Qcm Les Ondes

‫ض  ا  ت ‪ :‬أ ذات ارات ‪!"#‬دة‬

‫ء ا  ا ‬

‫ذ ‪ .‬ا &ال‬

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-./0‬‬ ‫ﻳﺘﻜﻮﻥ ﺍﻟﻔﺮﺽ ﻣﻦ ﺛﻤﺎﻧﻴﺔ ﺗﻤﺎﺭﻳﻦ ﻛﻞ ﺗﻤﺮﻳﻦ ﻳﺘﺄﻟﻒ ﻣﻦ ﺃﺭﺑﻌﺔ ﺃﺳﺌﻠﺔ ﻗﺪ ﺗﻜﻮﻥ ﻣﺴﺘﻘﻠﺔ ‪ .‬ﺃﺟﺐ ﻋﻦ ﻛﻞ ﺳﺆﺍﻝ ﺏ‬ ‫ﺻﺤﻴﺢ ﺃﻭ ﺧﻄﺄ ﺣﺴﺐ ﺗﻘﺪﻳﺮﻙ ﺍﻟﺸﺨﺼﻲ ﻓﻲ ﺍﻟﺨﺎﻧﺔ ﺍﻟﻤﻘﺎﺑﻠﺔ ﻟﻠﺴﺆﺍﻝ ﺍﻟﻤﺒﻴﻦ ﻓﻲ ﺍﻟﺠﺪﻭﻝ ﺃﺳﻔﻠﻪ‬ ‫‪ .‬ﻛﻞ ﺗﻤﺮﻳﻦ ﻳﻨﻘﻂ ﺏ ‪2,5pts‬‬ ‫ﻛﻞ ﺳﺆﺍﻝ ﺧﺎﻃﺊ ﺃﻭ ﺑﺪﻭﻥ ﺇﺟﺎﺑﺔ ﺗﺨﺼﻢ ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ ‪ 1pts :‬ﻭ ‪. 0,5pts‬‬ ‫ﺍﻧﻄﻼﻗﺎ ﻣﻦ ﺧﻄﺄﻳﻦ ﻓﻲ ﻧﻔﺲ ﺍﻟﺘﻤﺮﻳﻦ ﻳﻤﻨﺢ ﺻﻔﺮ ﻟﻠﺘﻤﺮﻳﻦ‬ ‫ا*) وا '‪.............................................. :‬‬

‫ا‪, -‬‬ ‫ا ر‪.‬‬

‫‪a‬‬

‫ا  ى‪...............................‬‬

‫‪b‬‬

‫‪c‬‬

‫‪d‬‬

‫‪1./‬‬ ‫‪2 ./‬‬ ‫‪3./‬‬ ‫‪4./‬‬ ‫‪5./‬‬ ‫‪6./‬‬ ‫‪7./‬‬ ‫‪8./‬‬

‫ا ‪../20 :89‬‬ ‫‪1‬‬ ‫‪www.ibnalkhatib2.canalblog.com‬‬

‫ء ا  ا ‬

‫ض  ا  ت ‪ :‬أ ذات ارات ‪!"#‬دة‬

‫ذ ‪ .‬ا &ال‬

‫ﺗﻤﺮﻳﻦ ‪1‬‬ ‫ﺃﻱ ﺍﻻﻗﺘﺮﺍﺣﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ ﺗﻨﻄﺒﻖ ﻋﻠﻰ ﺩﺭﺍﺳﺔ ﺍﻧﺘﺸﺎﺭ ﻣﻮﺟﺔ ﻣﻴﻜﺎﻧﻴﻜﻴﺔ ﻣﺘﻮﺍﻟﻴﺔ ؟‬ ‫‪ (a‬ﺍﻧﺘﺸﺎﺭ ﺍﻟﻤﻮﺟﺔ ﻫﻮ ﺍﻧﺘﻘﺎﻝ ﻟﻠﻤﺎﺩﺓ ﻭﺍﻟﻄﺎﻗﺔ‬ ‫‪ (b‬ﺳﺮﻋﺔ ﺍﻻﻧﺘﺸﺎﺭ ﺗﺘﻌﻠﻖ ﺑﻄﺒﻴﻌﺔ ﺍﻟﻮﺳﻂ ) ﻗﺼﻮﺭ ﺍﻟﻮﺳﻂ (‬ ‫‪ (c‬ﻟﺤﺴﺎﺏ ﺍﻟﺘﺄﺧﺮ ﺍﻟﺰﻣﻨﻲ ﻟﻤﻮﺟﺔ ﺻﻮﺗﻴﺔ ﻋﻨﺪ ﻧﻘﻄﺔ ﻣﺎ ﻋﻦ ﻣﻨﺒﻊ ﻧﺴﺘﻌﻤﻞ ﺭﺍﺳﻢ ﺍﻟﺘﺬﺑﺬﺏ ﻭﺯﺭ ﺍﻟﺤﺴﺎﺳﻴﺔ ﺍﻷﻓﻘﻴﺔ‬ ‫)ﺍﻟﻜﺴﺢ ( ﺃﻭ ﺯﺭ ﺍﻟﺤﺴﺎﺳﻴﺔ ﺍﻟﺮﺃﺳﻴﺔ ‪.‬‬ ‫‪ (d‬ﻋﻨﺪ ﺗﻼﻗﻲ ﻣﻮﺟﺘﻴﻦ ﻣﺘﻘﺎﺑﻠﺘﻴﻦ )ﺇﺷﺎﺭﺓ ﺍﺳﺘﻄﺎﻟﺘﻴﻬﻤﺎ ﻣﺘﻘﺎﺑﻠﺘﻴﻦ ( ﻳﺤﺪﺙ ﺇﺗﻼﻑ ﻟﻠﻤﻮﺟﺔ ‪.‬‬ ‫ﺗﻤﺮﻳﻦ ‪2‬‬ ‫ﻫﺬﻩ ﺃﺭﺑﻊ ﺍﻗﺘﺮﺍﺣﺎﺕ ﺗﻬﻢ ﺍﻧﺘﺸﺎﺭ ﺍﻟﺼﻮﺕ ﻓﻲ ﺍﻟﻬﻮﺍﺀ ‪:‬‬ ‫‪ (a‬ﻋﺒﺎﺭﺓ ﻋﻦ ﺍﻫﺘﺰﺍﺯ ﻗﺮﺏ ﺑﻘﺮﺏ ﻟﻠﺠﺰﻳﺌﺎﺕ ﺍﻟﻤﻜﻮﻧﺔ ﻟﻠﻬﻮﺍﺀ ‪.‬‬ ‫‪ (b‬ﻫﺬﺍ ﺍﻻﻫﺘﺰﺍﺯ ﻋﺒﺎﺭﺓ ﻋﻦ ﺍﻧﺘﻘﺎﻝ ﻣﺘﻌﺎﻣﺪ ﻣﻊ ﺍﺗﺠﺎﻩ ﺍﻻﻧﺘﺸﺎﺭ ‪.‬‬ ‫‪ (c‬ﻃﻮﻝ ﻣﻮﺟﺔ ﺻﻮﺗﻴﺔ ﺩﻭﺭﻳﺔ ﻻ ﺗﺘﻌﻠﻖ ﺑﺎﻟﺘﺮﺩﺩ ‪. N‬‬ ‫‪ (d‬ﻓﻲ ﻧﻔﺲ ﺍﻟﻮﺳﻂ ﻳﺴﺘﻤﻊ ﻣﻼﺣﻆ ﺍﻟﺼﻮﺕ ﺍﻟﺤﺎﺩ ﻗﺒﻞ ﺍﻟﺼﻮﺕ ﺍﻟﺨﻔﻴﺾ ﻭﺍﻟﻠﺬﻳﻦ ﻳﻨﺒﻌﺜﺎﻥ ﻣﻦ ﻧﻔﺲ ﺍﻟﻤﻨﺒﻊ ‪.