ÌâÄ¿ÄÚÈÝ

£¨1£©¢Ùͼ¼×ÖÐÓα꿨³ßµÄ¶ÁÊýÊÇ
2.98
2.98
cm£¬Í¼ÒÒÖÐÂÝÐý²â΢Æ÷µÄ¶ÁÊýÊÇ
5.680
5.680
mm£®

¢ÚÈçͼ2ÊÇijͬѧÓôòµã¼ÆʱÆ÷Ñо¿Ð¡³µ×öÔȱäËÙÖ±ÏßÔ˶¯Ê±´ò³öµÄÒ»ÌõÖ½´ø£¬A¡¢B¡¢C¡¢D¡¢EΪ¸ÃͬѧÔÚÖ½´øÉÏËùÑ¡µÄ¼ÆÊýµã£¬ÏàÁÚ¼ÆÊýµã¼äµÄʱ¼ä¼ä¸ôΪ0.1s£®ÓÉͼ¿ÉÖª£¬´òµã¼ÆʱÆ÷´òÏÂDµãʱС³µµÄ˲ʱËÙ¶ÈΪ
0.34
0.34
m/s£¬Ð¡³µµÄ¼ÓËÙ¶ÈΪ
0.39
0.39
m/s2£¨½á¹û¾ù±£ÁôÁ½Î»ÓÐЧÊý×Ö£©£®
£¨2£©ÎªÁ˲âÁ¿Ä³Ò»Î´Öª×èÖµµÄµç×èRx£¬Ä³ÊµÑéС×éÕÒÀ´ÒÔÏÂÆ÷²Ä£ºµçѹ±í£¨0¡«3V£¬ÄÚ×èÔ¼3k¦¸£©¡¢µçÁ÷±í£¨0¡«0.6A£¬ÄÚ×èÔ¼1¦¸£©¡¢»¬¶¯±ä×èÆ÷£¨0¡«15¦¸£¬2A£©¡¢µçÔ´£¨E=3V£¬ÄÚ×èºÜС£©¡¢¿ª¹ØÓëµ¼Ïߣ¬¸ÃС×éͬѧʵÑé²Ù×÷ÈçÏ£¬ÇëÄã°´ÒªÇóÌîд£º
¢Ù¶Ï¿ª¿ª¹Ø£¬Ç밴ͼ3¼×ËùʾµÄµç·ͼÔÚͼ3ÒÒÖÐÁ¬½ÓʵÎïͼ£»
¢ÚÈçͼ3ÒÒ£¬±ÕºÏ¿ª¹ØÇ°£¬Ó¦½«»¬¶¯±ä×èÆ÷µÄ»¬Æ¬ÖÃÓÚ
a
a
¶Ë£®
¢Û±ÕºÏ¿ª¹Ø£¬»ºÂýµ÷½Ú»¬¶¯±ä×èÆ÷£¬µÃµ½¶à×éµçѹ±íÓëµçÁ÷±íµÄ¶ÁÊý£¬½¨Á¢ºÏÊʵÄ×ø±êϵ×÷³öI-Uͼ±ûÈçÏ£»
¢ÜÕûÀíÒÇÆ÷
¸ù¾ÝI-Uͼ±û¿ÉµÃ£¬¸Ãδ֪µç×èµÄ×èֵΪRx=
5.0¦¸
5.0¦¸
£®£¨±£ÁôÁ½Î»ÓÐЧÊý×Ö£©
ÓÉÓÚʵÑéÖÐʹÓõĵç±í²»ÊÇÀíÏëµç±í£¬»á¶ÔʵÑé½á¹ûÔì³ÉÒ»¶¨µÄÓ°Ï죬Ôò¸ÃС×éͬѧʵÑé²â³öµÄµç×èÖµ
£¾
£¾
 RXµÄÕæʵֵ£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£®
ÀûÓÃÏÖÓеÄÒÇÆ÷£¬ÎªÁ˸ü¼Ó¾«È·µØ²âÁ¿Õâ¸öµç×èµÄ×èÖµ£¬Äã¸ø¸ÃʵÑéС×éµÄ½¨ÒéÊÇ£º
