
25 Bloque separatista.
170 escaños.
Ley: [ de financiación solidaria ]
No te puedes embuchacar los impuestos,
y se pierde menos dinero,
financiando Aragón, Valencia y Baleares,
no emitiendo cheques del Tesoro Catalán
porque compran en Catalunya,
con el dinero Catalán.
Te puedes embuchacar los impuestos,
y se pierde más dinero,
no financiando Aragón, Valencia y Baleares,
emitiendo cheques del Tesoro Catalán
aunque quizás compran en Catalunya,
con el dinero Catalán.
Ley-Anexa:
Es legal:
Comprar en Catalunya con el dinero Catalán.
No es legal:
No comprar en Catalunya con el dinero Catalán.
Ley:
Llueve y no me mojo
porque llevo paraguas.
Llueve y me mojo
porque no llevo paraguas.
Si llueve entonces me mojo
aunque quizás llevo paraguas.
Si llueve entonces no me mojo
aunque quizás no llevo paraguas.
Ley-Anexa:
Es legal:
Solgar de casa lloviendo,
sin viento.
Es legal:
No solgar de casa lloviendo,
con viento.
Ley:
El semáforo está en rojo y te tienes que parar.
El semáforo está en amarillo o no te tienes que parar.
Deducción:
Si el semáforo está en amarillo entonces te tienes que parar.
Si el semáforo está en azul entonces no te tienes que parar.
Ley:
Es legal:
el porno de no espectro de alma,
en el control de menores,
porque no son,
y se pueden desear,
infieles del prójimo.
Es ilegal:
el porno de espectro de alma,
en el control de mayores,
porque son,
y no se pueden desear,
fieles del prójimo.
Ley:
( 1/( 1+(-1)·(1/c)^{2}·(1/2)·m_{ij}·R_{iju}^{u} ) )·...
m·( d[u]d[u]+(-1)·(1/2)·m_{ij}·R_{iju}^{u} ) = U( S(u) )·d[t]d[t]
( 1/( 1+(-1)·(1/c)^{2}·(1/2)·m_{ij}·R_{ijv}^{v} ) )·...
m·( d[v]d[v]+(-1)·(1/2)·m_{ij}·R_{ijv}^{v} ) = U( S(v) )·d[t]d[t]
Ley:
( 1/( 1+(-1)·(1/c)^{2}·(1/2)·m_{ij}·R_{iju}^{u} ) )·...
(m/2)·c·( d[u]d[u]+(-1)·(1/2)·m_{ij}·R_{iju}^{u} ) = s·U( S(u) )·d[t]d[t]
( 1/( 1+(-1)·(1/c)^{2}·(1/2)·m_{ij}·R_{ijv}^{v} ) )·...
(m/2)·c·( d[v]d[v]+(-1)·(1/2)·m_{ij}·R_{ijv}^{v} ) = s·U( S(v) )·d[t]d[t]
Ley:
Sea f(u) = he^{iau} ==> ( S(u) )^{2} = h^{2}·(1/2)·e^{2iau}
Sea f(v) = he^{iav} ==> ( S(v) )^{2} = h^{2}·(1/2)·e^{2iav}
