Question

2£®Õæ¿Õʾ²¨¹ÜµÄ¹¤×÷Ô­ÀíÊǽ«´ÓµÆË¿K·¢³ö£¨³õËٶȲ»¼Æ£©µÄµç×Ó£¬¾­µÆË¿ÓëA°å¼äµÄ¼ÓËÙµçѹU₁¼ÓËÙºóÓÉA°åÖÐÐÄ¿×ÑØÖÐÐÄÏßKOÉä³ö£¬Ê¹µç×ÓÊø½øÈëM¡¢N¼äµç³¡Ê±µÄËÙ¶ÈÓëµç³¡·½Ïò´¹Ö±£¬Èç¹ûÔÚƫתµç¼«M¡¢N¼äûÓмӵçѹ£¬µç×ÓÊø½«´òÔÚÓ«¹âÆÁµÄÖÐÐÄOµã£¬²¢ÔÚÄÇÀï²úÉúÒ»¸öÁÁµã£®Èôijµç×Ó¾­¹ýƫתµç³¡£¨¿ÉÊÓΪÔÈÇ¿µç³¡£©ºó´òÔÚÓ«¹âÆÁÉϵÄPµã£¬ÈçͼËùʾ£¬ÒÑÖª¸Ãµç×Ó´©¹ýA°åʱµÄËÙ¶Èv₀=6.0¡Á10⁷m/s£¬M¡¢NÁ½°å¼äµÄµçѹU₂=2.0¡Á10²V£¬Á½°å¼äµÄ¾àÀëd=9¡Á10^-3m£¬°å³¤L₁=2.0¡Á10^-2m£¬°åÓҶ˵½Ó«¹âÆÁ¶î¾àÀëL₂=3.0¡Á10^-2m£¬È¡µç×ӵıȺÉ$\frac{e}{m}$=1.8¡Á10¹¹C/kg£®ÊÔÇó£º
£¨1£©¼ÓËÙµçѹU£»
£¨2£©µç×ÓÔÚƫתµç³¡ÖеļÓËٶȦÁ£»
£¨3£©Pµãµ½OµãµÄ¾àÀëy£®

Answer 1

·ÖÎö £¨1£©¸ù¾Ý¶¯Äܶ¨ÀíÇó³ö¼ÓËÙµçѹµÄ´óС£®
£¨2£©¸ù¾ÝÅ£¶ÙµÚ¶þ¶¨ÂÉÇó³öµç×ÓÔÚƫתµç³¡ÖеļÓËٶȣ®
£¨3£©¸ù¾ÝλÒÆʱ¼ä¹«Ê½Çó³öµç×ÓÔÚƫתµç³¡ÖеÄƫתλÒÆ£¬×¥×¡µç×Ó³öµç³¡Ê±Ëٶȵķ´ÏòÑÓ³¤Ïß¾­¹ýÖÐÖáÏßµÄÖе㣬ÔËÓÃÏàËÆÈý½ÇÐÎÇó³öPµãµ½OµãµÄ¾àÀ룮

½â´ð ½â£º£¨1£©¸ù¾Ý¶¯Äܶ¨ÀíµÃ£¬$eU=\frac{1}{2}m{{v}_{0}}^{2}$£¬
½âµÃU=$\frac{\frac{1}{2}m{{v}_{0}}^{2}}{e}=\frac{\frac{1}{2}¡Á36¡Á1{0}^{14}}{1.8¡Á1{0}^{11}}V$=10⁴V£®
£¨2£©µç×ÓÔÚƫתµç³¡ÖеļÓËÙ¶Èa=$\frac{eE}{m}=\frac{e{U}_{2}}{md}=\frac{1.8¡Á1{0}^{11}¡Á200}{9¡Á1{0}^{-3}}m/{s}^{2}$=4¡Á10¹⁵m/s²£¬
£¨3£©µç×ÓÔÚƫתµç³¡ÖеÄƫתλÒÆ${y}_{1}=\frac{1}{2}a{t}^{2}=\frac{1}{2}¡Á4¡Á1{0}^{15}¡Á£¨\frac{2¡Á1{0}^{-2}}{6¡Á1{0}^{7}}£©^{2}$=$\frac{2}{9}¡Á1{0}^{-3}$m£¬
ÒòΪµç×Ó³öµç³¡Ê±Ëٶȵķ´ÏòÑÓ³¤Ïß¾­¹ýÖÐÖáÏßµÄÖе㣬¸ù¾ÝÏàËÆÈý½ÇÐεã¬
$\frac{y}{{y}_{1}}=\frac{{L}_{2}+\frac{{L}_{1}}{2}}{\frac{{L}_{1}}{2}}$£¬
´úÈëÊý¾Ý½âµÃy=$\frac{8}{9}¡Á1{0}^{-3}m$¡Ö8.9¡Á10^-4m£®
´ð£º£¨1£©¼ÓËÙµçѹUΪ10⁴V£»
£¨2£©µç×ÓÔÚƫתµç³¡ÖеļÓËٶȦÁΪ4¡Á10¹⁵m/s²£»
£¨3£©Pµãµ½OµãµÄ¾àÀëyΪ8.9¡Á10^-4m£®

µãÆÀ ±¾Ì⿼²éÁË´øµçÁ£×ÓÔڵ糡ÖеļÓËÙºÍƫת£¬ÕÆÎÕ´¦Àíµç×Ó×öÀàƽÅ×Ô˶¯µÄ·½·¨£¬½áºÏÅ£¶ÙµÚ¶þ¶¨ÂɺÍÔ˶¯Ñ§¹«Ê½×ÛºÏÇó½â£®