mecanica fluidos.docx

# mecanica fluidos.docx - 5 Considerando el siguiente campo...

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5. Considerando el siguiente campo escalar φ = xy + z, obtener el vector unitario normal a la superficie determinada por φ = constante que pasa por el punto (2, 1, 0). Para poder hallar el vector unitario normal de φ = xy + 2 se considera: = ( d dx i + d dy j + d dz k ) . ( xy + z ) Entonces = ( d ( xy + z ) dx i + d ( xy + z ) dy j + d ( xy + z ) dz k ) Aplicando derivada parcial = ( yi + xj + k ) Remplazando los puntos en el vector resultante p = ( 2,1,0 ) [ φ ] ( 2,1,0 ) = [ yi + xj + k ] ( 2,1,0 ) = [ ( 1 ) i +( 2 ) j + k ] = [ i + 2 j + k ] En el caso del vector unitario normal se utiliza la siguiente ecuación ´ v = ´ v ¿ ´ v ¿ Hallando el modulo del vector 1 i + 2 j + k ¿ ¿ ¿¿ Luego reemplazando para hallar el vector unitario ´ v = 1 i + 2 j + k 6 = 1 6 i + 2 6 j + 1 6 k 7. Se desea calcular la masa específica de una pieza metálica, para esto se pesa en el aire dando como resultado 19 N y a continuación se pesa sumergida en agua dando un valor de 17 N.

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Datos: Paire = 19 N PH 2 O = 17 N E = 2 N ρ ab = ? ρ H 2 O = 1000 kg m 3 Solución:

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