Ley:
(m/2)·d_{t}[u(t)]^{2} = (pq)·k·(1/r)^{3}·(1/4)·h^{2}·e^{2i·au}
u(t) = (-1)·(1/i)·(1/a)·ln( ( (1/m)·(-1)·(pq)·k·(1/r)^{3}·h^{2}·(1/2) )^{(1/2)}·at )
(m/2)·d_{t}[v(t)]^{2} = (-1)·(pq)·k·(1/r)^{3}·(1/4)·h^{2}·e^{2i·av}
v(t) = (-1)·(1/i)·(1/a)·ln( ( (1/m)·(pq)·k·(1/r)^{3}·h^{2}·(1/2) )^{(1/2)}·at )
Ley:
(m/4)·c·d_{t}[u(t)]^{2} = s·(pq)·k·(1/r)^{3}·(1/4)·h^{2}·e^{2i·au}
u(t) = (-1)·(1/i)·(1/a)·ln( ( (s/c)·(1/m)·(-1)·(pq)·k·(1/r)^{3}·h^{2} )^{(1/2)}·at )
(m/4)·c·d_{t}[v(t)]^{2} = (-1)·s·(pq)·k·(1/r)^{3}·(1/4)·h^{2}·e^{2i·av}
v(t) = (-1)·(1/i)·(1/a)·ln( ( (s/c)·(1/m)·(pq)·k·(1/r)^{3}·h^{2} )^{(1/2)}·at )
Principio:
H(u,v,t) = (-1)·(c/l)·V·t^{2}·qE_{g}·( he^{iau}+re^{iav} )
H(v,u,t) = (c/l)·wr·t^{2}·qE_{e}·( he^{iav}+re^{iau} )
Principio:
(c/l)·V·t·d[t] = 2·d[u]
(c/l)·wr·t·d[t] = 2·d[u]
Ley:
d[x]d[u] = d[ (c/l)·V·t^{2} ]d[u] = (c/l)·V·2t·d[t]d[u] = 4·d[u]d[u]
d[x]d[u] = d[ (c/l)·wr·t^{2} ]d[u] = (c/l)·wr·2t·d[t]d[u] = 4·d[u]d[u]
Ley:
U(u) = d_{2u}[ H(u,v,t) ] = (-1)·(c/l)·V·t^{2}·qE_{g}·(1/2)·iah·e^{iau}
U(v) = d_{2v}[ H(v,u,t) ] = (c/l)·wr·t^{2}·qE_{e}·(1/2)·iah·e^{iav}
Ley:
F(u) = d_{2u}[ U(u) ] = (c/l)·V·t^{2}·qE_{g}·(1/4)·a^{2}·h·e^{iau}
F(v) = d_{2v}[ U(v) ] = (-1)·(c/l)·wr·t^{2}·qE_{e}·(1/4)·a^{2}·h·e^{iav}
Ley:
(m/2)·d_{t}[u(t)]^{2} = (-1)·(c/l)·V·t^{2}·qE_{g}·(1/2)·iah·e^{i·au}
u(t) = (-1)·4·(1/i)·(1/a)·ln( ( (1/m)·(c/l)·V·qE_{g}·hia^{3}·(1/16) )^{(1/4)}·t )
(m/2)·d_{t}[v(t)]^{2} = (c/l)·wr·t^{2}·qE_{e}·(1/2)·iah·e^{i·av}
v(t) = (-1)·4·(1/i)·(1/a)·ln( ( (1/m)·(c/l)·wr·(-1)·qE_{e}·hia^{3}·(1/16) )^{(1/4)}·t )
Principio:
H(u,v,t,s) = (-1)·s·(c/l)·V·t^{2}·qE_{g}·( he^{iau}+re^{iav} )
H(v,u,t,s) = s·(c/l)·wr·t^{2}·qE_{e}·( he^{iav}+re^{iau} )
Principio:
(c/l)·V·t·d[t] = 2·d[u]
(c/l)·wr·t·d[t] = 2·d[u]
Ley:
d[x]d[u] = d[ (c/l)·V·t^{2} ]d[u] = (c/l)·V·2t·d[t]d[u] = 4·d[u]d[u]
d[x]d[u] = d[ (c/l)·wr·t^{2} ]d[u] = (c/l)·wr·2t·d[t]d[u] = 4·d[u]d[u]
Ley:
U(u,s) = d_{2u}[ H(u,v,t,s) ] = (-1)·s·(c/l)·V·t^{2}·qE_{g}·(1/2)·iah·e^{iau}
U(v,s) = d_{2v}[ H(v,u,t,s) ] = s·(c/l)·wr·t^{2}·qE_{e}·(1/2)·iah·e^{iav}
