Download docx - PRUEBAs laboratorio corioles

ANÁLISIS E INTERPRETACIÓN DE LOS RESULTADOS

PRUEBA 01

0.00000 0.00500 0.01000 0.01500 0.02000 0.02500 0.03000 0.035000.000000.001000.002000.003000.004000.005000.006000.007000.008000.009000.01000

f(x) = 3.29363477310189 x² + 0.164478597533447 x − 0.000188667239896819R² = 0.998121880528957

PRUEBA 01

SABEMOS QUE PARA DETERMINAR EL COEFICIENTE DE CORIOLIS DEBEMOS SEGUIR LA SIGUIENTE FORMULA

α=∫V H3 dAV 3∗A

α=∫ (f ( x ) )3∗(f ( y ) )∗(dy)

V3xA

α=∫ (f ( y ) )3∗( f ( y ) )∗(dy)

V3xA

Tenemos que expresar la ecuación que describe la parábola en función de la variable “y”.

y=3.2936 x2+0.1645 x−0.0002

y3.2936

=x2+ 0.16453.2936

x−0.00023.2936

y3.2936

=(x+( 12∗0.16453.2936 ))2

−( 12∗0.16453.2936 )2

−0.00023.2936

x=√ y3.2936

+( 12∗0.16453.2936 )2

+ 0.00023.2936

−(

12∗0.1645

3.2936)

Reemplazando los valores en la formula

α=∫ (f ( y ) )3∗( f ( y ) )∗(dy)

V3∗A

α=∫0

0.009

(√ y3.2936

+( 12∗0.16453.2936 )2

+ 0.00023.2936

−(

12∗0.1645

3.2936))

4

0.0173 x0.0003

α=1.11

PRUEBA 02

0.00000 0.00500 0.01000 0.01500 0.02000 0.02500 0.030000.00000

0.00100

0.00200

0.00300

0.00400

0.00500

0.00600

0.00700

0.00800

0.00900

0.01000

f(x) = 5.20465778060914 x² + 0.206760590608011 x − 0.000188667239896814R² = 0.998121880528957

PRUEBA 02


y = 5.2047x2 + 0.2068x - 0.0002

y5.2047

=x2+ 0.20685.2047

x−0.00025.2047

y5.2047

=(x+( 12∗0.20685.2047 ))2

−( 12∗0.20685.2047 )2

−0.00025.2047

x=√ y5.2047

+( 12∗0.20685.2047 )2

+ 0.00025.2047

−(

12∗0.2068

5.2047)

Reemplazando los valores en la fórmula:

α=∫ (f ( y ) )3∗( f ( y ) )∗(dy)

V3∗A

α=∫0

0.009

(√ y5.2047

+( 12∗0.20685.2047 )2

+ 0.00025.2047

−(

12∗0.2068

5.2047))

4

(dy)

0.0133 x 0.0003

α=1.78

PRUEBA 03

0.00000 0.00500 0.01000 0.01500 0.02000 0.02500 0.03000 0.035000.00000

0.00100

0.00200

0.00300

0.00400

0.00500

0.00600

0.00700

0.00800

0.00900

0.01000

f(x) = 3.67053169170222 x² + 0.173634557144927 x − 0.000188667239896807R² = 0.998121880528957

PRUEBA 03


y = 3.6705x2 + 0.1736x - 0.0002

y3.6705

=x2+ 0.17363.6705

x−0.00023.6705

y3.6705

=(x+( 12∗0.17363.6705 ))2

−( 12∗0.17363.6705 )2

−0.00023.6705

x=√ y3.6705

+( 12∗0.17363.6705 )2

+ 0.00023.6705

−(

12∗0.1736

3.6705)


α=∫ (f ( y ) )3∗( f ( y ) )∗(dy)

V3∗A

α=∫0

0.009

(√ y3.6705

+( 12∗0.17363.6705 )2

+ 0.00023.6705

−(

12∗0.1736

3.6705))

4

(dy)

0.0163 x 0.0003

α=1.53

PRUEBA 05

0.00000 0.00500 0.01000 0.01500 0.02000 0.02500 0.03000 0.035000.00000

0.00100

0.00200

0.00300

0.00400

0.00500

0.00600

0.00700

0.00800

0.00900

0.01000

f(x) = 4.1316894321091 x² + 0.184219461857237 x − 0.000188667239896812R² = 0.998121880528957

PRUEBA 05


y = 4.1317x2 + 0.1842x - 0.0002

y4.1317

=x2+ 0.18424.1317

x−0.00024.1317

y4.1317

=(x+( 12∗0.18424.1317 ))2

−( 12∗0.18424.1317 )2

−0.00024.1317

x=√ y4.1317

+( 12∗0.18424.1317 )2

+ 0.00024.1317

−(

12∗0.1842

4.1317)


α=∫ (f ( y ) )3∗( f ( y ) )∗(dy)

V3∗A

α=∫0

0.009

(√ y4.1317

+( 12∗0.18424.1317 )2

+ 0.00024.1317

−(

12∗0.1842

4.1317))

4

(dy )

0.0153 x 0.0003

α=1.84

PRUEBA 06

0.00000 0.00500 0.01000 0.01500 0.02000 0.025000.00000

0.00100

0.00200

0.00300

0.00400

0.00500

0.00600

0.00700

0.00800

0.00900

0.01000

f(x) = 6.68698341714272 x² + 0.234361785335023 x − 0.00018866723989682R² = 0.998121880528957

