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andre [41]
3 years ago
11

Em muitas situações do dia a dia, temos que solucionar questões que exigem conhecimento das relações existentes entre as grandez

as físicas, seus múltiplos e submúltiplos. O exercício a seguir é um exemplo. Francisco, na sua chácara, próximo à sua casa de campo, separou uma área retangular de 150 m2 com 15 m de comprimento por 10 m de largura. Bem no centro dessa área será construída uma piscina. O fundo da piscina deve ser plano e horizontal com 1,5 m de profundidade. Observe o esquema que representa o projeto da construção da piscina e do seu entorno. Para tratar semanalmente a água da piscina, Francisco tem que utilizar 0,005 g de hipoclorito de sódio (cloro) a cada litro de água. Qual é o custo mensal com a cloração da água sabendo-se que um galão com 18 kg de hipoclorito custa R$135,00? Considere que um mês tenha 4,5 semanas.
Physics
1 answer:
VladimirAG [237]3 years ago
8 0

Answer:

R\$37,96

Explanation:

1) Olá, vamos calcular primeiramente o Volume da Piscina. Ela é uma piscina com uma forma tradicional, retangular. Temos a largura, o comprimento e a profundidade.

Volume=a.b.c\\\\Volume=largura*comprimento*altura\\Volume= 10*15*1,5\\Volume=225m^{3}

2) Precisamos fazer uma correlação entre a medida de capacidade (litro) e de volume, já que o custo é por litro d'água.

1 litro = 1dm^3\\1 m^{3}=1000 \:dm^{3}=1000 \:litros\\225\:m^{3}=225000\: litros

3) Observe que Francisco gasta 0,005 g de NaClO por semana.

Considerando que o mês tenha 4,5 semanas. E que quanto mais se trata a piscina mais se gasta Hipoclorito de Sódio. Temos uma relação diretamente proporcional.

Finalmente,

3.1) Rendimento do hipoclorito de produto em gramas por semana:

1 semana   \rightarrow      0,005g\\4,5 semanas  \rightarrow 0,0225g

3.2) Consumo mensal (y), considerando 4,5 semanas:

0,005g\rightarrow1 litro\\y\rightarrow225000\\y=1125 gramas \: para\: 225000 litros.\\\\1125gramas \:por\:litro\rightarrow 1 semana\\q \rightarrow 4,5\\q=5062,5 \:gramas=\:5,0625 Kg \:de \:NaClO\\\\

3.3)  O custo da manutenção, já que 18 kg de NaClO  custa R$ 135 e são grandezas diretamente proporcionais:

18 kg\rightarrow \$135\\5,0625  \:kg \:de \:NaOCl\rightarrow p\\p=\frac{5,0625*135}{18}=\frac{683.4375}{18} \therefore p=R\$37,96

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