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‫1- ﺣﺴﺎب ﺳﺮﻋﺔ اﻟﺠﺴﻢ ) 1‪ (S‬ﻓﻲ اﻟﻤﻮﺿﻌﻴﻦ 1‪ M‬و 3 ‪: M‬‬
‫2 ‪M 0M‬‬
‫= 1‪v‬‬         ‫- ﻓﻲ اﻟﻤﻮﺿﻊ 1‪: M‬‬
‫‪2τ‬‬
‫‪M 0 M 2 = 1, 4.10−2 m‬‬      ‫وﻣﻦ اﻟﺘﺴﺠﻴﻞ ﻧﺠﺪ‬
‫2−01.4 ,1‬
‫= 1‪ v‬أي ‪v1 = 0,35m.s‬‬
‫1−‬
‫وﻣﻨﻪ‬
‫3−01.02 × 2‬
‫‪M M‬‬
‫- ﻓﻲ اﻟﻤﻮﺿﻊ 3 ‪v3 = 2 4 : M‬‬
‫‪2τ‬‬
‫وﺣﺴﺐ اﻟﺘﺴﺠﻴﻞ ﻧﺠﺪ ‪M 2 M 4 = 3,8.10 m‬‬
‫2−‬

‫2−01.8,3‬
‫= 3‪ v‬أي 1− ‪v3 = 0,95m.s‬‬                  ‫وﻣﻨﻪ‬
‫3−01.02 × 2‬
‫* اﺳﺘﻨﺘﺎج ﺗﻐﻴﺮ آﻤﻴﺔ اﻟﺤﺮآﺔ ‪ Δp‬ﺑﻴﻦ 1‪ M‬و 3 ‪: M‬‬
‫1‪ Δp = p3 − p‬أي ) 1‪Δp = m1 (v3 − v‬‬   ‫ﻟﺪﻳﻨﺎ‬
‫ت.ع : )53 ,0 − 59,0(4 ,0 = ‪ Δp‬أي 1− ‪Δp = 0, 24kg.m.s‬‬
‫2- ﺣﺴﺎب ﺳﺮﻋﺔ اﻟﺠﺴﻢ ) 1‪ (S‬ﻓﻲ اﻟﻤﻮﺿﻌﻴﻦ 5 ‪ M‬و 6 ‪: M‬‬
‫6 ‪M 4M‬‬
‫= 5‪v‬‬           ‫- ﻓﻲ اﻟﻤﻮﺿﻊ 5 ‪: M‬‬
‫‪2τ‬‬
‫وﻣﻦ اﻟﺘﺴﺠﻴﻞ ﻧﺠﺪ ‪M 4 M 6 = 2.10 −2 m‬‬
‫2−01.2‬
‫= 5‪ v‬ﻳﻌﻨﻲ 1− ‪v5 = 0,5m.s‬‬                  ‫وﻣﻨﻪ :‬
‫3−01.02 × 2‬
‫7 ‪M 5M‬‬
‫= 6‪v‬‬          ‫- ﻓﻲ اﻟﻤﻮﺿﻊ 6 ‪: M‬‬
‫‪2τ‬‬
‫وﺣﺴﺐ اﻟﺘﺴﺠﻴﻞ ﻧﺠﺪ ‪M 5 M 7 = 2.10 −2 m‬‬
‫2−01.2‬
‫‪v6 = 0,5m.s‬‬     ‫1−‬
‫= 6‪ v‬أي‬             ‫وﻣﻨﻪ‬
‫3−01.02 × 2‬
‫* اﺳﺘﻨﺘﺎج ﺗﻐﻴﺮ آﻤﻴﺔ اﻟﺤﺮآﺔ ′‪ Δp‬ﺑﻴﻦ 5 ‪ M‬و 6 ‪: M‬‬
‫5‪ Δp′ = p6 − p‬إذن : ) 5‪Δp′ = m1 (v6 − v‬‬         ‫ﻧﻌﻠﻢ أن :‬
‫وﺑﻤﺎ ان 6‪ v5 = v‬ﻓﺈن 0 = ′‪Δp‬‬
‫3- اﻟﻤﺮﺣﻠﺔ اﻟﺘﻲ ﻳﺘﺤﻘﻖ ﻓﻴﻬﺎ ﻣﺒﺪا اﻟﻘﺼﻮر:‬
‫ﻣﺒﺪأ اﻟﻘﺼﻮر ﻳﺘﺤﻘﻖ ﻓﻲ اﻟﻤﺮﺣﻠﺔ اﻟﺜﺎﻧﻴﺔ ﺑﻴﻦ 4 ‪ M‬و 7 ‪ M‬ﻷن ﺣﺮآﺔ ) 1‪ (S‬ﻓﻲ هﺬﻩ اﻟﻤﺮﺣﻠﺔ ﻣﺴﺘﻘﻴﻤﻴﺔ‬
‫ﻣﻨﺘﻈﻤﺔ.‬
‫4- ﺗﻤﺜﻴﻞ اﻟﻤﺘﺠﻬﺔ ‪ Δp‬ﺑﻴﻦ 1‪ M‬و 3 ‪: M‬‬
‫اﻟﻤﺘﺠﻬﺔ ‪ Δp‬ﻟﻬﺎ ﻧﻔﺲ اﺗﺠﺎﻩ وﻧﻔﺲ ﻣﻨﺤﻰ ﺣﺮآﺔ ) 1‪ (S‬و ﻣﻨﻈﻤﻬﺎ هﻮ : 1− ‪. Δp = 0, 24kg.m.s‬‬
‫وﺣﺴﺐ اﻟﺴﻠﻢ ﻳﻜﻮن ﻃﻮل اﻟﻤﺘﺠﻬﺔ ‪ Δp‬هﻮ ‪. 4cm‬‬
‫اﻟﺘﻤﺜﻴﻞ:‬

‫ﻣﻦ اﻟﺘﻤﺜﻴﻞ اﻟﺴﺎﺑﻖ ﻧﺴﺘﻨﺘﺞ ان ﻟﻤﺠﻤﻮع ﻣﺘﺠﻬﺎت اﻟﻘﻮى اﻟﻤﻄﺒﻘﺔ ﻋﻠﻰ ) 1‪ (S‬ﻧﻔﺲ اﺗﺠﺎﻩ وﻧﻔﺲ ﻣﻨﺤﻰ اﻟﻤﺘﺠﻬﺔ‬
‫‪. Δp‬‬

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