‬‬ ‫ﺗﻤﺮﻳﻦ ‪3‬‬ ‫ﺳﺮﻋﺔ ﺍﻧﺘﺸﺎﺭ ﺍﻟﺼﻮﺕ ﻋﻠﻰ ﺳﻄﺢ ﺍﻷﺭﺽ ﺗﺴﺎﻭﻱ‪ ، 340m.s-1‬ﺑﻴﻨﻤﺎ ﺳﺮﻋﺔ ﻃﺎﺋﺮﺓ ﺍﻟﻜﻮﻧﻜﻮﺭﺩ ﻭﺍﻟﺘﻲ ﺗﻄﻴﺮ ﺑﺴﺮﻋﺔ‬ ‫ﺍﻟﺼﻮﺕ ﻋﻠﻰ ﺍﺭﺗﻔﺎﻉ ﻃﻴﺮﺍﻧﻬﺎ ﺍﻟﻌﺎﺩﻱ ﺑﺴﺮﻋﺔ ‪. (mach 1) 1100Km.h-1‬‬ ‫ﻧﻌﻄﻲ ‪ 3,6×3,4 = 12,24 :‬ﻭﺑﺎﺳﺘﻌﻤﺎﻝ ﺍﻟﻤﻌﻄﻴﺎﺕ ﺍﻟﺴﺎﺑﻘﺔ ‪.‬‬ ‫‪ (a‬ﺳﺮﻋﺔ ﺍﻧﺘﺸﺎ ﺭ ﺍﻟﺼﻮﺕ ﻻ ﺗﺘﻌﻠﻖ ﺑﺎﻻﺭﺗﻔﺎﻉ ‪.‬‬ ‫‪ (b‬ﺍﻧﺤﻔﺎﻅ ﺍﻟﻀﻐﻂ ﺍﻟﺠﻮﻱ ﻳﺴﺒﺐ ﺍﺯﺩﻳﺎﺩ ﺳﺮﻋﺔ ﺍﻟﺼﻮﺕ ‪.‬‬ ‫‪ (c‬ﻣﺪﺓ ﻃﻴﺮﺍﻥ ﺍﻟﻜﻮﻧﻜﻮﺭﺩ ﺑﻴﻦ ﺑﺎﺭﻳﺲ ﻭﻧﻴﻮﻳﻮﺭﻙ ) ‪ ( 6000Km‬ﻭﺍﻟﺘﻲ ﺗﻄﻴﺮ ﺑﺴﺮﻋﺔ )‪(mach2 =2mach1‬‬ ‫ﺃﻗﻞ ﻣﻦ ‪ 3H‬ﺛﻼﺙ ﺳﺎﻋﺎﺕ ‪.‬‬ ‫‪ '' (d‬ﺿﺠﻴﺞ '' ﻃﺎﺋﺮﺓ ﺍﻟﻜﻮﻧﻜﻮﺭﺩ ﻭﺍﻟﺘﻲ ﺗﻨﺘﻘﻞ ﺑﺴﺮﻋﺔ ‪ mach1‬ﻭﻋﻠﻰ ﺍﺭﺗﻔﺎﻉ ‪ 6,6Km‬ﻣﻦ ﻣﻼﺣﻆ ﻳﺴﻤﻊ ﺑﻌﺪ ‪6s‬‬ ‫ﺳﺘﺔ ﺛﻮﺍﻧﻲ ﻣﻦ ﻣﺮﻭﺭﻫﺎ ﻓﻮﻕ ﺭﺃﺳﻪ‪.‬‬ ‫ﺗﻤﺮﻳﻦ ‪4‬‬ ‫ﻧﺴﺘﻌﻤﻞ ﻣﻮﺟﺎﺕ ﻓﻮﻕ ﺻﻮﺗﻴﺔ ﺫﺍﺕ ﺗﺮﺩﺩ ‪ N = 40KHz‬ﺳﺮﻋﺔ ﺍﻧﺘﺸﺎﺭﻫﺎ ﻓﻲ ﺧﻼﻝ ﻫﺬﻩ ﺍﻟﻤﻼﺣﻈﺔ ﻫﻲ‪340ms-1 :‬‬ ‫‪ (a‬ﻃﻮﻝ ﻣﻮﺟﺔ ﺍﻟﻤﻮﺟﺎﺕ ﻓﻮﻕ ﺍﻟﺼﻮﺗﻴﺔ ‪. 8,5mm :‬‬ ‫‪ (b‬ﺍﻟﻤﺴﺎﻓﺔ ﺍﻟﻤﻘﻄﻮﻋﺔ ﺧﻼﻝ ﺩﻭﺭ ﻭﺍﺣﺪ ‪ T‬ﻫﻲ ‪. 8,5mm‬‬ ‫‪ (c‬ﻳﺘﻐﻴﺮ ﺍﻟﺘﺮﺩﺩ ﺇﺫﺍ ﻏﻴﺮﻧﺎ ﻏﺎﺯ ﺍﻟﻮﺳﻂ ﺍﻟﺬﻱ ﺗﻨﺘﺸﺮ ﻓﻴﻪ ﺍﻟﻤﻮﺟﺔ ﻓﻮﻕ ﺍﻟﺼﻮﺗﻴﺔ ‪.‬‬ ‫‪ (d‬ﺇﺫﺍ ﺗﻨﺎﻗﺺ ﺍﻟﺘﺮﺩﺩ ﺑﺎﻟﻨﺼﻒ ﺗﻨﺎﻗﺼﺖ ﺍﻟﺴﺮﻋﺔ ﺑﺎﻟﻨﺼﻒ ‪.‬‬ ‫‪2‬‬ ‫‪www.ibnalkhatib2.canalblog.com‬‬

‫ء ا  ا ‬

‫ض  ا  ت ‪ :‬أ ذات ارات ‪!"#‬دة‬

‫ذ ‪ .‬ا &ال‬

‫ﺗﻤﺮﻳﻦ‪5‬‬ ‫‪.‬‬

‫ﺗﻤﺜﻞ ﺍﻟﻮﺛﻴﻘﺔ ﺟﺎﻧﺒﻪ ﻣﻌﺎﻳﻨﺔ ﻣﻮﺟﺔ ﻓﻮﻕ ﺻﻮﺗﻴﺔ ﺑﻮﺍﺳﻄﺔ ﺭﺍﺳﻢ ﺍﻟﺘﺬﺑﺬﺏ‬ ‫ﺍﻟﻘﻴﻢ ﺍﻟﺘﻲ ﺿﺒﻂ ﻋﻠﻴﻬﺎ ﺍﻟﺠﻬﺎﺯ ﻫﻲ ‪:‬‬ ‫ﺍﻟﻜﺴﺢ ‪ 50µs/div :‬ﻭ ﺍﻟﺤﺴﺎﺳﻴﺔ ﺍﻟﺮﺃﺳﻴﺔ ‪ 2V/div :‬ﺗﻈﻬﺮ ﺍﻟﻤﻮﺟﺔ‬ ‫ﻋﻠﻰ ﺷﺎﺷﺔ ﺭﺍﺳﻢ ﺍﻟﺘﺬﺑﺬﺏ ﺑﺘﺄﺧﺮ ‪2div‬‬ ‫‪ (a‬ﺗﺮﺩﺩ ﺍﻟﻤﻮﺟﺔ ﺍﻟﻤﻌﺎﻳﻨﺔ ﻫﻮ ‪.40KHz‬‬ ‫‪ (b‬ﻟﻤﻌﺎﻳﻨﺔ ﺗﺬﺑﺬﺑﻴﻦ ﺃﻭ ﺛﻼﺛﺔ ﻓﻘﻂ ﻋﻠﻰ ﺷﺎﺷﺔ ﺭﺍﺳﻢ ﺍﻟﺘﺬﺑﺬﺏ ﻳﺠﺐ ﺃﻥ ﻧﺨﻔﺾ ﺍﻟﺤﺴﺎﺳﻴﺔ ﺍﻷﻓﻘﻴﺔ )ﻣﺜﻼ ‪(5µs/div‬‬ ‫‪ (c‬ﻭﺳﻊ ﺍﻟﻤﻮﺟﺔ ) ﺍﻻﺳﺘﻄﺎﻟﺔ ﺍﻟﻘﺼﻮﻳﺔ ( ﻳﺒﻘﻰ ﺛﺎﺑﺘﺎ ﺧﻼﻝ ﺍﻟﺰﻣﻦ‬ ‫‪ (d‬ﻭﺳﻊ ﺍﻟﻤﻮﺟﺔ ﻻ ﻳﺘﺠﺎﻭﺯ ‪. 2V‬‬ ‫ﺗﻤﺮﻳﻦ ‪6‬‬ ‫ﻧﺤﺪﺙ ﻓﻲ ﺣﻮﺽ ﻟﻠﻤﻮﺟﺎﺕ ﻣﻮﺟﺔ ﻣﻴﻜﺎﻧﻴﻜﻴﺔ ﺩﺍﺋﺮﻳﺔ‬ ‫ﺗﻨﺘﺸﺮ ﺑﺴﺮﻋﺔ ‪ ،V = 18,7cm.s-1‬ﺗﻤﺜﻞ‬ ‫ﺍﻟﺼﻮﺭﺓ ﺟﺎﻧﺒﻪ ﻟﺠﺰﺀ ﻣﻦ ﺍﻟﺤﻮﺽ ﻓﻲ ﺣﺎﻟﺔ ﺳﻜﻮﻥ ﻋﻨﺪ‬ ‫ﺗﺮﺩﺩ ﺍﻟﻮﻣﺎﺽ ﻋﻠﻰ ‪ . Ne‬ﺭﺳﻤﺖ ﻋﻠﻰ ﺍﻟﺼﻮﺭﺓ ﺇﺷﺎﺭﺗﻴﻦ‬ ‫ﺗﻔﺼﻞ ﺑﻴﻨﻬﻤﺎ ﻣﺴﺎﻓﺔ ‪ 10cm‬ﻛﺴﻠﻢ ﻟﻠﻘﻴﺎﺱ ‪.