785_Physics ProblemsTechnical Physics

785_Physics ProblemsTechnical Physics - Chapter 27 P27.69...

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Chapter 27 125 P27.69 Each speaker receives 60.0 W of power. Using P = I R 2 , we then have I R = = = P 60 0 3 87 . . W 4.00 A . The system is not adequately protected since the fuse should be set to melt at 3.87 A, or less . P27.70 V E = − A or dV E dx = − V IR E I dq dt E R A E A E A dV dx A dV dx = − = − = = = = = − = A A A A ρ ρ σ σ Current flows in the direction of decreasing voltage. Energy flows as heat in the direction of decreasing temperature. P27.71 R dx A dx wy = = z z ρ ρ where y y y y L x = + 1 2 1 R w dx y y y L x L w y y y y y L x R L w y y y y L L = + = + L N M O Q P = F H G I K J z ρ ρ ρ 1 2 1 0 2 1 1 2 1 0 2 1 2 1 b g b g b g ln ln FIG. P27.71 P27.72 From the geometry of the longitudinal section of the resistor shown in the figure, we see that b r y b a h = a f a f . From this, the radius at a distance y from the base is r a b y h b = + a f . For a disk-shaped element of volume
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