problem18_63

University Physics with Modern Physics with Mastering Physics (11th Edition)

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18.63: a) The tank is given as being “large,” so the speed of the water at the top of the surface in the tank may be neglected. The efflux speed is then obtained from , 2 1 2 p h ρg ρv + = or × + = + = ) m kg (1000 Pa) 10 20 . 3 ( ) m 50 . 2 ( ) s m 80 . 9 ( 2 2 3 5 2 ρ p h g v s. m 2 . 26 = b) Let m 50 . 3 0 = h and Pa. 10 20 . 4 5 0 × = p In the above expression for m 00 . 1 , - = h h v and . m 00 . 4 m 00 . 4 a 0 0 p h h p p - - - = Repeating the calculation for m 00 . 3 = h gives s m 1 . 16 = v and with s. m 5.44 m, 00 . 2 = = v h c) Setting 0 2 = v in the above expression gives a quadratic equation in h which may be re-expressed as . m 4.00 m 50 . 0 g m) 00 . 1 ( 0 a h ρg p ρ p h - - = - Denoting , m 43 . 21 m) 50 . 0 ( and m 204 . 10 g 2 2 0 a = = = = z ρg p y ρ p this quadratic becomes , 0 ) ) m 00 . 4 ( ) m 00 . 4 (( ) m 00 . 5 ( 2 2 2 = - + + + - z y h y h which has as its solutions
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Unformatted text preview: m. 47 . 13 and m 737 . 1 = = h h The larger solution is unphysical (the height is greater than the height of the tank), and so the flow stops when m. 74 . 1 = h Although use of the quadratic formula is correct, for this problem it is more efficient for those with programmable calculators to find the solution to the quadratic by iteration. Using m 00 . 2 = h (the lower height in part (b)) gives convergence to three figures after four iterations. (The larger root is not obtained by a convergent iteration.)...
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