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IDENTIFY: P = 2 t dt d 1/ 2 1 −1/ 2 2 t = 2t SET UP: . ∫ t1/ 2 dt = 3 t 3 / 2 . dt EXECUTE: (b) a = (a) The power P is related to the speed by Pt = K = 1 mv 2 , so v = 2 2Pt . m 2P d 2P 1 dv d 2 Pt P . = = t= = 2mt dt dt m m dt m 2 t 2P 1 2P 2 3 8P 3 ∫ t 2 dt = m 3 t 2 = 9m t 2 . m (c) x − x0 = ∫ v dt = EVALUATE: v, a, and x − x0 at a particular time are all proportional to P1/ 2 . The result in part (b) could also be obtained from P = Fv and a = F / m , so a = 6.93. P…

IDENTIFY: P = 2 t dt d 1/ 2 1 −1/ 2 2 t = 2t SET UP: . ∫ t1/ 2 dt = 3 t 3 / 2 . dt EXECUTE: (b) a = (a) The power P is related to the speed by Pt = K = 1 mv 2 , so v = 2 2Pt . m 2P d 2P 1 dv d 2 Pt P . = = t= = 2mt dt dt m m dt m 2 t 2P 1 2P 2 3 8P 3 ∫ t 2 dt = m 3 t 2 = 9m t 2 . m (c) x − x0 = ∫ v dt = EVALUATE: v, a, and x − x0 at a particular time are all proportional to P1/ 2 . The result in part (b) could also be obtained from P = Fv and a = F / m , so a = 6.93. P…

32 t 3 / 2 . dt EXECUTE: (b) a = (a) The power P is related to the speed by Pt = K = 12 mv 2 , so v = 2Pt . m 2P d 2P 1 dv d 2 Pt P . = = t= = 2mt dt dt m m dt m 2 t (c) x − x0 = ∫ v dt = EVALUATE: 2 P 12 2 P 2 32 8P 23 t dt = t = t . ∫ m m 3 9m v, a, and x − x0 at a particular time are all proportional to P1/ 2 . The result in part (b) could also be P . vm IDENTIFY and SET UP: For part (a) calculate m from the volume of blood pumped by…

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we expect the middle wire to have a net force of zero; the two on the outside will each be attracted toward the center, but the answers will be symmetrically distributed. For the wire which is the farthest left, F µ0 i2 = L 2π 1 1 1 1 + + + a 2a 3a 4a = 4π ××10−7 N/A2 (3.22 A)2 2π(0.083 m) 1+ 1 1 1 + + 2 3 4 = 5.21×10−5 N/m. For the second wire over, the contributions from the two adjacent wires should cancel. This leaves F µ0 i2 = L 2π E33-23 1 1 + + 2a 3a = 4π ××10−7 N/A2 (3.22 A)2 2π(0.083

we expect the middle wire to have a net force of zero; the two on the outside will each be attracted toward the center, but the answers will be symmetrically distributed. For the wire which is the farthest left, F µ0 i2 = L 2π 1 1 1 1 + + + a 2a 3a 4a = 4π ××10−7 N/A2 (3.22 A)2 2π(0.083 m) 1+ 1 1 1 + + 2 3 4 = 5.21×10−5 N/m. For the second wire over, the contributions from the two adjacent wires should cancel. This leaves F µ0 i2 = L 2π E33-23 1 1 + + 2a 3a = 4π ××10−7 N/A2 (3.22 A)2 2π(0.083

we expect the middle wire to have a net force of zero; the two on the outside will each be attracted toward the center, but the answers will be symmetrically distributed. For the wire which is the farthest left, F µ0 i2 = L 2π 1 1 1 1 + + + a 2a 3a 4a = 4π ××10−7 N/A2 (3.22 A)2 2π(0.083 m) 1+ 1 1 1 + + 2 3 4 = 5.21×10−5 N/m. For the second wire over, the contributions from the two adjacent wires should cancel. This leaves F µ0 i2 = L 2π E33-23 1 1 + + 2a 3a = 4π ××10−7 N/A2 (3.22 A)2 2π(0.083

we expect the middle wire to have a net force of zero; the two on the outside will each be attracted toward the center, but the answers will be symmetrically distributed. For the wire which is the farthest left, F µ0 i2 = L 2π 1 1 1 1 + + + a 2a 3a 4a = 4π ××10−7 N/A2 (3.22 A)2 2π(0.083 m) 1+ 1 1 1 + + 2 3 4 = 5.21×10−5 N/m. For the second wire over, the contributions from the two adjacent wires should cancel. This leaves F µ0 i2 = L 2π E33-23 1 1 + + 2a 3a = 4π ××10−7 N/A2 (3.22 A)2 2π(0.083