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Q1Phys131WI10SOL-E1%281D%20Motion%29

Q1Phys131WI10SOL-E1%281D%20Motion%29 - I. Traveling...

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Unformatted text preview: I. Traveling Electron [20 points] An electron in a cathode ­ray tube accelerates from rest with a constant acceleration of 5.33 x 1012 m/s2 for 0.150 µs, then drifts with a constant velocity for 0.200 µs, then slows to a stop with a negative acceleration of  ­2.67 x 1013 m/s2. (a) [8 pts] How far does the electron travel with the constant speed? v1 = ax t = (5.33 !1012 m/s 2 )(0.150 !10-6 s) = 8.0 !105 m/s "x2 = v1t2 = (8.0 !105 m/s)(0.150 !10-6 s) = 0.16 m (b) [4 pts] How far does the electron travel before it starts to slow down? !x1 = v0t + 1 ax t 2 = (0 m/s)(0.150 "10-6 s)+ 1 (5.33 "1012 m/s 2 )(0.150 "10-6 s) 2 = 0.06 m 2 2 Δx12 = Δx1 + Δx 2 = 0.06 m + 0.16 m = 0.22 m € (c) [8 pts] How far does the electron travel from start to stop? vx = v0 x 2 2 vx 2 − v0 x 2 (0 m/s) 2 − (8.0 ×105 m/s) 2 + 2ax Δx; Δx3 = = = 0.012 m 2a x 2(−2.67 ×1013 m/s 2 ) x = Δx1 + Δx 2 + Δx 3 = (0.06 m) + (0.16 m) + (0.012 m) = 0.232 m = 23.2 cm € (d) [5 pts Extra Credit] what is the electron’s average speed through out this journey?( no Partial credit) v 3 x = v 2 x + at 3 t total = t1 + t 2 + t 3 −v 2 x −8.0 × 10 5 m / s t total = 0.150µs + 0.200µs + 0.029µs t3 = = a −2.76 × 1013 m / s2 t total = 0.379µs t 3 = 2.89 × 10−8 = .029µs ; v 3x = 0 € € v ave = v ave v ave Δx 23.2 cm × 10−2 m = Δt 0.379 × 10 -6 s = 612,137, 203 m / s ≈ 75% C € ...
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