Opracownie wyników do ćwiczenia:

i

1 5

..

:=

j

1 11

..

:=

s

sec

:=

l

1265 mm

⋅

:=

c

226 mm

⋅

:=

dm

10

1

−

m

⋅

:=

Czasy przebiegu fal dla poszczególnych ilości cieczy:

t1

i

8.24 s

⋅

8.67 s

⋅

8.36 s

⋅

8.31 s

⋅

8.17 s

⋅

:=

t2

i

7.15 s

⋅

7.14 s

⋅

7.08 s

⋅

7.14 s

⋅

7.16 s

⋅

:=

t3

i

6.19 s

⋅

6.29 s

⋅

6.21 s

⋅

6.25 s

⋅

6.31 s

⋅

:=

t4

i

5.64 s

⋅

5.86 s

⋅

5.72 s

⋅

5.76 s

⋅

5.74 s

⋅

:=

t5

i

5.3 s

⋅

5.32 s

⋅

5.29 s

⋅

5.28 s

⋅

5.3 s

⋅

:=

t6

i

7.36 s

⋅

7.37 s

⋅

7.46 s

⋅

7.24 s

⋅

7.39 s

⋅

:=

t7

i

7 s

⋅

6.91 s

⋅

6.96 s

⋅

6.96 s

⋅

7.01 s

⋅

:=

t8

i

6.43 s

⋅

6.45 s

⋅

6.41 s

⋅

6.51 s

⋅

6.46 s

⋅

:=

t9

i

6.32 s

⋅

6.33 s

⋅

6.36 s

⋅

6.34 s

⋅

6.3 s

⋅

:=

t10

i

6.18 s

⋅

6.24 s

⋅

6.11 s

⋅

6.22 s

⋅

6.11 s

⋅

:=

t11

i

5.97 s

⋅

5.95 s

⋅

5.93 s

⋅

5.91 s

⋅

5.9 s

⋅

:=

Obliczenie wysokości poziomu cieczy z objętości cieczy i powierzchni zbiornika:

S

l c

⋅

:=

V

j

3 dm

3

⋅

4 dm

3

⋅

5 dm

3

⋅

6 dm

3

⋅

7 dm

3

⋅

8 dm

3

⋅

9 dm

3

⋅

10 dm

3

⋅

11 dm

3

⋅

12 dm

3

⋅

13 dm

3

⋅

:=

h

j

0.0105 m

⋅

0.014 m

⋅

0.0175 m

⋅

0.021 m

⋅

0.0245 m

⋅

0.028 m

⋅

0.0315 m

⋅

0.035 m

⋅

0.0385 m

⋅

0.042 m

⋅

0.0455 m

⋅

:=

S

0.286 m

2

=

Długości drogi poszczególnych pomiarów czasu.

l

j

2 l

⋅

2 l

⋅

2 l

⋅

2 l

⋅

2 l

⋅

3 l

⋅

3 l

⋅

3 l

⋅

3 l

⋅

3 l

⋅

3 l

⋅

:=

Wartości średnie czasów przebiegów fal:

tsr

1

1

5

i

t1

i

∑

=

5

:=

tsr

2

1

5

i

t2

i

∑

=

5

:=

tsr

3

1

5

i

t3

i

∑

=

5

:=

tsr

4

1

5

i

t4

i

∑

=

5

:=

tsr

5

1

5

i

t5

i

∑

=

5

:=

tsr

6

1

5

i

t6

i

∑

=

5

:=

tsr

7

1

5

i

t7

i

∑

=

5

:=

tsr

8

1

5

i

t8

i

∑

=

5

:=

tsr

9

1

5

i

t9

i

∑

=

5

:=

tsr

10

1

5

i

t10

i

∑

=

5

:=

tsr

11

1

5

i

t11

i

∑

=

5

:=

Wartości średnie czasów przebiegów fal:

Prędkości przebiegów poszczególnych fal:

tsr

j

8.35

7.134

6.25

5.744

5.298

7.364

6.968

6.452

6.33

6.172

5.932

s

=

v

j

l

j

tsr

j

:=

v

j

0.303

0.355

0.405

0.44

0.478

0.515

0.545

0.588

0.6

0.615

0.64

m s

1

−

⋅

=

Wykres do ćwiczenia:

lnh

j

ln h

j

m

1

−

⋅









:=

lnv

j

ln v

j

sec

⋅

m

1

−

⋅









:=

5

−

4.5

−

4

−

3.5

−

3

−

1.2

−

1

−

0.8

−

0.6

−

0.4

−

lnv

j

lnh

j

j

Opracowanie omyłek popełnionych w ćwiczeniu:

