Current and Resistance9517.63The power the beam delivers to the target is ()( )( )634.010 V2510A1.010 WVI−=∆=××=×P5The mass of cooling water that must flow through the tube each second if the rise in the water temperature is not to exceed 50°C is found from ( )(mtcT)=∆∆ ∆Pas 51.010 J s0.48 kg s4 186 J kgC50 CmtcT×∆===∆∆⋅°°√17.64The volume of the material is 336350.0 g1 m6.3610m7.86 g cm10 cmmassVdensity−=×L=⋅Since VA, the cross-sectional area of the wire is A VL=(a) From 2LLLRAVLVρρρ=, the length of the wire is given by -81.5 6.3610m9.3 m1110 mRVL−
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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.