To solve this problem it is necessary to apply the concepts related to the continuity of fluids in a pipeline and apply Bernoulli's balance on the given speeds.
Our values are given as


From the continuity equations in pipes we have to

Where,
= Cross sectional Area at each section
= Flow Velocity at each section
Then replacing we have,



From Bernoulli equation we have that the change in the pressure is

![7.3*10^3 = \frac{1}{2} (1000)([ \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1 ]^2-v_1^2)](https://tex.z-dn.net/?f=7.3%2A10%5E3%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%281000%29%28%5B%20%5Cfrac%7B%281.25%2A10%5E%7B-2%7D%29%5E2%20%7D%7B0.6%2A10%5E%7B-2%7D%29%5E2%7D%20v_1%20%5D%5E2-v_1%5E2%29)


Therefore the speed of flow in the first tube is 0.9m/s
1.5 with 8 exponent of 10.
1.5 x 100000000
150000000
Answer:
2.9Ω
Explanation:
Resistors are said to be in parallel when they are arranged side by side such that their corresponding ends are joined together at two common junctions. The combined resistance in such arrangement of resistors is given by;
1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn
Where;
Req refers to the equivalent resistance and R1, R2, R3 .......Rn refers to resistance of individual resistors connected in parallel.
Note that;
R1= 6.0Ω
R2 = 9.0Ω
R3= 15.0 Ω
Therefore;
1/Req = 1/6 + 1/9 + 1/15
1/Req= 0.167 + 0.11 + 0.067
1/Req= 0.344
Req= (0.344)^-1
Req= 2.9Ω
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