早稲田大学教育学部【物理】解答例 Ⅰ 問1 V1 = −e √ 2gh :::::::: √ 問2 V2 = −e 2gh + L 1+e √ √ g L ,w2 = 2gh + 2h 1+e :::::::::::::::::::::: L 1+e 問3 H1 = { e− L 4(1 + e)h √ g 2h ::::::::::::::::::: } 問4 w3 = ::::::::::::::::::::::: (m − M )w2 + 2M V2 m+M :::::::::::::::::::: 94 h 25 ::: 問5 H2 = 問6 小球 A:M aA = M g − kx ,小球 B:maB = kx + mg (M + m)k x Mm ::::::::::::: 問7 aG = g ,aBA = − : √ ( ) 2 Mg h− 問8 T1 = g k √ 問9 T2 = 2π ::::::::::::::: Mm (M + m)k :::::::::::::: Ⅱ 問1 右図 問2 E = 問3 I = eV V ,u = L kL :: ::: Sne2 V kL ::::::: 問6 IR = 問4 P = V2 V 2T [W] ,Q = [J] R R ::::::: :::::::: RV V ,VR = r::::: +R r::::: +R 問7 金属棒の消費電力は, IR VR = R V2 (r + R)2 = ( V2 √ r √ + R R )2 5 V2 4r 等号成立のとき,消費電力は最大となり,このとき √ r √ = R ゆえに R = :r R 4RV 2 4RV 2 問8 Pa = , P = b (2r + R)2 (2R + r)2 ::::::::: ::::::::: 問5 6 倍 :
© Copyright 2024 ExpyDoc