ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Ð¡ÃÎÀûÓÃÈçͼËùʾװÖ㬲âÁ¿Ð¡³µÔ˶¯µÄƽ¾ùËÙ¶È¡£ÈÃС³µ´ÓÐ±ÃæÉϵÄAµãÓɾ²Ö¹¿ªÊ¼Ï»¬£¬·Ö±ð²â³öС³µµ½´ïBµãºÍCµãµÄʱ¼ä£¬¼´¿É²â³ö²»Í¬½×¶ÎµÄƽ¾ùËÙ¶È¡£

£¨1£©Ó¦Ñ¡Ôñ½ÏСÆÂ¶ÈµÄÐ±Ãæ£¬ÕâÑùÉè¼ÆµÄÄ¿µÄÊÇÔÚʵÑéÖбãÓÚ²âÁ¿____________________¡£

£¨2£©Í¼ÖÐBC¶ÎµÄ·³ÌsBC=_______dm, BC ¶ÎµÄƽ¾ùËÙ¶ÈvBC=________________m/s.

£¨3£©ÔÚ²âÁ¿Ð¡³µµ½´ïBµãµÄʱ¼äʱ£¬Èç¹ûС³µ¹ýÁËBµã²ÅÍ£Ö¹¼ÆÊ±,²âµÃAB¶ÎµÄƽ¾ùËÙ¶ÈvAB»áÆ«____ (Ñ¡Ìî ¡°´ó¡±»ò¡°Ð¡¡±)

£¨4£©vAC________(Ñ¡Ìî¡°>¡±¡°<¡± »ò¡°=¡±) vAB

£¨5£©ÎªÁËʹС³µµ½´ïBºÍCµÄʱ¼ä²âÁ¿¸ü׼ȷ£¬Ð¡»ªÌá³öÔÚBºÍCÕâÀïÔö¼ÓÒ»¸öµ²°å¡£ÎªÁ˲âÁ¿Ð¡³µÔ˶¯¹ý³ÌÖÐϰë³ÌµÄƽ¾ùËÙ¶È£¬Ð¡»ªÈÃС³µ´ÓBµãÓɾ²Ö¹ÊÍ·Å,²â³öС³µµ½´ïCµãµÄʱ¼ä£¬´Ó¶ø¼ÆËã³öС³µÔÚϰë³ÌÔ˶¯µÄƽ¾ùËÙ¶È,ËûµÄ×ö·¨ÕýÈ·Âð?__________£¬ ÀíÓÉÊÇ:________________¡£

¡¾´ð°¸¡¿Ð¡³µµÄÔ˶¯Ê±¼ä 5 0.5 С > ²»ÕýÈ· ¼û½âÎö

¡¾½âÎö¡¿

(1)[1]ʵÑéÖУ¬Ð±ÃæÓ¦¾¡Á¿Ñ¡Ôñ½ÏСÆÂ¶È£¬ÕâÑùÉè¼ÆÊÇΪÁËʵÑéÖбãÓÚ²âÁ¿Ð¡³µµÄÔ˶¯Ê±¼ä£»

(2)[2]ÓÉͼʾ¿ÉÖª£º

sBC=60.0cm10.0cm=50.0cm=5dm£»

[3]С³µÍ¨¹ýBC¶ÎËùÓÃʱ¼ä£¬tBC=1s£¬BC¶ÎµÄƽ¾ùËÙ¶È£º

vBC===50cm/s=0.5m/s£»

(3)[4]Èç¹ûÈÃС³µ¹ýÁËBµã²ÅÍ£Ö¹¼ÆÊ±£¬»áµ¼ÖÂʱ¼äµÄ²âÁ¿½á¹ûÆ«´ó£¬Óɹ«Ê½v=Öª£¬Æ½¾ùËÙ¶È»áÆ«Ð¡£»

(4)[5] ÓÉÓÚС³µÔ˶¯Ô½À´Ô½¿ì£¬ËùÒÔAC¶ÎµÄƽ¾ùËÙ¶È»á´óÓÚAB¶ÎµÄƽ¾ùËÙ¶È£»

(5)[6][7]С³µÔÚÏ»¬¹ý³ÌÖУ¬µ½´ïBµãµÄËٶȲ¢²»ÎªÁ㣬ËùÒÔÈÃС³µ´ÓBµãÓɾ²Ö¹ÊÍ·Å£¬µ½´ïCµãµÄʱ¼ä£¬²¢²»µÈÓÚϰë³ÌµÄʱ¼ä£¬¹Ê²»ÕýÈ·¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø