Laboratorio de fisica 1 ensayo
Enviado por Helena • 19 de Diciembre de 2018 • 2.456 Palabras (10 Páginas) • 257 Visitas
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RESOLUCION DE LAS ECUACIONES DEL TARUGO CON PIE DE REY
*Diametro
Promedio de Diametro del Tarugo:[pic 75]
[pic 76]
[pic 77]
[pic 78]
Error sistemático
[pic 79]
Error Aleatorio [pic 80]
[pic 81]
[pic 82]
[pic 83]
[pic 84]
Error Total o absoluto[pic 85]
[pic 86]
[pic 87]
[pic 88]
[pic 89]
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*Altura
Promedio de la altura del Tarugo:
[pic 90]
[pic 91]
[pic 92]
[pic 93]
Error sistemático
[pic 94]
Error Aleatorio [pic 95]
[pic 96]
[pic 97]
[pic 98]
[pic 99]
Error Total o absoluto[pic 100]
[pic 101]
[pic 102]
[pic 103]
[pic 104]
VOLUMEN DEL TARUGO CON PIE DE REY
D= (17.922 mm = (1.7922 cm[pic 105][pic 106]
H=(20.86) mm = (2.086) cm[pic 107][pic 108]
Vt = π(2H[pic 109]
Vt = (1.79222(2.086)[pic 110][pic 111][pic 112]
Vt = K(A)nB
Z = K(A)n = 0.7854 (1.7922)2 = 2.52268975[pic 113][pic 114]
= 2 () ( ) = 0.03568388[pic 115][pic 116][pic 117]
Vt =(2.52268975( 2.086) cm3[pic 118][pic 119]
Vt = AxB
t= = 2.086 = 5.26233082 y[pic 120][pic 121][pic 122]
t) = 0.1251316[pic 123][pic 124]
Vt = (5.26233082 0.1251316) cm3 [pic 125]
Densidad del Tarugo con pie de rey
= [pic 126][pic 127]
= = = 0.60429496[pic 128][pic 129][pic 130]
= [pic 131][pic 132][pic 133]
= 0.60429496 x [pic 134]
= 0.0189724
= = (0.60429496 0.0189724) g/cm3 [pic 135][pic 136][pic 137][pic 138][pic 139]
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RESOLUCION DE LAS ECUACIONES DEL TARUGO CON MICROMETRO
*Diámetro
Promedio de Diámetro del Tarugo:[pic 140]
[pic 141]
[pic 142]
[pic 143]
Error sistemático
[pic 144]
Error Aleatorio [pic 145]
[pic 146]
[pic 147]
[pic 148]
[pic 149]
Error Total o absoluto[pic 150]
[pic 151]
[pic 152]
[pic 153]
[pic 154]
*Altura
Promedio de la altura del Tarugo:[pic 155]
[pic 156]
[pic 157]
[pic 158]
Error sistemático
[pic 159]
Error Aleatorio [pic 160]
[pic 161]
[pic 162]
[pic 163]
[pic 164]
Error Total o absoluto[pic 165]
[pic 166]
[pic 167]
[pic 168]
[pic 169]
VOLUMEN DEL TARUGO CON MICROMETRO:
D= (18.078 mm = (1.8078 cm[pic 170][pic 171]
H=(20.288) mm = (2.0288) cm[pic 172][pic 173]
Vt = π(2H[pic 174]
Vt = (1.80782(2.0288)[pic 175][pic 176][pic 177]
Vt = K(A)nB
Z = K(A)n = 0.7854 (1.8078)2 = 2.56679782[pic 178][pic 179]
= 2 () ( ) = 0.04794293[pic 180][pic 181][pic 182]
Vt =(2.56679782( 2.0288) cm3[pic 183][pic 184]
Vt
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