ardakaraca Posted November 22, 2016 Report Share Posted November 22, 2016 Hi, I'm confused with this question. Does anyone help? Reply Link to post Share on other sites More sharing options...
IBTopper Posted November 23, 2016 Report Share Posted November 23, 2016 Hi, sure let me try this one. Btw which book does this come from? Reply Link to post Share on other sites More sharing options...
ardakaraca Posted November 23, 2016 Author Report Share Posted November 23, 2016 It is from Cambridge option book. A friend of mine solved the question as in the image. Reply Link to post Share on other sites More sharing options...
kw0573 Posted December 8, 2016 Report Share Posted December 8, 2016 In the case, it would be easier to use conservation of energy: gain in kinetic energy = loss of gravitational potential energy. Because linear speed v = Volume flow rate / Area (V / A), you could rearrange for Volumetric flow rate. V should have units m3 / s. The expression you have provided, using dimension (or unit) analysis, yields (m4 (ms-2) (m) / m2)1/2 = (m4 s-2)1/2=m2s-1, which is not correct. The correct solution is as follows. 1/2 m(v22 - v12) = mgh (v22 - v12) = 2gh (V / A2)2 - (V / A1)2 = 2gh V2 (1/A22 - 1/A12) = 2gh V2 (A12 - A22) / (A22A12) = 2gh V2 = 2gh*(A22A12) / (A12 - A22) V = A1A2 sqrt (2gh / (A12 - A22)), which has units m2m2 ((ms-2)(m) / (m4))1/2 = m4 (s-2 m-2)1/2 = m4 (s-1m-1) which has the desired units m3s-1. Remember to plug in all numbers using meter as base units. and 1 m2 = 10 000 cm2 Reply Link to post Share on other sites More sharing options...
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