Question

16£®£¨1£©¹ØÓÚ´òµã¼ÆʱÆ÷£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇBD
A£®´òµãµÄÖÜÆÚÓɽ»Á÷µçѹ¾ö¶¨µÄ
B£®´òµãµÄÖÜÆÚÓɽ»Á÷µçµÄƵÂʾö¶¨µÄ£»
C£®µç´Å´òµã¼ÆʱÆ÷ËùʹÓõĵçѹΪ220VµÄ½»Á÷µç£»
D£®µç»ð»¨´òµã¼ÆʱÆ÷ËùʹÓõĵçѹΪ220VµÄ½»Á÷µç£»
£¨2£©³¤ÀÖijУÎïÀí̽¾¿ÊµÑéС×飬ʹÓÃ50HZ½»Á÷µç£¬ÀûÓôòµã¼ÆʱÆ÷´ò³öµÄÒ»ÌõÖ½´øÈçͼËùʾ£®A¡¢B¡¢C¡¢
D¡¢EΪÎÒÃÇÔÚÖ½´øÉÏËùÑ¡µÄ¼ÆÊýµã£®Ã¿ÏàÁÚµÄÁ½¸ö¼ÆÊýµãÖ®¼äÓÐËĸöСµãδ»­³ö£®Ôò£º

¢Ù¼ÆÊýµã¼äµÄʱ¼ä¼ä¸ôΪ0.1s£»
¢Ú¸ù¾ÝÖ½´øÉÏËù²âµÃÊý¾Ý£¬´òµã¼ÆʱÆ÷´òÏÂCµãʱС³µµÄ˲ʱËٶȣ¬v_C=0.34m/s£»£¨¼ÆËã½á¹û±£ÁôÁ½Î»ÓÐЧÊý×Ö£©
¢ÛС³µµÄ¼ÓËÙ¶Èa=0.40 m/s²£®£¨¼ÆËã½á¹û±£ÁôÁ½Î»ÓÐЧÊý×Ö£©

Answer 1

·ÖÎö £¨1£©¸ù¾ÝʵÑéµÄÔ­ÀíÒÔ¼°²Ù×÷ÖеÄ×¢ÒâÊÂÏîÈ·¶¨ÕýÈ·µÄ²Ù×÷²½Ö裮
£¨2£©¸ù¾Ýij¶Îʱ¼äÄÚµÄƽ¾ùËٶȵÈÓÚÖмäʱ¿ÌµÄ˲ʱËÙ¶ÈÇó³öCµãµÄ˲ʱËٶȣ¬¸ù¾ÝÁ¬ÐøÏàµÈʱ¼äÄÚµÄλÒÆÖ®²îÊÇÒ»ºãÁ¿Çó³öС³µµÄ¼ÓËٶȣ®

½â´ð ½â£º£¨1£©A¡¢´òµã¼ÆʱÆ÷µÄ´òµãÖÜÆÚÓɽ»Á÷µçµÄƵÂʾö¶¨£¬¹ÊA´íÎó£¬BÕýÈ·£®
C¡¢µç´Å´òµã¼ÆʱÆ÷ʹÓÃ4-6VµÄ½»Á÷µçѹ£¬µç»ð»¨´òµã¼ÆʱÆ÷ʹÓÃ220VµÄ½»Á÷µçѹ£¬¹ÊC´íÎó£¬DÕýÈ·£®
¹ÊÑ¡£ºBD£®
£¨2£©Ã¿ÏàÁÚµÄÁ½¸ö¼ÆÊýµãÖ®¼äÓÐËĸöСµãδ»­³ö£¬¿ÉÖªÏàÁÚ¼ÆÊýµã¼äµÄʱ¼ä¼ä¸ôΪ0.1s£»
CµãµÄ˲ʱËٶȵÈÓÚBD¶ÎµÄƽ¾ùËٶȣ¬Ôò${v}_{C}=\frac{{x}_{BD}}{2T}=\frac{£¨3.20+3.60£©¡Á1{0}^{-2}}{0.2}$m/s=0.34m/s£¬
Á¬ÐøÏàµÈʱ¼äÄÚµÄλÒÆÖ®²î¡÷x=0.40cm£¬¸ù¾Ý¡÷x=aT²µÃ£¬¼ÓËÙ¶ÈΪ£º
a=$\frac{¡÷x}{{T}^{2}}=\frac{0.4¡Á1{0}^{-2}}{0.01}m/{s}^{2}=0.40m/{s}^{2}$£®
¹Ê´ð°¸Îª£º£¨1£©BD£»£¨2£©0.1£¬0.34£¬0.40£®

µãÆÀ ½â¾ö±¾ÌâµÄ¹Ø¼üÕÆÎÕÖ½´øµÄ´¦Àí·½·¨£¬»áͨ¹ýÖ½´øÇó½â˲ʱËٶȺͼÓËٶȣ¬¹Ø¼üÊÇÔȱäËÙÖ±ÏßÔ˶¯ÍÆÂÛµÄÔËÓã®