Dokładność stopera:

Niepewność systematyczna związana
z włączaniem i wyłączaniem stopera:

Dt

0.01 s

⋅

:=

Dlt

0.3 s

⋅

:=

f

1.1414

:=

Standardowe odchylenia czasów przebiegów fal od wartości średniej:

St

1

f

1

5

i

t1

i

tsr

1

−

(

)

2

1

5 5

1

−

(

)

⋅

⋅









∑

=

⋅

:=

St

1

0.098 s

=

St

2

f

1

5

i

t2

i

tsr

2

−

(

)

2

1

5 5

1

−

(

)

⋅

⋅









∑

=

⋅

:=

St

2

0.016 s

=

St

3

f

1

5

i

t3

i

tsr

3

−

(

)

2

1

5 5

1

−

(

)

⋅

⋅









∑

=

⋅

:=

St

3

0.026 s

=

St

4

f

1

5

i

t4

i

tsr

4

−

(

)

2

1

5 5

1

−

(

)

⋅

⋅









∑

=

⋅

:=

St

4

0.04 s

=

St

5

f

1

5

i

t5

i

tsr

5

−

(

)

2

1

5 5

1

−

(

)

⋅

⋅









∑

=

⋅

:=

St

5

7.571

10

3

−

×

s

=

St

6

f

1

5

i

t6

i

tsr

6

−

(

)

2

1

5 5

1

−

(

)

⋅

⋅









∑

=

⋅

:=

St

6

0.041 s

=

St

7

f

1

5

i

t7

i

tsr

7

−

(

)

2

1

5 5

1

−

(

)

⋅

⋅









∑

=

⋅

:=

St

7

0.02 s

=

St

8

f

1

5

i

t8

i

tsr

8

−

(

)

2

1

5 5

1

−

(

)

⋅

⋅









∑

=

⋅

:=

St

8

0.019 s

=

St

9

f

1

5

i

t9

i

tsr

9

−

(

)

2

1

5 5

1

−

(

)

⋅

⋅









∑

=

⋅

:=

St

9

0.011 s

=

St

10

f

1

5

i

t10

i

tsr

10

−

(

)

2

1

5 5

1

−

(

)

⋅

⋅









∑

=

⋅

:=

St

10

0.031 s

=

St

11

f

1

5

i

t11

i

tsr

11

−

(

)

2

1

5 5

1

−

(

)

⋅

⋅









∑

=

⋅

:=

St

11

0.015 s

=

Odchylenia standardowe całkowite dla poszczególnych czasów przebiegów fal:

Stcal

j

St

j

( )

2

Dt

2

Dlt

2

+

3

+

:=

Wartości do wykresu

Stcal

j

0.248

0.229

0.23

0.232

0.228

0.232

0.229

0.229

0.228

0.23

0.229

s

=

lnh

j

-4.556

-4.269

-4.046

-3.863

-3.709

-3.576

-3.458

-3.352

-3.257

-3.17

-3.09

=

lnv

j

-1.194

-1.037

-0.904

-0.82

-0.739

-0.663

-0.608

-0.531

-0.512

-0.486

-0.447

=

a

5

1

5

j

v

j

h

j

⋅

( )

∑

=

⋅

1

5

j

v

j

∑

=

1

5

j

h

j

∑

=

⋅

−

5

1

5

j

v

j

( )

2

∑

=

⋅

1

5

j

v

j

∑

=

















2

−

:=

b

1

5

1

5

j

h

j

∑

=

a

1

5

j

v

j

∑

=

⋅

−

















⋅

:=

b

0.014

−

m

=

a

0.08 s

=

Niepewności

ua

5

5

2

−

1

5

j

h

j

( )

2

∑

=

a

1

5

j

v

j

h

j

⋅

( )

∑

=

⋅

−

b

1

5

j

h

j

∑

=

⋅

−

5

1

5

j

v

j

( )

2

∑

=

⋅

1

5

j

v

j

∑

=

















2

−

⋅

4.089

10

3

−

×

s

=

:=

ub

ua

1

5

1

5

j

v

j

( )

2

∑

=

⋅

⋅

1.639

10

3

−

×

m

=

:=