ÌâÄ¿ÄÚÈÝ

3£®Ä³¿ÎÍâ»î¶¯Ð¡×éÀûÓÃСÇòµÄÊúÖ±ÉÏÅ×Ô˶¯ÑéÖ¤»úеÄÜÊØºã¶¨ÂÉ£®Èçͼ¼×£¬µ¯Éä×°Öý«Ð¡ÇòÊúÖ±ÏòÉϵ¯³ö£¬ÏȺóͨ¹ý¹âµçÃÅA¡¢B£¬¹âµç¼ÆÊ±Æ÷²â³öСÇòÉÏÉý¹ý³ÌÖÐͨ¹ýA¡¢BµÄʱ¼ä·Ö±ðΪ¡÷tA¡¢¡÷tB£¬Óÿ̶ȳ߲â³ö¹âµçÃÅA¡¢B¼äµÄ¾àÀëΪh£¬ÓÃÂÝÐý²â΢Æ÷²âÁ¿Ð¡ÇòµÄÖ±¾¶d£¬Ä³´Î²âÁ¿½á¹ûÈçͼÒÒ£¬Æä¶ÁÊýd=12.987mm£®µ±µØµÄÖØÁ¦¼ÓËÙ¶ÈΪg£®ÔÚÎó²î·¶Î§ÄÚ£¬ÈôСÇòÉÏÉý¹ý³ÌÖлúеÄÜÊØºã£¬ÔòÌâÖиø³öµÄÎïÀíÁ¿d¡¢¡÷tA¡¢¡÷tB¡¢g¡¢hÖ®¼äÓ¦Âú×ãµÄ¹ØÏµÊ½Îª${£¨\frac{d}{{¡÷{t_A}}}£©^2}-{£¨\frac{d}{{¡÷{t_B}}}£©^2}=2gh$£®

·ÖÎö ÂÝÐý²â΢Æ÷µÄ¶ÁÊýµÈÓڹ̶¨¿Ì¶È¶ÁÊý¼ÓÉϿɶ¯¿Ì¶È¶ÁÊý£¬Ðè¹À¶Á£®¸ù¾Ý¼«¶Ìʱ¼äÄ򵀮½¾ùËٶȵÈÓÚ˲ʱËÙ¶ÈÇó³öСÇòͨ¹ý¹âµçÃÅA¡¢BµÄËÙ¶È£¬½áºÏÖØÁ¦ÊÆÄܵÄÔö¼ÓÁ¿ºÍ¶¯ÄܵļõСÁ¿ÏàµÈµÃ³ö»úеÄÜÊØºãÂú×ãµÄ¹ØÏµÊ½£®

½â´ð ½â£ºÂÝÐý²â΢Æ÷µÄ¹Ì¶¨¿Ì¶È¶ÁÊýΪ12.5mm£¬¿É¶¯¿Ì¶È¶ÁÊýΪ0.01¡Á48.7mm=0.487mm£¬Ôòd=12.987mm£®
СÇòͨ¹ý¹âµçÃÅA¡¢BµÄËÙ¶È·Ö±ðΪ${v}_{A}=\frac{d}{¡÷{t}_{A}}$£¬${v}_{B}=\frac{d}{¡÷{t}_{B}}$£¬
Ôò¶¯ÄܵļõСÁ¿$¡÷{E}_{k}=\frac{1}{2}m£¨{{v}_{A}}^{2}-{{v}_{B}}^{2}£©$£¬
ÖØÁ¦ÊÆÄܵÄÔö¼ÓÁ¿Îª¡÷Ep=mgh£¬
¸ù¾Ý¡÷Ek=¡÷EpÖª£¬${£¨\frac{d}{{¡÷{t_A}}}£©^2}-{£¨\frac{d}{{¡÷{t_B}}}£©^2}=2gh$£®
¹Ê´ð°¸Îª£º12.987£¬${£¨\frac{d}{{¡÷{t_A}}}£©^2}-{£¨\frac{d}{{¡÷{t_B}}}£©^2}=2gh$£®

µãÆÀ ½â¾ö±¾ÌâµÄ¹Ø¼üÖªµÀʵÑéµÄÔ­Àí£¬±¾Ìâץס֨Á¦ÊÆÄܵÄÔö¼ÓÁ¿ºÍ¶¯ÄܵļõСÁ¿ÊÇ·ñÏàµÈÑéÖ¤»úеÄÜÊØºã£¬ÖªµÀ¼«¶Ìʱ¼äÄ򵀮½¾ùËٶȵÈÓÚ˲ʱËÙ¶È£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø