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TARUGO ESFERA PLACA

Medid

a

dt(mm)

H

(mm)

mt

(g)

de(mm)

me(g)

a(mm)

hp(mm)

l(mm)

m p

(g)

01 15.06 101.65 15.6 15.07 5.4 50.0 0.15 54.045 0.7

0! 15.0! 101.0" 15.6 15.06 5.5 51.015 0.1 54.04 0."0 15.05 101.07 15.6 15.07 5. 50.005 0.1 54.05 0.7

04 15.06 101.06 15.7 15.065 5.4 50.0! 0.1 54.0 0."

05 15.04 101.05 15.7 15.06 5. 50.01 0.! 54.0 0."´ x 15.05 101.1"! 15.64 15.065 5." 50.016 0.1 54.06 0.76

Es= El 0.01 0.01 0.05 0.01 0.05 0.01 0.01 0.01 0.05

σ 0.01! 0.! 0.04 0.004 0.07 0.00" 0.04 0.005 0.04

Ea 0.006 0.4 0.07 0.006 0.105 0.1! 0.06 0.075 0.06

∆ X 0.006 0.4 0.0" 0.01 0.11 0.1! 0.06 0.075 0.07

Medid

a´ x ± ∆ x

(mm)

15.046± 0.0

!!

101.1"!± 0.

4

15.64± 0.

0"

15.065± 0.01

5."± 0.1

1

50.016± 0.1!

0.1± 0.

06

54.06 ±

0.01!5

0.76± 0

.07

#$%&me' (#t)

(c m3)

Maa

mt

(g)

#$%&me'

'

(m)

me(g)

#$%&me' (#*)

(c m3)Maa

m*

Medida

́z ± ∆ z

71.++"+ ± 0.6!!

+740"

15.64± 0.

0"

0.7+0!1"4 ± 0.00564 0.5146

± 0.160+160.76± 0

.07

Medid

a´ ρ ± ∆ ρ

( g

c m3)

0.!117 ± 1.0"!"

71.++"+ ± 0.6!!+

!.166 ± 0.4565

Solución:

Tarugo:

Diámetro:


2/12

ERROR ,EL -STRUMETO ( Ei )

Ei=1

2(0.02 )=0.01mm

PROME,-O ( ´ x)

´ x=15.06+15.02+15.05+15.06+15.04

5=15.05

,ES#-AC-O ESTA,AR ,E LA ME,-A ( σ )

σ =√ (15.05−15.06)2+(15.05−15.02)2+(15.05−15.05)2+(15.05−15.06)2+(15.05−15.04)2

5

σ =√ (−0.01)2+(0.03)2+(0)2+(−0.01)2+(0.01)2

5

σ =√0.0001+0.0009+0.0001+0.0001

5=√

0.0012

5=√ 0.00024=0.015491

σ =0.02

ERROR ALEATOR-O ( Ea )

Ea= 3σ

√ 5−1=

3(0.0)2

Ea=0.03

ERROR A/SOLUTO ( ∆ x )


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∆ x=√ (0.01)2+(0.03)2=0.031622766

∆ x=0.03

ME,-,A

´ x ± ∆ x=17.47±0.03

Altura:


Ei=1

2(0.02 )=0.01mm

PROME,-O (´ x)

´ x=101.65+101.08+101.07+101.06+101.05

5=101.182


σ =√(101.182−101.65)2+(101.182−101.08)2+(101.182−101.07)2+(101.182−101.06)2+(101.182−

5

σ =√ (−0.468)2+(0.102)2+(−0.112)2+(0.122)2+(0.112)2

5

σ = 0.219+0.01+0.0125+0.0148+0.0125

5=

0.2688

5=√ 0.05376=0.02318

σ =0.23



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Ea= 3σ

√ 5−1=

3(0.2318)2

=0.3477

Ea=0.34


∆ x=√ (0.01 )2

+(0.34 )2

=0.34

∆ x=0.34

ME,-,A

´ x ± ∆ x=101.182±0.34

Masa del tarugo:


Ei=1

2(0.1)=0.05g

PROME,-O (´ x)

´ x=15.6+15.6+15.6+15.7+15.7

5=15.64


σ =

√(15.64−15.6)2+(15.64−15.6)2+(15.64−15. 6)2+(15.64−15.7)2+(15.64−15.7)2

5

σ =√ (0.04)2+(0 .04)2+(0.04)2+(0 .06)2+(0.06)2

5


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σ =√ 0.0016+0.0016+0.0016+0 .0036+0.00365 =√ 0.125 =√ 0.0024=0.0489σ =0.04


Ea= 3σ

√ 5−1=

3(0.0489)2

=0.07335

Ea=0.07


∆ x=√ (0.05)2+(0.07)2=0.086

∆ x=0.08

ME,-,A

´ x ± ∆ x=15.64±0.08

VOLUMEN DEL TARUGO:

,imet$2 15.046 ± 030!! mm 1.5046 ± 0.00!! m

A%t&a2 101.1"! ± 0.4 mm 10.11"! ± 0.04 m

V t =(1 .5046±0.0022)2

× π

4×(10.1182±0.034)

