ÌâÄ¿ÄÚÈÝ

ÈçͼËùʾΪµçÁ÷±íµÄ¹¹Ô죬ͼ¼×ΪÌãÐδÅÌúÓëÌúÐļä´Å³¡µÄʾÒâͼ£¬Í¼ÒÒÖÐabcd±íʾµÄÊǵçÁ÷¼ÆÖеÄͨµçÏßȦ¡£ab=cd=1cm£¬ad=bc=0.9cm£¬¹²ÓÐ50ÔÑ£¬ÏßȦÁ½±ßËùÔÚλÖõĴŸÐӦǿ¶ÈΪ0.5ÌØ£¬ÒÑÖªÏßȦÿת1¡ã£¬µ¯»É²úÉúµÄ×è°­ÏßȦƫתµÄÁ¦¾ØΪ2.5¡ÁN?M¡£
(1)µ±ÏßȦÖеçÁ÷Ϊ0.6ºÁ°²Ê±£¬Ö¸Õ뽫ת¹ý¶àÉÙ¶È?
(2)Èç¹ûÖ¸ÕëµÄ×î´óƫת½ÇΪ90¡ã£¬ÔòÕâÖ»µçÁ÷¼ÆÁ¿³ÌÊǶàÉÙ?
(3)µ±Ö¸Õëƫת½ÇΪ40¡ã½Çʱ£¬Í¨ÈëÏßȦµÄµçÁ÷¶à´ó?
(¿¼²é¶ÔµçÁ÷±íÄÚ²¿½á¹¹µÄÈÏʶ£¬¿¼²éµçÁ÷±íµÄ¹¤×÷Ô­Àí£¬¿¼²éµçÁ÷±íÄÚµç´ÅÁ¦¾ØµÄ¼ÆËã¡£¿¼²éÔËÓÿα¾ÖªÊ¶½â¾öʵ¼ÊÎÊÌâµÄÄÜÁ¦)
¼×                          ÒÒ
(1)£½NBS=1.35¡ÁN?M ¦È1£½54¡ã(2)Imax=Mmax/NBS=1¡Á (A)=1.0(mA)(3)I2=M2/NBS=0.44¡Á=0.44(mA)
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø