ÌâÄ¿ÄÚÈÝ

4£®ÎªÁ˶ԻðÐǼ°ÆäÖÜΧµÄ¿Õ¼ä»·¾³½øÐÐ̽²â£¬ÎÒ¹úÔÚ2011Äê10Ô·¢ÉäµÚÒ»¿Å»ðÐÇ̽²âÆ÷¡°Ó©»ðÒ»ºÅ¡±£¬Ì½²âÆ÷ÔÚÀë»ðÐDZíÃæ¸ß¶È·Ö±ðΪh1ºÍh2µÄÔ²¹ìµÀÉÏÔ˶¯Ê±£¬ÖÜÆÚ·Ö±ðΪT1ºÍT2£¬»ðÐÇ¿ÉÊÓΪÖÊÁ¿·Ö²¼¾ùÔȵÄÇòÌ壬ÇÒºöÂÔ»ðÐÇ×ÔתµÄÓ°Ï죬ÍòÓÐÒýÁ¦³£Á¿ÎªG£¬½öÀûÓÃÒÔÉÏÊý¾Ý£¬¿ÉÒÔ¼ÆËã³ö£¨¡¡¡¡£©
A£®»ðÐDZíÃæµÄÖØÁ¦¼ÓËٶȺ͡°Ó©»ðÒ»ºÅ¡±µÄ¼ÓËÙ¶È
B£®»ðÐǵÄÖÊÁ¿ºÍ»ðÐǶԡ°Ó©»ðÒ»ºÅ¡±µÄÒýÁ¦
C£®»ðÐǵÄÃܶȺ͡°Ó©»ðÒ»ºÅ¡±µÄÖÊÁ¿
D£®»ðÐDZíÃæµÄÖØÁ¦¼ÓËٶȺͻðÐǶԡ°Ó©»ðÒ»ºÅ¡±µÄÒýÁ¦

·ÖÎö ¸ù¾ÝÍòÓÐÒýÁ¦Ìṩ̽²âÆ÷×öÔ²ÖÜÔ˶¯µÄÏòÐÄÁ¦£¬ÁгöµÈʽ£®
¸ù¾ÝÃܶȹ«Ê½Çó³öÃܶȣ®
ÔËÓÃÍòÓÐÒýÁ¦µÈÓÚÖØÁ¦±íʾ³ö»ðÐDZíÃæµÄÖØÁ¦¼ÓËÙ¶È£®

½â´ð ½â£ºA¡¢ÓÉÓÚÍòÓÐÒýÁ¦Ìṩ̽²âÆ÷×öÔ²ÖÜÔ˶¯µÄÏòÐÄÁ¦£¬ÔòÓУº
$G\frac{Mm}{£¨R+{h}_{1}£©^{2}}=m\frac{4{¦Ð}^{2}}{{T}^{2}}£¨R+{h}_{1}£©$£»$G\frac{Mm}{{£¨R+{h}_{2}£©}^{2}}=m\frac{4{¦Ð}^{2}}{{T}^{2}}£¨R+{h}_{2}£©$£¬
¿ÉÇóµÃ»ðÐǵÄÖÊÁ¿Îª£ºM=$\frac{4{¦Ð}^{2}£¨R+{h}_{1}£©^{3}}{G{{T}_{1}}^{2}}$=$\frac{4{¦Ð}^{2}{£¨R+{h}_{2}£©}^{3}}{G{{T}_{2}}^{2}}$
»ðÐǵİ뾶Ϊ£º$R=\frac{\root{3}{£¨\frac{{T}_{2}}{{T}_{1}}£©^{2}{h}_{2}}-{h}_{1}}{1-\root{3}{£¨\frac{{T}_{2}}{{T}_{1}}£©^{2}}}$£¬
ÔÚ»ðÐDZíÃæµÄÎïÌåÓУº$G\frac{Mm}{{R}^{2}}=mg$£¬
¿ÉµÃ»ðÐDZíÃæµÄÖØÁ¦¼ÓËÙ¶ÈΪ£ºg=$\frac{GM}{{R}^{2}}$£¬
¡°Ó©»ðÒ»ºÅ¡±ÔÚÀë»ðÐDZíÃæ¸ß¶È·Ö±ðΪh1ºÍh2µÄÔ²¹ìµÀÉÏÔ˶¯Ê±¼ÓËÙ¶È·Ö±ðΪ£º
${a}_{1}=\frac{GM}{£¨R+{h}_{1}£©^{2}}$£¬${a}_{2}=\frac{GM}{{£¨R+{h}_{2}£©}^{2}}$£®
¹ÊAÕýÈ·£®
B¡¢´ÓAÑ¡Ïî·ÖÎöÖªµÀ¿ÉÒÔÇó³ö»ðÐǵÄÖÊÁ¿£¬ÓÉÓÚ²»ÖªµÀ¡°Ó©»ðÒ»ºÅ¡±µÄÖÊÁ¿£¬ËùÒÔ²»ÄÜÇó³ö»ðÐǶԡ°Ó©»ðÒ»ºÅ¡±µÄÒýÁ¦£¬¹ÊB´íÎó£®
C¡¢´ÓAÑ¡Ïî·ÖÎöÖªµÀ¿ÉÒÔÇó³ö»ðÐǵÄÖʰ뾶£¬²»ÄÜÇó³ö¡°Ó©»ðÒ»ºÅ¡±µÄÖÊÁ¿£¬¹ÊC´íÎó£®
D¡¢´ÓAÑ¡Ïî·ÖÎöÖªµÀ¿ÉÒÔÇó³ö»ðÐǵıíÃæµÄÖØÁ¦¼ÓËÙ¶È£¬ÓÉÓÚ²»ÖªµÀ¡°Ó©»ðÒ»ºÅ¡±µÄÖÊÁ¿£¬ËùÒÔ²»ÄÜÇó³ö»ðÐǶԡ°Ó©»ðÒ»ºÅ¡±µÄÒýÁ¦£¬¹ÊD´íÎó£®
¹ÊÑ¡£ºA£®