Ley:
F(u,s) = d_{2u}[ U(u,s) ] = s·(c/l)·V·t^{2}·qE_{g}·(1/4)·a^{2}·h·e^{iau}
F(v,s) = d_{2v}[ U(v,s) ] = (-1)·s·(c/l)·wr·t^{2}·qE_{e}·(1/4)·a^{2}·h·e^{iav}
Ley:
(m/4)·c·d_{t}[u(t)]^{2} = (-1)·s·(c/l)·V·t^{2}·qE_{g}·(1/2)·iah·e^{i·au}
u(t) = (-1)·4·(1/i)·(1/a)·ln( ( (s/c)·(1/m)·(c/l)·V·qE_{g}·hia^{3}·(1/8) )^{(1/4)}·t )
(m/4)·c·d_{t}[v(t)]^{2} = s·(c/l)·wr·t^{2}·qE_{e}·(1/2)·iah·e^{i·av}
v(t) = (-1)·4·(1/i)·(1/a)·ln( ( (s/c)·(1/m)·(c/l)·wr·(-1)·qE_{e}·hia^{3}·(1/8) )^{(1/4)}·t )
Principio:
H(u,v,t) = (-1)·(c/l)^{2}·V·t^{3}·qE_{g}·( he^{(1/2)·iau}+re^{(1/2)·iav} )
H(v,u,t) = (c/l)^{2}·wr·t^{3}·qE_{e}·( he^{(1/2)·iav}+re^{(1/2)·iau} )
Principio:
(c/l)^{2}·V·t^{2}·d[t] = (25/12)·d[u]
(c/l)^{2}·wr·t^{2}·d[t] = (25/12)·d[u]
Ley:
d[x]d[u] = d[ (c/l)^{2}·V·t^{3} ]d[u] = (c/l)^{2}·V·3t^{2}·d[t]d[u] = (25/4)·d[u]d[u]
d[x]d[u] = d[ (c/l)^{2}·wr·t^{3} ]d[u] = (c/l)^{2}·wr·3t^{2}·d[t]d[u] = (25/4)·d[u]d[u]
Ley:
U(u) = d_{(5/2)·u}[ H(u,v,t) ] = (-1)·(c/l)·V·t^{2}·qE_{g}·(1/5)·iah·e^{(1/2)·iau}
U(v) = d_{(5/2)·v}[ H(v,u,t) ] = (c/l)·wr·t^{2}·qE_{e}·(1/5)·iah·e^{(1/2)·iav}
Ley:
F(u) = d_{(5/2)·u}[ U(u) ] = (c/l)·V·t^{2}·qE_{g}·(1/25)·a^{2}·h·e^{(1/2)·iau}
F(v) = d_{(5/2)·v}[ U(v) ] = (-1)·(c/l)·wr·t^{2}·qE_{e}·(1/25)·a^{2}·h·e^{(1/2)·iav}
Ley:
(m/2)·d_{t}[u(t)]^{2} = (-1)·(c/l)^{2}·V·t^{3}·qE_{g}·(1/5)·iah·e^{(1/2)·i·au}
u(t) = (-1)·10·(1/i)·(1/a)·ln( ( (1/m)·(c/l)^{2}·V·qE_{g}·hia^{3}·(1/250) )^{(1/5)}·t )
(m/2)·d_{t}[v(t)]^{2} = (c/l)^{2}·wr·t^{3}·qE_{e}·(1/5)·iah·e^{(1/2)·i·av}
v(t) = (-1)·10·(1/i)·(1/a)·ln( ( (1/m)·(c/l)^{2}·wr·(-1)·qE_{e}·hia^{3}·(1/250) )^{(1/5)}·t )
Ley:
(m/4)·c·d_{t}[u(t)]^{2} = (-1)·s·(c/l)^{2}·V·t^{3}·qE_{g}·(1/5)·iah·e^{(1/2)·i·au}
u(t) = (-1)·10·(1/i)·(1/a)·ln( ( (s/c)·(1/m)·(c/l)^{2}·V·qE_{g}·hia^{3}·(1/125) )^{(1/5)}·t )
(m/4)·c·d_{t}[v(t)]^{2} = s·(c/l)^{2}·wr·t^{3}·qE_{e}·(1/5)·iah·e^{(1/2)·i·av}
v(t) = (-1)·10·(1/i)·(1/a)·ln( ( (s/c)·(1/m)·(c/l)^{2}·wr·(-1)·qE_{e}·hia^{3}·(1/125) )^{(1/5)}·t )
Experimento:
Con el triángulo funcionó con la electricidad,
porque solgó una cuerda eléctrica.
