Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
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
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
