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英語 高校生

⑹赤本の模範解答 Jillian and Martin have decided to get married in June. They are both looking forward to their wedding day. だったのですが、2文目 例えば The... 続きを読む

英語 (80 分) 次の英文を読んで, 下記の英語による設問に答えなさい。 (*印の付いた語に は注があります。) I Martin looked quickly up the street and down to make sure that nobody was watching him, then he turned round, pushed open the door of the jeweller's shop and walked smartly up to the man who was standing behind the counter. He had been passing this shop for many months, because it was near Jillian's *flat. But he had only been looking in the window for a few days. Once he and Jillian had started to talk about getting married, he knew he ought to buy her an engagement ring. So he had been window-shopping in all the jewellers' shops he could find. But since he had seen one particular ring in the shop nearest Jillian's flat, he had not been thinking seriously about any others. This,/ Martin decided,/ was the ring for Jillian. And so he had gone into the shop to see it more( a ). “I've been looking at a lot of rings lately," he told the assistant, “but I'm sure this is the one for my young lady. Yes, she ought to like it," he said, looking at the price ticket. The man behind the counter smiled. “T'm sure she will, sir. But if the size isn't right, or if she wants to change it, please bring it back." At half past ten that night Martin found himself walking in the rain towards Jillian's flat. At his side were Jillian and Philippa, talking noisily. But Martin was (b ). He had turned the collar of his raincoat up and had pushed his hands deep into his pockets. In his right hand was a little square

未解決 回答数: 1
物理 高校生

物理の単振動での質問です。 問142の(2)(3)がわかりません。

0 139 センサー1) センサー42 けた旦 142 m(N/m] (2) ==2d cos 、Cm) 142 センサー 41 2d 変えて のセンサー 42 2d 3Vg 1 の指針) めると、式で表すことができる。 (3) 周期の何倍になるかを考えるとよ い。また,単振動では速さが変化するので, 変位が時間に比例するわけ ではないことに注意する。 解説(1) 斜面に平行な方向の力のつり合いより. (1) 力のつり合いから求める。 (2) 振幅 A, 角振動数 を求 3 振 kx k2x N fe (1) 物体 A, Bを質量2m の1つの物体として考える。 kd - 2mg sin30° =0 86 第1部 様 C00000 )30° mg ゆえに、k= d N/m)] *2mg O 2m (2) T=2r、k,の= k より、 @=, (2)別解 振幅を2d, 2m 000 30° h 1 初期位相を-号と考えて, mg Chapler これに(1)の結果を用いると, ω =, d g V 2d 2m 10 x=2d sin t=0[s]のとき物体 A, Bは振動の最下端にあり,そのとき の変位は-2dであるので, 振幅A[m]は2dである。した 2d g t [m] 2d (1 )hd: 2 =ー2d cos, ョセンサー41 がって,x= -Acosot =ー2d cos. g t[m] V 2d センサー 44 (3) =d のとき, t= そなので、 T 2m 2 2m 2元 2d 04 3 3 3V g mg d -2d コー T=2TU 別解 初めてェ=4となるときに物体Bが物体Aから離れる から,(2)の結果より, 120° 4--2d cos, ゆえに, cosに一言 よって、に等となるから, (=等、) d=-2d cos\ 2d t= T より I-T」 2d 2元 3 2x A0s wtの接 V 2d 3V g センサ

未解決 回答数: 1