Con el hexágono no funcionó con la electricidad,
porque debe ser una cuerda gravitatoria,
Ley:
Debe ser cuerda eléctrica con polígono regular impar = 2n+1
Debe ser cuerda gravitatoria con polígono regular par = 2n+2
Ley: [ del DNA de la gente que es ]
( CH=CH-CH=CH-C=C )-O-( C=C-C=C-CH=CH )-( CH=C=CH )
( N=N-N=N-C=C )-O-( C=C-C=C-N=N )-( N=C=N )
Hexágono-Pentágono-Hexágono-Pentágono = 2 puertas de alma
[2·[A]·H_{2}]·[2·[B]·H_{2}]·[O_{2}]·[5e] = [16e]·[2·[A]-O-[B]]·[2·H{2}]
Ley: [ del DNA de la gente que no es ]
( CH=CH-CH=CH-C=C )-O-O-( C=C-C=C-CH=CH )-( CH=C=C=CH )
( N=N-N=N-C=C )-O-O-( C=C-C=C-N=N )-( N=C=C=N )
Hexágono-Hexágono-Hexágono-Hexágono = 0 puertas de alma
[2·[A]·H_{2}]·[2·[B]·H_{2}]·[2·O_{2}]·[5e] = [16e]·[2·[A]-O-O-[B]]·[2·H{2}]
Ley: [ de plasma celular ]
C=C=C=C=C=C=CH-CH <==> 1 Constructor
C=C=C=C=CH-CH_{2}-CH_{2}-CH <==> 1 Destructor
Ley: [ de infección de plasma celular ]
[Constructor]·[2·H_{2}] = [2e]·[Destructor]
Ley:
La fiebre química de calor,
es electricidad positiva en el plasma,
para la des-infección de las células.
La fiebre química de hielor,
es electricidad negativa en el plasma,
de la infección de las células.
Anexo:
La fiebre química es una entalpía,
que impide la reproducción de los virus y no es una guerra.
Anexo:
El paracetamol y el ibuprofeno son anticuerpos,
porque bajan la fiebre.
Ley: [ de retículos endoplasmáticos eléctricos ]
De Núcleo:
De protones:
TACCCCCATACCCCCCATACCCCCAT
SCAAAAACSCAAAAAACSCAAAAACS
De Plasma:
De electrones:
TACCCCCATACCCCATACCCCCAT
SCAAAAACSCAAAACSCAAAAACS
Ley:
En Núcleo tiene,
un campo eléctrico positivo
El Plasma tiene,
un campo eléctrico negativo.
Ley: [ del aparato reproductor del Golgi ]
De Núcleo:
TACCCCCAT
SCAAAAACS
Clásico:
Lluir [o] Lucir [o] Lutzire-dom
f(l,u,0) = g(0,u,c)
Creuar [o] Cruzar [o] Crutzare-dom
f(e,u,0) = g(0,u,z)
Llum [o] Luz [o] Lum
Crum [o] Cruz [o] Crum
Ley: [ de Trans-Warp potencia 2 ]
(c/l)^{2}·V·t^{3} = n·cT
V = (0.01) ( Metro / Segundo ) & T = ( 31,536,000 ) Segundo = 1 año
t = ( n·l^{2}·(1/V)·(T/c) )^{(1/3)}
n·l^{2}·(1/V)·(T/c) = ( n·10.512 ) ( Segundo )^{3}
Anexo:
Debe haber 4 híper-espacios y 4 potencias de velocidad,
porque las distancias no son tan grandes.
