217_Physics ProblemsTechnical Physics

217_Physics ProblemsTechnical Physics - Chapter 8 P8.2 (a)...

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Chapter 8 219 P8.2 (a) We take the zero configuration of system potential energy with the child at the lowest point of the arc. When the string is held horizontal initially, the initial position is 2.00 m above the zero level. Thus, Um g y g == = 400 2 00 800 N m J a fa f .. (b) From the sketch, we see that at an angle of 30.0° the child is at a vertical height of 200 1 300 .c o s . m af a f −° above the lowest point of the arc. Thus, FIG. P8.2 g y g ° = 400 2 00 1 30 0 107 m J a fa fa f o s . . (c) The zero level has been selected at the lowest point of the arc. Therefore, U g = 0 at this location. *P8.3 The volume flow rate is the volume of water going over the falls each second: 30 5 1 2 1 8 m m m s m s 3 . bg = The mass flow rate is m t V t = ρ 1 000 1 8 1 800 kg m m s kg s 33 ej e j . If the stream has uniform width and depth, the speed of the water below the falls is the same as the speed above the falls. Then no kinetic energy, but only gravitational energy is available for conversion into internal and electric energy. The input power is P in 2 energy kg s m s m J s = = = × t mgy t m t gy 1 800 9 8 5 8 82 10 4 The output power is
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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