AN LISIS CUANTITATIVO

ANÁLISIS CUANTITATIVO.
Ecuaciones a emplear: X=X0tm
m = b =
a) Promedio de tiempo:
DATOS:
X(m)
0.1
0.2
0.3
0.4
0.5
0.60.7
0.8
0.9
1.0
T1(s)
0.22
0.37
0.62
0.78
0.84
0.82
1.00
1.09
0.10
0..97
T2(s)
0.23
0.35
0.69
0.75
0.81
0.75
0.94
1.00
1.00
1.03
T3(s)
0.22
0.31
0.57
0.72
0.82
0.84
0.97
0.90
1.03
1.15
Prom.(t)
0.220.34
0.63
0.75
0.82
0.80
0.97
0.99
1.04
1.05
(t)2
0.0.5
0.116
0.397
0.563
0.67
0.64
0.94
0.98
1.08
1.10

b) Aceleración usando la formula a=(2x/t2):

a(m/s2)
4.13
3.46
1.51
1.42
1.49
1.86
1.49
1.631.66
0.91

Aceleración promedio= 2.79 m/s2

TABLA DE DATOS N°1
∑x
∑y
∑xy

logt
logX
logt . logX

Log0.22
Log0.1
(-0.66)(-1)
0.44
Log0.34
Log0.2
(-0.47)(-0.69)
0.22
Log0.63
Log0.3
(-0.20)(-0.52)
0.04Log0.75
Log0.4
(-0.13)(-0.39)
0.02
Log0.82
Log0.5
(-0.09)(-0.30)
0.01
Log0.80
Log0.6
(-0.097)(-0.22)
0.01
Log0.97
Log0.7
(-0.013)(-0.15)
0.00
Log0.99
Log0.8
(0.004)(-0.09)
0.00
Log1.04
Log0.9(0.02)(-0.04)
0.00
Log1.05
Log1.0
(0.021)(0.00)
0.00

-1.62

-3.40

1.18

0.74

m =
m = = = = 1.32
m= 1.32
b =
b= = = -0.126
b= -0.126
b = logX0 → = X0 → 10-0.126= X0 →X0 = 0.748
X0= 0.748

•LA ECUACIÓN EXPERIMENTAL DE X=X(t) SEGÚN LA TABLA DE DATOS N° 1
X = X0tm
X= (0,748)t1.32

TABLA DE DATOS N° 2
∑x
∑y
∑xy

Log(t2)
logX
Log(t2) . logX
(Log(t2))2
Log(0.22)2Log0.1
(-1.32)(-1)
1.74
Log(0.34)2
Log0.2
(-0.94)(-0.69)
0.88
Log(0.63)2
Log0.3
(-0.4)(-0.52)
0.16
Log(0.75)2
Log0.4
(-0.26)(-0.39)
0.07
Log(0.82)2
Log0.5
(-0.18)(-0.30)
0.03
Log(0.80)2
Log0.6(-0.19)(-0.22)
0.04
Log(0.97)2
Log0.7
(-0.03)(-0.15)
0.00
Log(0.99)2
Log0.8
(0.008)(-0.09)
0.00
Log(1.04)2
Log0.9
(0.04)(-0.04)
0.00
Log(1.05)2
Log1.0
(0.04)(0.00)
0.00

-3.23

-3.40

2.38

2.92

m =
m = == = 0.68
m= 0,68
b =
b = = = = -0.323
b= -0.32
b = logX0 → = Xo → = X0 → X0 = 0.4786
X0= 0.48
•LA ECUACION EXPERIMENTAL DE X=X(t2) SEGÚN LA TABLA DE DATOS N° 2
X = X0tm...