La velocidad máxima de la luz debe ser = c^{5}
Ley:
int[ pq·k·(1/r) ][ 2pi·r·au·(1+(-1)·au) ] = 2pi·(pq)·k·ln(r)·au·(1+(-1)·au)
int[ (-1)·pq·k·(1/r) ][ 2pi·r·av·(1+(-1)·av) ] = (-1)·2pi·(pq)·k·ln(r)·av·(1+(-1)·av)
Ley:
Si f(u) = h·au·(1+(-1)·au) ==> ( S(u) )^{2} = h^{2}·(1/24)·(1+(-2)·au)^{4}
Si f(u) = h·av·(1+(-1)·av) ==> ( S(v) )^{2} = h^{2}·(1/24)·(1+(-2)·av)^{4}
Ley:
(m/2)·d_{t}[u(t)]^{2} = (pq)·k·(1/r)^{3}·(1/48)·h^{2}·(2au)^{4·[1:(-1)]}
u(t) = ( ( (2/m)·(pq)·k·(1/r)^{3}·(1/48) )^{(1/2)}·(2a)^{2·[1:(-1)]}·ht )^{( (-1)/2·[1:(-1)]+(-1))}
(m/2)·d_{t}[v(t)]^{2} = (-1)·(pq)·k·(1/r)^{3}·(1/48)·h^{2}·(2av)^{4·[1:(-1)]}
v(t) = ( ( (2/m)·(pq)·k·(1/r)^{3}·(1/48) )^{(1/2)}·(2a)^{2·[1:(-1)]}·hit )^{( (-1)/2·[1:(-1)]+(-1))}
Examen de teoría de cuerdas:
Ley:
(m/2)·d_{t}[u(t)]^{2} = ...
... (-1)·mc^{2}·( 1/( 1+(-1)·( (d_{t}[w]·r)/c )^{2} )^{(1/2)} )·(1/2)·iah·e^{(1/2)·iau}
u(t) = ?
(m/2)·d_{t}[v(t)]^{2} = ...
... mc^{2}·( 1/( 1+(-1)·( (d_{t}[w]·r)/c )^{2} )^{(1/2)} )·(1/2)·iah·e^{(1/2)·iav}
v(t) = ?
Ley:
(m/2)·d_{t}[u(t)]^{2} = ...
... (-1)·mc^{2}·( 1/( 1+(-1)·( (d_{tt}^{2}[w]·rt)/c )^{2} )^{(1/2)} )·...
... (1/2)·iah·e^{(1/2)·( (-1)·(1/4)·[2:(-1)]+1 )·iau}
u(t) = (-1)·4·(1/i)·(1/a)·( 1/( (-1)·(1/4)·[2:(-1)]+1 ) )·...
... ln( (-1)·c·( (d_{tt}^{2}[w]·r)/c )^{(-1)·(1/4)·[2:(-1)]}·( iha·(1/16) )^{(1/2)}·at^{(-1)·(1/4)·[2]+1} )
Ley: [ de orientación vectorial en profundidad ]
Si ( Fr = qg·(d/2)·sin(s) & d[F]·r = mu^{2}·(1/d)·cos(s)·x^{2}·d[x] ) ==> ...
... tan(s) = (2/3)·(m/q)·(1/g)·du^{2}
Ley: [ de orientación vectorial en profundidad ]
Si ( p = Mv & d[p]·r = mu·(1/d)·x^{2}·d[x] ) ==> u = ( 3·(M/m)·vr·(1/d)^{2} )
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