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Sagot :
Answer:
1.-dh/dt = 5.31*10⁻⁵ m/seg
2.-dh/dt = 1.99*10⁻⁵ m/seg
3.-dh/dt = 1.59*10⁻⁴ m/seg
Step-by-step explanation:PREGUNTA INCOMPLETA NO SE INDICAN LAS FORMAS DE LOS TANQUES.
Asumiremos que los tres tanques son:
el primero cilindro recto de Vc = π*r²*h ( r es radio de la base y h la altura)
el segundo asumiremos que es eliptico recto de Ve = π*a*b*h aqui a y b son los ejes de la elipse y h la altura
El tercero es un cono invertido Vco = 1/3 *π*r²*h ( r es el radio de la base.
1.-Caso del cilindro
Vc = π*r²*h
Derivando en ambos miembros de la expresión tenemos:
dV(c) / dt = π*r²*dh/dt
Sustituyendo
1.5 Lts/seg = 3.14 * (3)²*dh/dt
1.5/1000 m³/seg = 28.26 m² dh/dt
1.5/ 28260 m = dh/dt
Despejando dh/dt
dh/dt = 1.5 / 28260 = 5.31*10⁻⁵ m/seg
dh/dt = 5.31*10⁻⁵ m/seg
2.-La elipse
Ve = π*a*b*h
Aplicando el mismo procedimiento tenemos:
DVe/dt = 1.5 Lts/seg = π* 6*4* dh/dt
1.5 /1000 = 75.36 *dh/dt
dh/dt = 1.5 / 75360 m/seg
dh/dt = 1.99*10⁻⁵ m/seg
3. El cono invertido
Vco = (1/3)*π*r²*h
DVco/dt = (1/3)*π*r²*dh/dt
1.5/1000 = 9.42 *dh/dt
dh/dt = 1.5/9420
dh/dt = 1.59*10⁻⁴ m/seg
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