‬‬ ‫‪ (a‬ﻃﻮﻝ ﻣﻮﺟﺔ ‪ λ = 1,7cm‬ﺑﺘﻘﺮﻳﺐ ‪. 0,1cm‬‬ ‫‪ (b‬ﺗﺮﺩﺩ ﺍﻟﻤﻨﺒﻊ ﻫﻮ ‪ N = 11Hz‬ﺑﺘﻘﺮﻳﺐ ‪. 1 Hz‬‬ ‫‪ (c‬ﻋﻨﺪ ﺿﺒﻂ ﺗﺮﺩﺩ ﺍﻟﻮﻣﺎﺽ ﻋﻠﻰ ﺗﺮﺩﺩ ‪ N’e = 10 Hz‬ﻧﻌﺎﻳﻦ ﺣﺮﻛﺔ ﺑﻄﻴﺌﺔ ﻓﻲ ﺍﻟﻤﻨﺤﻰ ﺍﻟﻤﻌﺎﻛﺲ ‪.‬‬ ‫‪ (d‬ﺳﺮﻋﺔ ﺍﻧﺘﺸﺎﺭ ﺍﻟﺤﺮﻛﺔ ﺍﻟﺒﻄﻴﺌﺔ ﺍﻟﺴﺎﺑﻘﺔ ﺑﻴﺮﻋﺔ ﻇﺎﻫﺮﻳﺔ ‪Va = 1,7 cm.s-1‬‬ ‫ﺗﻤﺮﻳﻦ ‪7‬‬ ‫ﻫﺬﻩ ﺃﺭﺑﻊ ﺍﻗﺘﺮﺍﺣﺎﺕ ﺗﻬﻢ ﺍﻟﻀﻮﺀ ‪:‬‬ ‫‪ (a‬ﺍﻟﻀﻮﺀ ﻋﺒﺎﺭﺓ ﻋﻦ ﻣﻮﺟﺔ ﻣﺴﺘﻌﺮﺿﺔ ‪ ،‬ﺗﻨﺘﺸﺮ ﻓﻲ ﺟﻤﻴﻊ ﺍﻷﻭﺳﺎﻁ ﺑﻨﻔﺲ ﺍﻟﺴﺮﻋﺔ ‪.‬‬ ‫‪ (b‬ﺍﻟﻀﻮﺀ ﺍﻷﺣﺎﺩﻱ ﺍﻟﻠﻮﻥ ﺍﻟﻤﻨﺒﻌﺚ ﻣﻦ ﺟﻬﺎﺯ ﺍﻟﻠﻴﺰﺭ ‪ Laser‬ﻋﺒﺎﺭﺓ ﻋﻦ ﺇﺷﻌﺎﻋﺎﺕ ﻟﻬﺎ ﻧﻔﺲ ﻃﻮﻝ ﺍﻟﻤﻮﺟﺔ ﻟﻜﻦ‬ ‫ﺑﺘﺮﺩﺩﺍﺕ ﻣﺨﺘﻠﻔﺔ‪.‬‬ ‫‪ (c‬ﺗﺒﺪﺩ ﺍﻟﻀﻮﺀ ﺍﻷﺑﻴﺾ ﺑﻮﺍﺳﻄﺔ ﻣﻮﺷﻮﺭ ﻳﻈﻬﺮ ﺃﻥ ﻣﻌﺎﻣﻞ ﺍﻧﻜﺴﺎﺭ ﻭﺳﻂ ﻳﺘﻐﻴﺮ ﺑﺘﻐﻴﺮ ﺍﻟﺘﺮﺩﺩ ‪.‬‬ ‫‪ (d‬ﺍﻟﻈﺎﻫﺮﺓ ﺍﻟﻔﻴﺰﻳﺎﺋﻴﺔ ﺍﻟﻤﻌﺎﻳﻨﺔ ﻋﻨﺪ ﺍﺟﺘﻴﺎﺯ ﺿﻮﺀ ﺍﻟﻠﻴﺰﺭ ﻓﺘﺤﺔ ﻫﻲ ﻇﺎﻫﺮﺓ ﺍﻹﻧﻜﺴﺎﺭ‪.‬‬

