·ÖÎö ¸ù¾ÝÏàÁÚµÄÏàµÈʱ¼ä¼ä¸ôλÒÆÖ®²îÊÇ·ñÏàµÈÀ´ÅжÏС³µÊÇ·ñ×öÔȱäËÙÖ±ÏßÔ˶¯£®
Ö½´øʵÑéÖУ¬ÈôÖ½´øÔȱäËÙÖ±ÏßÔ˶¯£¬²âµÃÖ½´øÉϵĵã¼ä¾à£¬ÀûÓÃÔȱäËÙÖ±ÏßÔ˶¯µÄÍÆÂÛ£¬¿É¼ÆËã³ö´ò³öijµãʱֽ´øÔ˶¯µÄ˲ʱËٶȺͼÓËٶȣ®
½â´ð ½â£º£¨1£©ÏàÁÚµÄÏàµÈʱ¼ä¼ä¸ôλÒÆÖ®²îÏàµÈ£¬¼´¡÷x=at2=³£Êý£¬ËùÒÔС³µ×öÔȼÓËÙÖ±ÏßÔ˶¯£®
£¨2£©ÀûÓÃÔȱäËÙÖ±ÏßÔ˶¯µÄÍÆÂÛ£ºÖмäʱ¿ÌµÄ¼´Ê±ËٶȵÈÓÚÕâ¶Îʱ¼äÄÚµÄƽ¾ùËٶȣ¬µÃ£º
vB=$\frac{{x}_{AC}^{\;}}{{t}_{AC}^{\;}}$=$\frac{17.50}{2¡Á0.1}¡Á1{0}_{\;}^{-2}m/s$=0.875m/s
${v}_{D}^{\;}=\frac{{x}_{CE}^{\;}}{{t}_{CE}^{\;}}=\frac{49.00-17.50}{2¡Á0.1}¡Á1{0}_{\;}^{-2}m/s=1.575m/s$
£¨3£©ÓÉÓÚÏàÁڵļÆÊýµãµÄ¾àÀëÖ®²îÏàµÈ£¬¼´XBC-XAB=XCD-XBC=XDE-XCD=3.5cm
¸ù¾ÝÔ˶¯Ñ§¹«Ê½µÃ£º¡÷x=at2£¬
a=$\frac{{x}_{BC}^{\;}-{x}_{AB}^{\;}}{{T}_{\;}^{2}}$=3.5m/s2
¹Ê´ð°¸Îª£º£¨1£©¡÷x=at2=³£Êý£¬ÔȱäËÙÖ±Ïߣ»£¨2£©Öмäʱ¿ÌµÄ¼´Ê±ËٶȵÈÓÚÕâ¶Îʱ¼äÄÚµÄƽ¾ùËٶȣ¬0.875£¬1.575£¬£¨3£©3.5$m/{s}_{\;}^{2}$£®
µãÆÀ ҪעÒⵥλµÄ»»Ë㣮¶ÔÓÚÖ½´øµÄÎÊÌ⣬ÎÒÃÇÒªÊìϤÔȱäËÙÖ±ÏßÔ˶¯µÄÌصãºÍһЩ¹æÂÉ£®
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£º¶àÑ¡Ìâ
A£® | ÎÞÂÛµ¼ÌåÅõÈçºÎÔ˶¯£¬×îÖÕµç·ÖеĵçÁ÷I=$\frac{mg}{BL}$ | |
B£® | µ¼Ìå°ô¼õÉÙµÄÖØÁ¦ÊÆÄÜÈ«²¿±ä³Éµç×èR²úÉúµÄ½¹¶úÈÈ | |
C£® | µ¼Ìå°ôµÄ¼ÓËٶȿÉÄÜÏÈÔö¼Óºó¼õС£¬×îÖÕ¼õСµ½Áã | |
D£® | µ¼Ìå°ôµÄËٶȿÉÄÜÏÈÔö¼Ó£¬Ò²¿ÉÄÜÏȼõС£¬Ò²¿ÉÄÜÒ»Ö±²»±ä |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£º¼ÆËãÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | ¢Ù¢Ú | B£® | ¢Û¢Ü | C£® | ¢Ù¢Û | D£® | ¢Ú¢Ü |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£º¼ÆËãÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | 45¡ã | B£® | 30¡ã | C£® | 60¡ã | D£® | 75¡ã |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | С»·AµÄ¼ÓËٶȴóСΪ$\frac{\sqrt{3}k{q}^{2}}{2m{l}^{2}}$ | B£® | С»·AµÄ¼ÓËٶȴóСΪ$\frac{\sqrt{3}k{q}^{2}}{3m{l}^{2}}$ | ||
C£® | ºãÁ¦FµÄ´óСΪ$\frac{\sqrt{3}k{q}^{2}}{3{l}^{2}}$ | D£® | ºãÁ¦FµÄ´óСΪ$\frac{\sqrt{3}k{q}^{2}}{2{l}^{2}}$ |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£º¶àÑ¡Ìâ
A£® | ÐÇÇòAµÄÖÊÁ¿Ò»¶¨´óÓÚBµÄÖÊÁ¿ | |
B£® | ÐÇÇòAµÄÏßËÙ¶ÈÒ»¶¨´óÓÚBµÄÏßËÙ¶È | |
C£® | Ë«ÐǼä¾àÀëÒ»¶¨£¬Ë«ÐǵÄ×ÜÖÊÁ¿Ô½´ó£¬Æäת¶¯ÖÜÆÚԽС | |
D£® | Ë«ÐǵÄ×ÜÖÊÁ¿Ò»¶¨£¬Ë«ÐÇÖ®¼äµÄ¾àÀëÔ½´ó£¬Æäת¶¯ÖÜÆÚԽС |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | ÔÚÐÇÇò±íÃæËùÄÜ´ïµ½µÄ×î´ó¸ß¶ÈÊÇ480m | |
B£® | ÊúÖ±Éý¿ÕʱλÒÆ×î´óµÄʱ¿ÌÊÇ8sÄ© | |
C£® | ÊúÖ±Éý¿ÕºóËÙ¶È×î´óµÄʱ¿ÌÊÇ8sÄ©ºÍ40sÄ© | |
D£® | ̽²âÆ÷ÔÚ48sÄ©·µ»ØÐÇÇò±íÃæ |
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com