Diametro

Hallar las formulas experimentales:

Obtenga las formulas experimentales usando el método de regresión lineal para las graficas obtenidas en los casos a),d),e) y f).
Caso d) t vs. D

m = p*( log (d) * log (t)) - ( log (d) * log (t))

p ((log (d))2-((log(d))2

b= ((log (d))2* log (t) -  log (d)* ( log(d) *log(t))

p ((log (d))2-((log (d))2


h=20 cm
Diámetro
(cm) Tiempo (s) D=log(d) T=log(t) D*T= log(d) * log(t) D2=(log(d))2
1.5 59.5 log⁡(1.5)=0.176 log⁡(59.5)=1.775 log⁡〖(1.5)*〗 log⁡(59.5)=0.3124 0.031
2.0 33.7 log⁡(2.0)=0.301 log⁡(33.7)=1.528 log⁡(2.0)*log⁡(33.7)=0.459 0.091
3.0 14.9 log⁡(3.0)=0.477 log⁡(14.9)=1.173log⁡(3.0)*log⁡(14.9)=0.559 0.228
5.0 5.3 log⁡(5.0)=0.698 log⁡(5.3)=0.724 log⁡(5.0)* log⁡(5.3)=0.505 0.487
7.0 2.7 log⁡(7.0)=0.845 log⁡〖(2.7)〗=0.431 log⁡(7.0)* log⁡〖(2.7)〗=0.364 0.714


m= 5*(2.1994)-(2.497*5.631)
= -2.020
5*1.5502-2.4972

b= 1.5502*5.631-2.497*2.1994 =2.135 k=10b=102.135=136.45

5*1.5502-2.4972

Ecuación de la función potencial Y=kxmy =136.45x-2.020

Ecuación de la función lineal y=-2.020x+2.135






h=30 cm

Diámetro
(cm) Tiempo (s) D=log(d) T=log(t) D*T= log(d) * log(t) D2=(log(d))2
1.5 73.0 log⁡(1.5)=0.176 log⁡(73.0)=1.863 log⁡〖(1.5)*〗 log⁡(73.0)=0.328 0.031
2.0 41.2 log⁡(2.0)=0.301 log⁡(41.2)=1.615 log⁡(2.0)*log⁡(41.2)=0.486 0.091
3.0 18.4log⁡(3.0)=0.477 log⁡(18.4)=1.265 log⁡(3.0)*log⁡(18.4)=0.603 0.228
5.0 6.8 log⁡(5.0)=0.698 log⁡(6.8)=0.833 log⁡(5.0)* log⁡(6.8)=0.581 0.487
7.0 3.2 log⁡(7.0)=0.845 log⁡〖(3.2)〗=0.505 log⁡(7.0)* log⁡〖(3.2)〗=0.427 0.714


m= 5*(2.425)-( 2.497*6.081)
= -2.013
5*1.551-2.4972

b= 1.5502*6.081-2.497*2.425 =2.224 k=10b=102.224=167.49

5*1.5502-2.4972


Ecuación de lafunción potencial Y=kxm y =167.49x-2.013

Ecuación de la función lineal y= -2.013x+2.224





h=10 cm

Diámetro
(cm) Tiempo (s) D=log(d) T=log(t) D*T= log(d) * log(t) D2=(log(d))2
1.5 43.0 log⁡(1.5)=0.176 log⁡(43.0)=1.633 log⁡〖(1.5)*〗 log⁡(43.0)=0.287 0.031
2.0 23.7 log⁡(2.0)=0.301 log⁡(23.7)=1.374 log⁡(2.0)*log⁡(23.7)0.414 0.091
3.0 10.5log⁡(3.0)=0.477 log⁡(10.5)=1.021 log⁡(3.0)*log⁡(10.5)0.487 0.228
5.0 3.9 log⁡(5.0)=0.698 log⁡(3.9)=0.591 log⁡(5.0)*log⁡(3.9)=0.412 0.487
7.0 2.0 log⁡(7.0)=0.845 log⁡〖(2.0)〗=0.301 log⁡(7.0)* log⁡〖(2.0)〗=0.254 0.714
m= 5*(1.854)-( 2.497*4.92)
= -1.984
5*1.551-2.4972




b= 1.5502*4.92-2.497*1.854 =1.977 k=10b=101.977=94.84

5*1.5502-2.4972

Ecuación de la funciónpotencial Y=kxm y =94.84x-1.984

Ecuación de la función lineal y= = -1.984x+1.977





h=4 cm
Diámetro
(cm) Tiempo (s) D=log(d) T=log(t) D*T= log(d) * log(t) D2=(log(d))2
1.5 26.7 log⁡(1.5)=0.176 log⁡(26.7)=1.427 log⁡〖(1.5)*〗 log⁡(26.7)=0.251 0.031
2.0 15.0 log⁡(2.0)=0.301 log⁡(15.0)=1.176 log⁡(2.0)*log⁡(15.0)= 0.354 0.091
3.0 6.8 log⁡(3.0)=0.477log⁡(6.8)=0.833 log⁡(3.0)*log⁡(6.8)=0.398 0.228
5.0 2.6 log⁡(5.0)=0.698 log⁡(2.6)=0.415 log⁡(5.0)*log⁡(3.9)=0.289 0.478
7.0 1.3 log⁡(7.0)=0.845 log⁡〖(1.3)〗=0.114 log⁡(7.0)* log⁡〖(2.0)〗=0.096 0.714

m= 5*(1.388)-( 2.497*3.965)
= -1.948
5*1.551-2.4972

b= 1.5502*3.965-2.497*1.388 =1.764 k=10b=101.764=58.076

5*1.5502-2.4972

Ecuación de la funciónpotencial Y=kxm y =58.076 x-1.948

Ecuación de la función lineal y= = -1.948x+1.764







h=1 cm

Diámetro
(cm) Tiempo (s) D=log(d) T=log(t) D*T= log(d) * log(t) D2=(log(d))2
1.5 13.5 log⁡(1.5)=0.176 log⁡(13.5)=1.130 log⁡〖(1.5)*〗 log⁡(13.5)=0.198 0.031
2.0 7.8 log⁡(2.0)=0.301 log⁡(7.8)=0.892 log⁡(2.0)*log⁡(7.8)= 0.268 0.091
3.0...