ÌâÄ¿ÄÚÈÝ

14£®Èçͼ£¬ÖÊÁ¿m=1¡Á10-3kg¡¢´øµçÁ¿q=1¡Á10-2CµÄ´øµçÁ£×Ó´ÓÊúÖ±·ÅÖõÄÁ½µçÈÝÆ÷¼«°åABÖ®¼äÌù×ÅA¼«°åÒÔËÙ¶Èvx=4m/sƽÐм«°å·ÉÈëÁ½¼«°å¼ä£¬Ç¡´Ó¼«°åBÉϱßÔµOµã·É³ö£¬ÒÑÖª¼«°å³¤L=0.4m£¬¼«°å¼ä¾àd=0.15m£®µçÈÝÆ÷¼«°åÉÏ·½Óпí¶ÈΪx=0.3mµÄÇøÓò±»Æ½¾ù·ÖÎªÇøÓò¢ñ¡¢¢ò¡¢¢ó£¬ÆäÖТñ¡¢¢óÓÐÔÈÇ¿´Å³¡£¬ËüÃǵĴŸÐÇ¿¶È´óСÏàµÈ£¬¾ù´¹Ö±Ö½ÃæÇÒ·½ÏòÏà·´£¬OΪDC±ßÖе㣬PΪDC±ßÖд¹ÏßÉÏÒ»µã£¬´øµçÁ£×Ó´ÓOµãÀ뿪µç³¡£¬Ö®ºó½øÈë´Å³¡£¬Ô˶¯¹ì¼£¸ÕºÃÓëÇøÓò¢óµÄÓұ߽çÏàÇУ¬²»¼ÆÁ£×ÓµÄÖØÁ¦£®Çó£º
£¨1£©¸ÃµçÈÝÆ÷¼«°åABËù¼ÓµçѹU´óС£»
£¨2£©ÔÈÇ¿´Å³¡µÄ´Å¸ÐӦǿ¶È´óСB£»
£¨3£©ÈôÏÖÔÚ¢ñ¡¢¢óÇøÓòËù¼Ó´Å¸ÐӦǿ¶È´óСB¡ä=2T£¬Á£×ÓÉäÈëOµãºó¾­¹ý3´Îƫת´òµ½Pµã£¬ÔòOPµÄ¾àÀëΪ¶àÉÙ£¿

·ÖÎö £¨1£©´øµçÁ£×ÓÔÚAB¼ä×öÀàÆ½Å×Ô˶¯Ç¡´ÓB°å±ßÔµÉä³ö£¬¸ù¾ÝÁ£×Ó×öÀàÆ½Å×Ô˶¯µÄˮƽ¡¢ÊúÖ±Î»ÒÆºÍ³õËٶȾÍÄÜÇóµÃ¼ÓËÙ¶È£¬´Ó¶øÇó³öƫתµçѹµÄ´óС£®
£¨2£©ÓÉÀàÆ½Å×Ô˶¯µÄÌØÕ÷£¬´øµçÁ£×Ӵӵ糡Éä³öʱºÃÏóÊÇ´ÓÔÈËÙÎ»ÒÆµÄÖеãÉä³ö£¬ÓÉ´Ë¿ÉÒԺܼòµ¥µØÇó³öƫת½ÇµÄ´óС£¬ÔÙ¸ù¾ÝÌâÒâÓë±ß½çÏàÇдӶøÖªµÀ¾ÍÄÜÇó³ö´øµçÁ£×ÓÔڴų¡ÖÐÔÈËÙÔ²ÖÜÔ˶¯µÄ°ë¾¶£¬ÓÉÂåÂØ×ÈÁ¦ÌṩÏòÐÄÁ¦¾ÍÄÜÇó³ö´Å¸ÐӦǿ¶È´óС£®
£¨3£©Óɼ¸ºÎ¹ØÏµ£¬´øµçÁ£×Ó¾­¹ýÈý´ÎÔ²ÖÜÔ˶¯ºÍÈý´ÎÔÈËÙÖ±ÏßÔ˶¯µ½´ïPµã£¬ÓÉÈý½Çº¯ÊýºÜÈÝÒ×±íʾ³öOPÖ®¼äµÄ¾àÀ룮

½â´ð ½â£º£¨1£©ÔÚAB¼«°å¼äÀàÆ½Å×£¬
  L=vxt
  $d=\frac{1}{2}a{t}^{2}=\frac{1}{2}¡Á\frac{Uq}{dm}¡Á£¨\frac{L}{{v}_{x}}£©^{2}$
 ´úÈëÊý¾ÝÓУºU=0.45V
£¨2£©ÉèÁ£×Ó³ö¼«°åºóËÙ¶È´óСΪv£¬Óëˮƽ¼Ð½Ç¦Á
  $tan¦Á=\frac{\frac{L}{2}}{d}=\frac{4}{3}$     ËùÒÔ£º$v=\frac{{v}_{x}}{sin¦Á}=\frac{4}{\frac{4}{5}}m/s=5m/s$
  ½øÈëÓұߴų¡Ç¡ÓëÓұ߽çÏàÇУ¬ÉèÔڴų¡ÖÐÔ²Ô˶¯°ë¾¶Îªr
   ¹ÊÓУº$sin¦Á=\frac{r-0.1}{r}=\frac{4}{5}$   
   ½âµÃ£ºr=0.5
   ¶ÔÁ£×Ó£º$Bqv=m\frac{{v}^{2}}{r}$
  ËùÒÔ£º$B=\frac{mv}{qr}$   ´úÈëÊý¾ÝµÃ£ºB=1T
£¨3£©µ±B¡ä=2Tʱ£¬$r¡ä=\frac{r}{2}=0.25m$
   Á£×ÓÉäÈëOµãºó¾­¹ý3´Îƫת´òµ½Pµã¹ÊÓÐ
  OP=$3¡Á2r¡äcos¦Á+3¡Á\frac{x}{3}tan¦Á$=1.3m
´ð£º£¨1£©¸ÃµçÈÝÆ÷¼«°åABËù¼ÓµçѹU´óСΪ0.45V£®
£¨2£©ÔÈÇ¿´Å³¡µÄ´Å¸ÐӦǿ¶È´óСBΪ1T£®
£¨3£©ÈôÏÖÔÚ¢ñ¡¢¢óÇøÓòËù¼Ó´Å¸ÐӦǿ¶È´óСB¡ä=2T£¬Á£×ÓÉäÈëOµãºó¾­¹ý3´Îƫת´òµ½Pµã£¬ÔòOPµÄ¾àÀëΪ1.3m£®

µãÆÀ ±¾ÌâµÄ¼¼ÇÉÔÚÓÚ£¬ÔÚÀàÆ½Å×Ô˶¯ÖÐËÙ¶È·½ÏòÓëÎ»ÒÆ·½ÏòµÄ¼Ð½ÇÓÐÒ»¸ö¶¨Á¿¹ØÏµ£¬¼´tan¦È=tan¦Á  ´Ó¸Ã¹ØÏµÖªµÀÁ£×ÓÀ뿪ÔÈÇ¿µç³¡Ê±Ëٶȵķ´ÉäÑÓ³¤Ïßͨ¹ýÔÈËÙÎ»ÒÆµÄÖе㣬´Ó¶ø´ÓÎ»ÒÆµÄ·½Ïò±íʾ³öËÙ¶È·½Ïò£¬ÓÚÊÇÔڴų¡ÖÐ×öÔÈËÙÔ²ÖÜÔ˶¯µÄ°ë¾¶Ò²ÄÜÇó³ö£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø