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1 (c) Masaya Kasuga Shaltics 2011
2 1
3 2
4 TeX OHP 3
5 4
6 BLACK 5
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8 7
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11 2.3 10
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14 3.2 13
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17 3.4 16
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20 5 E = hν h 1 2 mv2 mc 2 mgh 2 2 mv 1 hν/c 19
1
(c) Masaya Kasuga Shaltics 2010 1 1 1 2 1 1.1 2 3 2 3 3 4 1.2 4 4 5 1.3 5 5 1.4 6 6 6 1.5 7 7 1.6 7 8 8 8 9 2 2.1 10 9 10 9 10 11 2.2 11 11 12 2.3 13 2.4 14 12 12 15 3 3.1 13 13 16 14 15 3.2 14 15 17 16
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00 OHP ( OHP ) Code L00 L00 L0 L0 L00 L0 L00 L00 L00 L0 L0 L00 L00 L00 L00 L0 L00 L00+L00 L00 L0 L00 L00+L00 L00 K K K K K K L0 L0 L0 L0 L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L00
MC-440 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67
1 NASA 2 PLANET-C 2 ISAS No.320 2007.11
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脳を鍛えるチャレンジド学習プリント
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d (K + U) = v [ma F(r)] = (2.4.4) t = t r(t ) = r t 1 r(t 1 ) = r 1 U(r 1 ) U(r ) = t1 t du t1 = t F(r(t)) dr(t) r1 = F dr (2.4.5) r F 2 F ( F) r A r
2.4 ( ) U(r) ( ) ( ) U F(r) = x, U y, U = U(r) (2.4.1) z 2 1 K = mv 2 /2 dk = d ( ) 1 2 mv2 = mv dv = v (ma) (2.4.2) ( ) U(r(t)) r(t) r(t) + dr(t) du du = U(r(t) + dr(t)) U(r(t)) = U x = U(r(t)) dr(t)
hirameki_09.dvi
2009 July 31 1 2009 1 1 e-mail: [email protected] 2 SF 2009 7 31 3 1 5 1.1....................... 5 1.2.................................. 6 1.3..................................... 7 1.4...............................
ma22-9 u ( v w) = u v w sin θê = v w sin θ u cos φ = = 2.3 ( a b) ( c d) = ( a c)( b d) ( a d)( b c) ( a b) ( c d) = (a 2 b 3 a 3 b 2 )(c 2 d 3 c 3 d
A 2. x F (t) =f sin ωt x(0) = ẋ(0) = 0 ω θ sin θ θ 3! θ3 v = f mω cos ωt x = f mω (t sin ωt) ω t 0 = f ( cos ωt) mω x ma2-2 t ω x f (t mω ω (ωt ) 6 (ωt)3 = f 6m ωt3 2.2 u ( v w) = v ( w u) = w ( u v) ma22-9
1.1 ft t 2 ft = t 2 ft+ t = t+ t 2 1.1 d t 2 t + t 2 t 2 = lim t 0 t = lim t 0 = lim t 0 t 2 + 2t t + t 2 t 2 t + t 2 t 2t t + t 2 t 2t + t = lim t 0
A c 2008 by Kuniaki Nakamitsu 1 1.1 t 2 sin t, cos t t ft t t vt t xt t + t xt + t xt + t xt t vt = xt + t xt t t t vt xt + t xt vt = lim t 0 t lim t 0 t 0 vt = dxt ft dft dft ft + t ft = lim t 0 t 1.1
m d2 x = kx αẋ α > 0 (3.5 dt2 ( de dt = d dt ( 1 2 mẋ kx2 = mẍẋ + kxẋ = (mẍ + kxẋ = αẋẋ = αẋ 2 < 0 (3.6 Joule Joule 1843 Joule ( A B (> A ( 3-2
3 3.1 ( 1 m d2 x(t dt 2 = kx(t k = (3.1 d 2 x dt 2 = ω2 x, ω = x(t = 0, ẋ(0 = v 0 k m (3.2 x = v 0 ω sin ωt (ẋ = v 0 cos ωt (3.3 E = 1 2 mẋ2 + 1 2 kx2 = 1 2 mv2 0 cos 2 ωt + 1 2 k v2 0 ω 2 sin2 ωt = 1
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21 2 27 Bethe-Bloch Bethe-Bloch (stopping range) Bethe-Bloch FNAL (Fermi National Accelerator Laboratory) - (SciBooNE ) SciBooNE Bethe-Bloch FNAL - (SciBooNE ) Bethe-Bloch 1 0.1..............................
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op-amp-v1.dvi
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物体の自由落下の跳ね返りの高さ 要約 物体の自由落下に対する物体の跳ね返りの高さを測定した 自由落下させる始点を高くするにつれ 跳ね返りの高さはただ単に始点の高さに比例するわけではなく 跳ね返る直前の速度に比例することがわかった
物体の自由落下の跳ね返りの高さ 要約 物体の自由落下に対する物体の跳ね返りの高さを測定した 自由落下させる始点を高くするにつれ 跳ね返りの高さはただ単に始点の高さに比例するわけではなく 跳ね返る直前の速度に比例することがわかった (1) 目的球技において必ず発生する球の跳ね返りとはどのような規則性に基づいて発生しているのかを調べるために 4 種類の物体を用い様々な床の上で実験をして跳ね返りの規則性を測定した
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