‫‪3‬‬ ‫‪www.ibnalkhatib2.canalblog.com‬‬

‫ء ا  ا ‬

‫ض  ا  ت ‪ :‬أ ذات ارات ‪!"#‬دة‬

‫ذ ‪ .‬ا &ال‬

‫ﺗﻤﺮﻳﻦ ‪8‬‬ ‫ﻧﻀﻲﺀ ﺷﻘﺎ ﻋﺮﺿﻪ ‪ a‬ﺑﻀﻮﺀ ﺃﺣﻤﺮ ﻃﻮﻝ ﻣﻮﺟﺘﻪ ‪ ، λ =690 nm‬ﻧﺸﺎﻫﺪ ﻋﻠﻰ ﺷﺎﺷﺔ ﺗﺒﻌﺪ ﻋﻨﻪ ﺏ ‪ D = 2 m‬ﺑﻘﻌﺔ‬ ‫ﻣﺮﻛﺰﻳﺔ ﻋﺮﺿﻬﺎ ‪ L‬ﺣﻮﻟﻬﺎ ﺑﻘﻊ ﻋﺮﺿﻬﺎ ‪ . L / 2‬ﻗﻴﺎﺱ ‪10L /2 =2,3 cm :‬‬ ‫‪ (a‬ﺍﻟﻔﺮﻕ ﺍﻟﺰﺍﻭﻱ ‪ θ‬ﻳﺰﺩﺍﺩ ﺑﺎﺯﺩﻳﺎﺩ ﻋﺮﺽ ﺍﻟﺸﻖ ‪. a‬‬ ‫‪ (b‬ﺍﻟﻔﺮﻕ ﺍﻟﺰﺍﻭﻱ ‪ θ‬ﻳﺰﺩﺍﺩ ﺑﺎﺯﺩﻳﺎﺩ ﺍﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ ﺍﻟﺸﻖ ﻭﺍﻟﺸﺎﺷﺔ ‪. D‬‬ ‫‪ (c‬ﻋﺮﺽ ﺍﻟﺸﻖ ﻳﺴﺎﻭﻱ ‪. a = 0,6 cm :‬‬ ‫‪ (d‬ﺍﻟﻔﺮﻕ ﺍﻟﺰﺍﻭﻱ ‪ θ‬ﻳﺰﺩﺍﺩ ﺑﺎﺳﺘﻌﻤﺎﻝ ﺍﻟﻀﻮﺀ ﺍﻷﺯﺭﻕ ﺑﺪﻝ ﺍﻟﻀﻮﺀ ﺍﻷﺣﻤﺮ ‪.‬‬

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‫‪4‬‬ ‫‪www.ibnalkhatib2.canalblog.com‬‬

‫ض  ا  ت ‪ :‬أ ذات ارات ‪!"#‬دة‬

‫ء ا  ا ‬

‫ذ ‪ .‬ا &ال‬

‫ﺍﻟﺘﺼﺤﻴﺢ‬ ‫ا‪, -‬‬

‫‪a‬‬

‫ا ر‪.‬‬ ‫‪1./‬‬

‫‪b‬‬

‫=<;‬

‫‪:8‬‬

‫‪d‬‬

‫‪c‬‬

‫‪:8‬‬

‫‪:8‬‬

‫‪:8‬‬

‫‪:8‬‬

‫‪:8‬‬

‫‪3./‬‬

‫‪:8‬‬

‫‪:8‬‬

‫=<;‬

‫‪:8‬‬

‫‪4./‬‬

‫=<;‬

‫=<;‬

‫‪:8‬‬

‫‪:8‬‬

‫‪5./‬‬

‫=<;‬

‫=<;‬

‫‪:8‬‬

‫‪:8‬‬

‫‪6./‬‬

‫‪:8‬‬

‫‪:8‬‬

‫=<;‬

‫‪:8‬‬

‫‪7./‬‬

‫‪:8‬‬

‫‪:8‬‬

‫=<;‬

‫‪:8‬‬

‫‪8./‬‬

‫‪:8‬‬

‫‪:8‬‬

‫=<;‬

‫‪:8‬‬

‫‪2 ./‬‬

‫=<;‬

‫ﺍﻧﺘﻈﺮﻭﺍ ﻻﺣﻘﺎ ﺇﺟﺎﺑﺎﺕ ﻣﻌﻠﻠﺔ ﻭﻣﻔﺼﻠﺔ ﺑﻨﻔﺲ ﺍﻟﻤﻮﻗﻊ‬

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