120_Physics ProblemsTechnical Physics

120_Physics - 122 P5.7 The Laws of Motion(a F = ma and v 2 = vi2 2 ax f or a = f v 2 vi2 f 2x f Therefore F = m ev 2 f vi2 j 2x f F = 9.11 10 31(b

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122 The Laws of Motion P5.7 (a) Fm a = and vv a x fi f 22 2 =+ or a x f = 2 . Therefore, x F f = ×− × L N M O Q P 31 5 2 5 2 18 2 911 10 7 00 10 3 00 10 200500 364 10 ej bg . .. . kg ms m N (b) The weight of the electron is g g == × = × −− 980 893 10 31 30 . kg m s N 2 ch c h The accelerating force is 408 10 11 . × times the weight of the electron. P5.8 (a) g g = ( ) = 120 4 448 120 534 lb N lb lb N . af (b) m F g g = 534 54 5 N 9.80 m s kg 2 . P5.9 g g 900 N , m 900 91 8 N 9.80 m s kg 2 . F g c h on Jupiter 2 kg m s kN 918 259 238 . P5.10 Imagine a quick trip by jet, on which you do not visit the rest room and your perspiration is just canceled out by a glass of tomato juice. By subtraction, g g p p = and g g C C = give
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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