V t =[(1 .5046)2× π 4 ±(1.5046)2× π 4 ( 0.00221.5046 )×2]×(10.1182±0.034 )

V t =(1.778±0.00519952 ) ×(10.1182±0.034)


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V t =(1.778)(10.1182)±(1.778)(10.1182)√(0.005199521.778 )2

+( 0.03410.1182)2

V t =(17.99±0.058 ) c m3

DENSDAD DEL TARUGO:

ρ=m

V = 15.64 ±0.08

17.99±0.058

ρ=0.869±0.869

√( 0.0815.64 )

2

+(0.05817.99 )2

=0.2117±0.006

ρ=0.2±0.006 g

c m3

!laca:

Longitud:


Ei=1

2(0.02 )=0.01mm

PROME,-O (´ x)

´ x=54.045+54.04+54.035+54.03+54.03

5=54.036



7/12

σ =√ (54.036−54.045)2+(54.036−54.04)2+(54.036−54.035)2+(54.036−54.03)2+(54.036−54.03

5

σ =

√(−0.009)2+(−0.004)2+(0.001)2+(0.006)2+(0.006)2

5

σ = 0. 000081+0.000016+0.000001+0.000036+0.000036

5=

0.00017

5=√ 0.000034

σ =0.00583

σ =0.005


Ea= 3σ

√ 5−1=

3(0.005)2

=0.0075

Ea=0.0075


∆ x=√ (0.01)2+(0.0075)2=0.0125

∆ x=0.0125

ME,-,A

´ x ± ∆ x=54.036±0.0125

Anc"o:


Ei=1

2(0.02 )=0.01mm


8/12

PROME,-O (´ x)

´ x=50.03+50.015+50.005+50.02+50.01

5=50.016


(50.016−50.015)¿

(50.016−50.005)¿

(50.016−50.02)¿

(50.016−50.01)

¿¿2¿¿¿¿

(50.016−50.03)2+¿¿

σ =√ ¿

σ =

√

(−0.014 )2+(0.001)2+(0 .011)2+(−0.004)2+(0.006)2

5

σ =√ 0.000196+0.000001+0.000121+0.000016+0.0000365 =√ 0.000375 =√ 0.000074σ =0.0086

σ =0.008


Ea= 3σ

√ 5−1=

3(0.008)2

=0.012

Ea=0.02


9/12


∆ x=√ (0.005)2+(0.012)2=0.12

∆ x=0.012

ME,-,A

´ x ± ∆ x=50.016±0.012

Altura:


Ei=1

2(0.02 )=0.01mm

PROME,-O (´ x)

´ x=0.15+0.1+0.1+0.1+0.2

5=0.13


σ =√ (0.13−0.15)2+(0.13−0.1)2+(0.13−0.1)2+(0.13−0.1)2+(0.13−0.1)2

5

σ =√ (−0 .02)2+(0.03)2+(0.03)2+(0.03)2+(0.07)2

5

σ =√ 0 .0004+0.0009+0.0009+0.0009+0.00495 =√ 0.0085 =√ 0.0016=0.04σ =0.04


10/12


Ea= 3σ

√ 5−1=

3(0.04)2

=0.06

Ea=0.06


∆ x=√ (0.01)2+(0.06)2=0.06

∆ x=0.06

ME,-,A

´ x ± ∆ x=0.13±0.06

Masa de la #laca:


Ei=1

2

(0.1)=0.05mm

PROME,-O (´ x)

´ x=0.7+0.8+0.8+0.7+0.8

5=0.76


σ =√ (0.76−0.7)2+(0.76−0.8)2+(0.76−0.8)2+(0.76−0.7)2+(0.76−0.8)2

5


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σ =√ (0.06)2+(0 .04)2+(0.04)2+(0 .06)2+(0.04)2

5

σ =

√0.0036+0.0016+0.0036+0 .0036+0.0016

5

=

√0.012

5

=√ 0.0024=0.04

σ =0.4


Ea= 3σ

√ 5−1=

3(0.4)2

=0.06

Ea=0.6


∆ x=√ (0.05)2+(0.06)2=0.07

∆ x=0.07

ME,-,A

´ x ± ∆ x=0.76±0.7

VOLUMEN DE LA !LA$A:

V p=(50.016±0.12 )× (0.13±0.06 ) ×(54.036±0.0125)

V p=[6.50208±6.50208√( 0.1250.016 )2

+(0.060.13 )2

]×(54.036±0.0125)V p=(6.50208±2.99 )×(54.036±0.0125)


12/12

V p=[351.346±351.346√( 2.996.50208 )2

+( 0.012554.036 )2

]V p=351.346±160.916mm

3

Saem$ &e2 1cm 10 mm

1c m3=1000m m3

E't$'e2

V p=0.351346±0.160916c m3

DENSDAD DE LA !LA$A:

ρ=m

V =

0.76±0.7

0.3513±0.16

ρ=2.1714 ±2.1714√( 0.070.76 )2

+( 0.160.3513 )2

=2.1633±0.987

ρ=2.2±1 g

c m3