PRUEBA 06


y = 6.687x2 + 0.2344x - 0.0002

y6.687

=x2+ 0.23446.687

x−0.00026.687

y6.687

=( x+( 12∗0.23446.687 ))2

−( 12∗0.23446.687 )2

−0.00026.687

x=√ y6.687

+( 12∗0.23446.687 )2

+ 0.00026.687

−(

12∗0.2344

16.687)


α=∫ (f ( y ) )3∗( f ( y ) )∗(dy)

V3∗A

α=∫0

0.009

(√ y6.687

+( 12∗0.23446.687 )2

+ 0.00026.687

−(

12∗0.2344

6.687))

4

(dy )

0.0123 x 0.0003

α=¿1.37

PRUEBA 07

0.00000 0.00500 0.01000 0.01500 0.02000 0.02500 0.03000 0.035000.00000

0.00100

0.00200

0.00300

0.00400

0.00500

0.00600

0.00700

0.00800

0.00900

0.01000

f(x) = 3.62652234435141 x² + 0.172590486708841 x − 0.000188667239896807R² = 0.998121880528957

PRUEBA 07


y = 3.6265x2 + 0.1726x - 0.0002

y3.6265

=x2+ 0.17263.6265

x−0.00023.6265

y3.6265

=(x+( 12∗0.17263.6265 ))2

−( 12∗0.17263.6265 )2

−0.00023.6265

x=√ y3.6265

+( 12∗0.17263.6265 )2

+ 0.00023.6265

−(

12∗0.1726

3.6265)


α=∫ (f ( y ) )3∗( f ( y ) )∗(dy)

V3∗A

α=∫0

0.009

(√ y3.6265

+( 12∗0.17263.6265 )2

+ 0.00023.6265

−(

12∗0.1726

3.6265))

4

(dy)

0.0163 x 0.0003

α=1.97

PRUEBA 08

0.00000 0.00500 0.01000 0.01500 0.02000 0.02500 0.030000.00000

0.00100

0.00200

0.00300

0.00400

0.00500

0.00600

0.00700

0.00800

0.00900

0.01000

f(x) = 5.43349242916279 x² + 0.211257048432405 x − 0.000188667239896837R² = 0.998121880528957

PRUEBA 08

y = 5.4335x2 + 0.2113x - 0.0002

y5.4335

=x2+ 0.21135.4335

x−0.00025.4335

y5.4335

=(x+( 12∗0.21135.4335 ))2

−( 12∗0.21135.4335 )2

−0.00025.4335

x=√ y5.4335

+( 12∗0.21135.4335 )2

+ 0.00025.4335

−(

12∗0.2113

5.4335)


α=∫ (f ( y ) )3∗( f ( y ) )∗(dy)

V3∗A

α=∫0

0.009

(√ y5.4335

+( 12∗0.21135.4335 )2

+ 0.00025.4335

−(

12∗0.2113

5.4335))

4

(dy )

0.0133 x 0.0003

α=1.63

PRUEBA 09

0.00000 0.00500 0.01000 0.01500 0.02000 0.025000.00000

0.00100

0.00200

0.00300

0.00400

0.00500

0.00600

0.00700

0.00800

0.00900

0.01000

f(x) = 8.96127023995809 x² + 0.271304165488993 x − 0.000188667239896821R² = 0.998121880528957

PRUEBA 09

y = 8.9613x2 + 0.2713x - 0.0002

y8.9613

=x2+ 0.27138.9613

x−0.00028.9613

y8.9613

=(x+( 12∗0.27138.9613 ))2

−( 12∗0.27138.9613 )2

−0.00028.9613

x=√ y8.9613

+( 12∗0.27138.9613 )2

+ 0.00028.9613

−(

12∗0.2713

8.9613)


α=∫ (f ( y ) )3∗( f ( y ) )∗(dy)

V3∗A

α=∫0

0.009

(√ y8.9613

+( 12∗0.27138.9613 )2

+ 0.00028.9613

−(

12∗0.2713

8.9613))

4

(dy)

0.0103 x 0.0003

α=1.32

PRUEBA 10

0.00000 0.00500 0.01000 0.01500 0.02000 0.02500 0.030000.00000

0.00100

0.00200

0.00300

0.00400

0.00500

0.00600

0.00700

0.00800

0.00900

0.01000

f(x) = 5.78127062284989 x² + 0.217913092641587 x − 0.00018866723989682R² = 0.998121880528957

PRUEBA 10

y = 5.7813x2 + 0.2179x - 0.0002

y5.7813

=x2+ 0.21795.7813

x−0.00025.7813

y5.7813

=(x+( 12∗0.21795.7813 ))2

−( 12∗0.21795.7813 )2

−0.00025.7813

x=√ y5.7813

+( 12∗0.21795.7813 )2

+ 0.00025.7813

−(

12∗0.2179

5.7813)


α=∫ (f ( y ) )3∗( f ( y ) )∗(dy)

V3∗A

α=∫0

0.009

(√ y5.7813

+( 12∗0.21795.7813 )2

+ 0.00025.7813

−(

12∗0.2179

5.7813))

4

(dy )

0.0133 x0.0003

α=1.44