µãÆÀ ±¾ÌâÒÔ¡°Ó©»ðÒ»ºÅ¡±»ðÐÇ̽²âÆ÷Ϊ±³¾°²ÄÁÏ£¬ÌåÏÖÁËÏÖ´úº½Ìì¼¼ÊõʼÖÕÊǸ߿¼µÄÒ»¸öÈȵ㣮Ö÷Òª¿¼²é¶ÔÍòÓÐÒýÁ¦¶¨ÂÉ¡¢Å£¶ÙµÚ¶þ¶¨ÂÉ¡¢ÔÈËÙÔ²ÖÜÔ˶¯µÈ֪ʶµãµÄ×ÛºÏÔËÓÃÄÜÁ¦£®ÏòÐÄÁ¦µÄ¹«Ê½Ñ¡È¡Òª¸ù¾ÝÌâÄ¿ÌṩµÄÒÑÖªÎïÀíÁ¿»òËùÇó½âµÄÎïÀíÁ¿Ñ¡È¡Ó¦Óã®ÒªÇó½âÒ»¸öÎïÀíÁ¿´óС£¬ÎÒÃÇÓ¦¸Ã°ÑÕâ¸öÎïÀíÁ¿Ïȱíʾ³öÀ´£¬ÔÙ¸ù¾ÝÒÑÖªÁ¿½øÐÐÅжϣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
9£®Ä³Í¬Ñ§Òª²âÁ¿ÓÉijÖÖвÄÁÏÖÆ³ÉµÄ´Öϸ¾ùÔȵÄÔ²ÖùÐε¼ÌåµÄµç×èÂʦѲ½ÖèÈçÏ£º

£¨1£©ÓÃ20·Ö¶ÈµÄÓα꿨³ß²âÁ¿Æä³¤¶ÈÈçͼ¼×Ëùʾ£¬ÓÉͼ¿ÉÖªÆä³¤¶ÈΪ5.015cm£»
£¨2£©ÓÃÂÝÐý²â΢Æ÷²âÁ¿ÆäÖ±¾¶ÈçͼÒÒËùʾ£¬ÓÉͼ¿ÉÖªÆäÖ±¾¶Îª4.700mm£»
£¨3£©ÓöàÓõç±íµÄµç×è¡°¡Á10¡±µ²£¬°´ÕýÈ·µÄ²Ù×÷²½Öè²â´ËÔ²ÖùÐε¼ÌåµÄµç×裬±íÅ̵ÄʾÊýÈçͼ±ûËùʾ£¬Ôò¸Ãµç×èµÄ×èֵԼΪ220¦¸£®
£¨4£©¸ÃͬѧÏëÓ÷ü°²·¨¸ü¾«È·µØ²âÁ¿Æäµç×èR£¬ÏÖÓÐµÄÆ÷²Ä¼°Æä´úºÅºÍ¹æ¸ñÈçÏ£º
´ý²âÔ²ÖùÐε¼Ì壨µç×èΪR£©
µçÁ÷±íA1£¨Á¿³Ì4mA£¬ÄÚ×èԼΪ50¦¸£©
µçÁ÷±íA2£¨Á¿³Ì10mA£¬ÄÚ×èԼΪ30¦¸£©
µçѹ±íV1£¨Á¿³Ì3V£¬ÄÚ×èԼΪ10k¦¸£©
µçѹ±íV2£¨Á¿³Ì15V£¬ÄÚ×èԼΪ25k¦¸£©
Ö±Á÷µçÔ´E£¨µç¶¯ÊÆ4V£¬ÄÚ×è²»¼Æ£©
»¬¶¯±ä×èÆ÷R1£¨×èÖµ·¶Î§0¡«15¦¸£¬¶î¶¨µçÁ÷2.0A£©
»¬¶¯±ä×èÆ÷R2£¨×èÖµ·¶Î§0¡«2k¦¸£¬¶î¶¨µçÁ÷0.5A£©
¿ª¹ØS£¬µ¼ÏßÈô¸É£®
Ϊ¼õСʵÑéÎó²î£¬ÒªÇó²â¾¡¿ÉÄܵĶà×éÊý¾Ý½øÐзÖÎö£¬ÇëÔÚͼ¶¡µÄÐéÏß¿òÖл­³öºÏÀíµÄʵÑéÔ­Àíµç·ͼ£¬²¢±êÃ÷ËùÓÃÆ÷²ÄµÄ´úºÅ£®

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

¾«Ó¢¼Ò½ÌÍø