Answer:
a) 
Explanation:
a) the cross-sectional area of the hose would be the square of radius times pi. And since the sectional radius is half of its diameter d. We can express the cross-sectional area A1 in term of diameter d1

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Answer:
x=0.154kg
Explanation:
(x*L)+(0.5kg*4200*50)+(x*4200*(-50)=0
(x*333 000J/kg*c)+(0.5kg*4200J/kg*C*(-40C))+(x*4200J/kg*C*50C)=0