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英文 國中

大概有七題的英文文法問題 (有打勾勾的地方✔️(我把他整理成一張圖片了 感謝善心人士解答💗💗

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物理 高中

想請問為什麼角動量方向是逆時針而不是順時針?是不是和位置向量的方向有關?因為我不是很確定位置向量是由質點指向質心還是由質心指向質點😅🥺

故 範例 2 系統總角動量 雙質系統,質系到货心让和質量成反比 (F), 小將正 於0 故力 點,其 82 L=2 mA MIW 質量為m的甲球與質量為5m的乙球分別固定在一長為L的細桿兩端,並繞其質心以 角速率(W逆時針旋轉,轉軸與細桿垂直,旋轉時細桿長度不變;設細桿極輕,其質量可 以忽略不計,且兩球的直徑與桿長相比極小,也可以忽略不計。則相對於質心, (1)系統的總動量大小為 0 (2)此轉動系統的角動量的量值為 / MLW 答 M₁ 連心力, CM 為L,行 L = 4r (1) 1a F +2= L Fcos a MI 盲向圓 (E)w => => ₁ = Ma m₁ +m₂ 5 13 = $ 1 Ilw V= Ilw 、方向 逆时针。 (2)有動量方向同 4,+L₂ = rip,+₂ p₂ L = m,r,³w, + m₂r, w₂ (m, ri²+ m²) w miw P₁ + P₂ = (-5m x = lw) + (mx lw) 【96.指考改】 學習概念 2 角動量變化與力矩的 1. 力矩與角動量的關係:質點所受的合力矩=質 由牛頓第二運動定律: 切線力=p: At At 再由力矩定義:c=rxF=rxp. 在極短時間內,若,固定或r的變化可結 則角動量變化△Z=A(rxp)=Arxp_ 即力矩7=7×F=rx Ap AL At At 角動量和力矩 範例 3 如右圖所示,單擺懸掛在固定的O點上, m,擺錘的半徑遠小於ㄥ,設重力加速度” 自由釋放,當擺線和鉛直方向的夾角為二 (1)擺錘所受的重力對點的力矩量值為 (2)擺錘所受的擺線張力對點的力矩量 (3)擺錘相對於點的角動量對時間的三 合

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