Question

7£®Èçͼ¼×Ëùʾ£¬×ã¹»³¤µÄ´Ö²ÚбÃæÓëˮƽÃæ³É¦È=37¡ã¹Ì¶¨·ÅÖã¬Ð±ÃæÉÏƽÐÐÐéÏßaa¡äºÍbb¡äÖ®¼äÓд¹Ö±Ð±ÃæÏòÉϵÄÓнçÔÈÇ¿´Å³¡£¬¼ä¾àΪd=1m£¬´Å¸ÐӦǿ¶ÈΪBËæʱ¼ät±ä»¯¹æÂÉÈçͼÒÒËùʾ£®ÏÖÓÐÒ»ÖÊÁ¿Îªm=0.1Kg£¬×ܵç×èΪR=10¦¸£¬±ß³¤Ò²Îªd=1mµÄÕý·½ÐνðÊôÏßȦMNPQ£¬Æä³õʼλÖÃÓÐÒ»°ëÃæ»ýλÓڴų¡ÖУ¬ÔÚt=0ʱ¿Ì£¬ÏßȦǡºÃÄܱ£³Ö¾²Ö¹£¬´ËºóÔÚt=0.25sʱ£¬ÏßȦ¿ªÊ¼ÑØбÃæÏ»¬£¬Ï»¬¹ý³ÌÖÐÏßȦMN±ßʼÖÕÓëÐéÏßaa¡ä±£³ÖƽÐУ®ÒÑÖªÏßȦÍêÈ«½øÈë´Å³¡Ç°ÒѾ­¿ªÊ¼×öÔÈËÙÖ±ÏßÔ˶¯£®Ç󣺣¨È¡sin37¡ã=0.6£¬cos37¡ã=0.8£¬g=10m/s²£©
£¨1£©Ç°0.25sÄÚͨ¹ýÏßȦijһ½ØÃæµÄµçÁ¿£»
£¨2£©ÏßȦÓëбÃæ¼äµÄ¶¯Ä¦²ÁÒòÊý£»
£¨3£©ÏßȦ´Ó¿ªÊ¼Ô˶¯µ½Í¨¹ýÕû¸ö´Å³¡µÄ¹ý³ÌÖУ¬µç×èÉϲúÉúµÄ½¹¶úÈÈ£®

Answer 1

·ÖÎö £¨1£©¸ù¾Ý·¨À­µÚµç´Å¸ÐÓ¦¶¨ÂÉÇó³ö¸ÐÓ¦µç¶¯ÊÆ£¬Óɱպϵç·µÄÅ·Ä·¶¨ÂÉÇó³ö¸ÐÓ¦µçÁ÷£¬¸ù¾Ý$q=\overline{I}•¡÷t$Çóͨ¹ýÏßȦijһ½ØÃæµÄµçÁ¿
£¨2£©t=0.25sʱ¿Ì£¬ÏßȦ¿ªÊ¼ÑØбÃæÏ»¬£¬¶ÔÏß¿òÊÜÁ¦·ÖÎö£¬¸ù¾ÝƽºâÌõ¼þÁÐʽÇó³ö¦Ì
£¨3£©ÏßȦ×îºóÔÈËÙͨ¹ý´Å³¡£¬¸ù¾ÝƽºâÇó³öÔÈËÙÔ˶¯µÄËٶȣ¬ÓÉÄÜÁ¿Êغ㶨ÂÉÇó³öµç×èÉϲúÉúµÄ½¹¶úÈÈ

½â´ð ½â£º£¨1£©0.25sÄÚ£¬¸ù¾Ý·¨À­µÚµç´Å¸ÐÓ¦¶¨ÂÉ£¬ÓÐ
$E=\frac{¡÷B}{¡÷t}•\frac{{d}_{\;}^{2}}{2}=\frac{2-1}{0.25}¡Á\frac{1}{2}=2V$
¸ÐÓ¦µçÁ÷$I=\frac{E}{R}=\frac{2}{10}A=0.2A$
0.25sÄÚͨ¹ýÏßȦijһ½ØÃæµÄµçÁ¿q=I•¡÷t=0.2¡Á0.25C=0.05C
£¨2£©¸ù¾ÝÀã´Î¶¨ÂÉ£¬ÏßȦÄÚ¸ÐÓ¦µçÁ÷Ϊ˳ʱÕë·½Ïò
0.25sʱ£¬ÏßȦ¿ªÊ¼ÑØбÃæÏ»¬
mgsin37¡ã=¦Ìmgcos37¡ã+BId
´úÈëÊý¾Ý£º0.6=¦Ì¡Á0.8+2¡Á0.2¡Á1
½âµÃ£º¦Ì=0.25
£¨3£©ÉèÏßȦ×îºóÔÈËÙÔ˶¯Ê±µÄËÙ¶ÈΪv£¬
¸ÐÓ¦µç¶¯ÊÆE=Bdv¢Ù
¸ÐÓ¦µçÁ÷$I=\frac{E}{R}$¢Ú
°²ÅàÁ¦${F}_{°²}^{\;}=BId$¢Û
ÁªÁ¢¢Ù¢Ú¢ÛµÃ${F}_{°²}^{\;}=\frac{{B}_{\;}^{2}{d}_{\;}^{2}v}{R}$
ÔÈËÙÔ˶¯Ê±£¬¶ÔÏßȦ¸ù¾ÝƽºâÌõ¼þµÃ
$mgsin37¡ã=¦Ìmgcos37¡ã+\frac{{B}_{\;}^{2}{d}_{\;}^{2}v}{R}$
´úÈëÊý¾Ý£º$0.6=0.2+\frac{{2}_{\;}^{2}¡Á{1}_{\;}^{2}v}{10}$
½âµÃv=1m/s
¸ù¾ÝÄÜÁ¿Êغ㶨ÂÉ£¬ÓÐ
$mgsin37¡ã£¨d+\frac{d}{2}£©=\frac{1}{2}m{v}_{\;}^{2}+¦Ìmgcos37¡ã$$£¨d+\frac{d}{2}£©$+Q
´úÈëÊý¾Ý£º$0.6¡Á\frac{3}{2}=\frac{1}{2}¡Á0.1¡Á{1}_{\;}^{2}+0.2¡Á\frac{3}{2}+Q$
½âµÃ£ºQ=0.55J
´ð£º£¨1£©Ç°0.25sÄÚͨ¹ýÏßȦijһ½ØÃæµÄµçÁ¿0.05C£»
£¨2£©ÏßȦÓëбÃæ¼äµÄ¶¯Ä¦²ÁÒòÊý0.25£»
£¨3£©ÏßȦ´Ó¿ªÊ¼Ô˶¯µ½Í¨¹ýÕû¸ö´Å³¡µÄ¹ý³ÌÖУ¬µç×èÉϲúÉúµÄ½¹¶úÈÈ0.55J

µãÆÀ ±¾ÌâÖÐÏȲúÉú¸ÐÉúµç¶¯ÊÆ£¬Òª»áÔËÓ÷¨À­µÚ¶¨ÂÉÇó¸ÐÓ¦µç¶¯ÊÆ£®ÏßȦ½øÈë´Å³¡ºó£¬²úÉú¶¯Éúµç¶¯ÊÆ£¬ÓÉE=BLvÇó½â¸ÐÓ¦µç¶¯ÊÆ£®