Trabajo De Fisica Cuestionario.docx

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Cuestionario 2.Con los datos de la tabla 02 y 03, determine el valor verdadero del volumen del paralelepΓ­pedo, halle el error relativo y porcentual. Compare sus resultados y explique. 2.1 Tabla 02: Utilizando regla graduada Medida π‘₯1 π‘₯2 π‘₯3 π‘₯4 π‘₯5 Promedio: 𝑋̅

L(mm) 98mm 100mm 97mm 98mm 99mm ̅𝐿=98.4mm

H(mm) 35mm 36mm 37mm 37mm 35mm Μ… 𝐻=36mm

A(mm) 36mm 38mm 38mm 38mm 37mm Μ… 𝐴=37.4mm

a) Valor verdadero del volumen del paralelepΓ­pedo La desviaciΓ³n del volumen de cada medida es: 𝑉1 βˆ’ 𝑉̅ =123480-132531=-9051π‘šπ‘š3 𝑉2 βˆ’ 𝑉̅ =136800-132531=4269π‘šπ‘š3 𝑉3 βˆ’ 𝑉̅ =136382-132531=3851π‘šπ‘š3 𝑉4 βˆ’ 𝑉̅ =137788-132531=5257π‘šπ‘š3 𝑉5 βˆ’ 𝑉̅ =128205-132531=-4326π‘šπ‘š3 La varianza del volumen es: 2 π‘’π‘š =

(βˆ’9051)2 +(4269)2 +(3851)2 +(5257)2 +(βˆ’4326)2 5

2 π‘’π‘š =32265097.6

La desviaciΓ³n estΓ‘ndar 2 𝜎 = βˆšπ‘’π‘š

𝜎 = √32265097.6 𝜎 = 5680.237 La incertidumbre o error estΓ‘ndar del volumen es: βˆ†π‘‰ =

𝜎 βˆšπ‘›

=

5680.237 √5

= 2540.279

Valor verdadero del volumen del paralelepΓ­pedo 𝑉 = 𝑉̅ Β± βˆ†π‘‰ 𝑉 = (132531 Β± 2540.279) π‘šπ‘š3

V(π‘šπ‘š3 ) 123480π‘šπ‘š3 136800π‘šπ‘š3 136382π‘šπ‘š3 137788π‘šπ‘š3 128205π‘šπ‘š3 𝑉̅ = 132531π‘šπ‘š3

b) Error relativo y porcentual -Error relativo πΈπ‘Ÿ =

𝐸𝑇 𝑉̅

-Error porcentual πΈπ‘Ÿ% = 100πΈπ‘Ÿ 2.2 Tabla 03: Utilizando regla graduada Medida L(mm) H(mm) 99.22mm 36.26mm π‘₯1 99.22mm 36.28mm π‘₯2 99.24mm 36.30mm π‘₯3 99.20mm 36.30mm π‘₯4 99.24mm 36.26mm π‘₯5 Μ… =36.28mm Promedio: 𝑋̅ 𝐿̅=99.224mm 𝐻 a) Valor verdadero del volumen del paralelepΓ­pedo

A(mm) 37.36mm 37.34mm 37.32mm 37.30mm 37.32mm 𝐴̅=37.328mm

La desviaciΓ³n del volumen de cada medida es: 𝑉1 βˆ’ 𝑉̅ =123480-132531=-9051π‘šπ‘š3 𝑉2 βˆ’ 𝑉̅ =136800-132531=4269π‘šπ‘š3 𝑉3 βˆ’ 𝑉̅ =136382-132531=3851π‘šπ‘š3 𝑉4 βˆ’ 𝑉̅ =137788-132531=5257π‘šπ‘š3 𝑉5 βˆ’ 𝑉̅ =128205-132531=-4326π‘šπ‘š3 La varianza del volumen es: 2 π‘’π‘š =

(βˆ’9051)2 +(4269)2 +(3851)2 +(5257)2 +(βˆ’4326)2 5

2 π‘’π‘š =32265097.6

La desviaciΓ³n estΓ‘ndar 2 𝜎 = βˆšπ‘’π‘š

𝜎 = √32265097.6 𝜎 = 5680.237 La incertidumbre o error estΓ‘ndar del volumen es: βˆ†π‘‰ =

𝜎 βˆšπ‘›

=

5680.237 √5

= 2540.279

Valor verdadero del volumen del paralelepΓ­pedo 𝑉 = 𝑉̅ Β± βˆ†π‘‰

V(π‘šπ‘š3 ) 134410.715π‘šπ‘š3 134412.860π‘šπ‘š3 134442.016π‘šπ‘š3 134315.808π‘šπ‘š3 134293.870π‘šπ‘š3 𝑉̅ = 134375.054π‘šπ‘š3

𝑉 = (132531 Β± 2540.279) π‘šπ‘š3

b) Error relativo y porcentual -Error relativo πΈπ‘Ÿ =

𝐸𝑇 𝑉̅

-Error porcentual πΈπ‘Ÿ% = 100πΈπ‘Ÿ 3.Con los datos de la Tabla 04 determine el valor del volumen verdadero del cilindro, halle el error relativo y porcentual. Tabla 04: Utilizando el vernier Medida H(mm) 69.88mm π‘₯1 69.90mm π‘₯2 69.88mm π‘₯3 69.92mm π‘₯4 69.92mm π‘₯5 Μ… =69.9mm Promedio: 𝑋̅ 𝐻 Volumen del cilindro

D(mm) 34.36mm 34.30mm 34.30mm 34.32mm 34.34mm Μ… =34.324mm 𝐷

𝑉=

πœ‹π· 2 𝐻 4

a) Valor verdadero del volumen del cilindro La desviaciΓ³n del volumen de cada medida es: 𝑉1 βˆ’ 𝑉̅ = 64796.284 βˆ’ 64679.114=117.17π‘šπ‘š3 𝑉2 βˆ’ 𝑉̅ = 64588.666 βˆ’ 64679.114=-90.448π‘šπ‘š3 𝑉3 βˆ’ 𝑉̅ = 64570.185 βˆ’ 64679.114=-108.929π‘šπ‘š3 𝑉4 βˆ’ 𝑉̅ = 64682.512 βˆ’ 64679.114=3.398π‘šπ‘š3 𝑉5 βˆ’ 𝑉̅ = 64757.921 βˆ’ 64679.114=78.807π‘šπ‘š3 La varianza del volumen es: 2 π‘’π‘š =

(117.17)2 +(βˆ’90.448)2 +(βˆ’108.929)2 +(3.398)2 +(78.807)2 5

2 π‘’π‘š =7999.453

La desviaciΓ³n estΓ‘ndar 2 𝜎 = βˆšπ‘’π‘š

V(π‘šπ‘š3 ) 64796.284π‘šπ‘š3 64588.666π‘šπ‘š3 64570.185π‘šπ‘š3 64682.512π‘šπ‘š3 64757.921π‘šπ‘š3 𝑉̅ = 64679.114π‘šπ‘š3

𝜎 = √7999.453 𝜎 = 89.440 La incertidumbre o error estΓ‘ndar del volumen es: βˆ†π‘‰ =

𝜎 βˆšπ‘›

=

89.440 √5

= 39.999

Valor verdadero del volumen del paralelepΓ­pedo 𝑉 = 𝑉̅ Β± βˆ†π‘‰ 𝑉 = (132531 Β± 39.999) π‘šπ‘š3

b) Error relativo y porcentual -Error relativo πΈπ‘Ÿ =

𝐸𝑇 𝑉̅

-Error porcentual πΈπ‘Ÿ% = 100πΈπ‘Ÿ

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