Problem 111
Two resistors are connected in series to a battery with an emf of 12 V. The voltage across the rst resistor is 2.7 V and the current through the second resistor is 0.15 A. Find the resistance of the two resistors.
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**Workischangeintotalenergyamount. **Kindsofenergycanchangeevenifnoworkisdone. • GravitationalPotentialEnergy o Lifting/loweringthings;“height” o U =Gmgh § U G=(mass)(gravity)(height) § Example:Massof60kgliftedfromheight3mà4m.Beforeand afterGPE,howmuchwork 2 U Gi(60kg)(9.8m/s )(3m)=1764J 2 U Gf(60kg)(9.8m/s )(4m)=2352J W=U GfU =Gi88J • KineticEnergy“motion” 2 o K=½mv Example:Howmuchenergydoesa0.25kgballhaveat5m/s K=½mv =½(0.25kg)(5m/s) 2
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Chapter 21, Problem 111 is Solved
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