µçÁ÷±í²ÉÓÃÍâ½Ó·¨
µçÁ÷±í²ÉÓÃÍâ½Ó·¨
£®
·ÖÎö£º£¨1£©¢Ù¸ù¾ÝͼʾÓα꿨³ß¡¢ÂÝÐý²â΢Æ÷¶ÔÆä¶ÁÊý£»¢Ú¸ù¾ÝÔȱäËÙÖ±ÏßÔ˶¯ÖÐʱ¼äÖеãµÄËٶȵÈÓڸùý³ÌÖеÄƽ¾ùËٶȣ¬¿ÉÒÔÇó³ö´òÖ½´øÉÏDµãʱС³µµÄ˲ʱËٶȴóС£®¸ù¾ÝÔȱäËÙÖ±ÏßÔ˶¯µÄÍÆÂÛ¹«Ê½¡÷x=aT2¿ÉÒÔÇó³ö¼ÓËٶȵĴóС£®
£¨2£©¢Ù¸ù¾ÝʵÑéµç·ͼÁ¬½ÓʵÎïµç·ͼ£»
¢Ú±ÕºÏ¿ª¹ØÇ°£¬»¬Æ¬Ó¦ÖÃÓÚ×î´ó×èÖµ´¦£¬¸ù¾Ýµç·ͼȷ¶¨»¬Æ¬µÄλÖã»
¢Ü¸ù¾ÝͼÏóÓÉÅ·Ä·¶¨ÂÉÇó³öµç×è×èÖµ£»¸ù¾Ýµç·ͼÓëÅ·Ä·¶¨ÂÉ·ÖÎöʵÑéÎó²î£»¸ù¾Ý´ý²âµç×è×èÖµÓëµç±íÄÚ×èµÄ¹Øϵȷ¶¨µçÁ÷±íµÄ½Ó·¨£¬´Ó¶ø¸Ä½øʵÑ飮
½â´ð£º½â£º£¨1£©¢ÙÓÉͼ1ËùʾÓα꿨³ß¿ÉÖª£¬Ö÷³ßʾÊýΪ2.9cm£¬Óαê³ßʾÊýΪ8¡Á0.1mm=0.8mm=0.08cm£¬ÔòÓα꿨³ßʾÊýΪ2.9cm+0.08cm=2.98cm£»
ÓÉͼ2ËùʾÂÝÐý²â΢Æ÷¿ÉÖª£¬¹Ì¶¨¿Ì¶ÈʾÊýΪ5.5mm£¬¿É¶¯¿Ì¶ÈʾÊýΪ18.0¡Á0.01mm=0.180mm£¬ÂÝÐý²â΢Æ÷ËùʾΪ5.5mm+0.180mm=5.680mm£¨5.679-5.681¾ùÕýÈ·£©£®
¢Ú×öÔȱäËÙÖ±ÏßÔ˶¯ÖÐʱ¼äÖеãµÄËٶȵÈÓڸùý³ÌÖеÄƽ¾ùËٶȣ¬DµãµÄ˲ʱËÙ¶ÈvD=
CE
2t
=
AE-AC
2t
=
0.1198-0.0522
2¡Á0.1
=0.34m/s£»
ÓÉÔȱäËÙÖ±ÏßÔ˶¯µÄÍÆÂÛ¹«Ê½¡÷x=aT2¿ÉÖª£¬¼ÓËÙ¶Èa=
DE-AE
3t2
=
AE-AD-AB
3t2
=
0.1198-0.0840-0.0241
3¡Á0£®12
=0.39m/s2£®
£¨2£©¢Ù¸ù¾Ýµç·ͼÁ¬½ÓʵÎïµç·ͼ£¬ÊµÎïµç·ͼÈçͼËùʾ£»
¢ÚÓÉͼʾµç·ͼ¿ÉÖª£¬±ÕºÏ¿ª¹ØÇ°£¬»¬Æ¬Ó¦ÖÃÓÚa¶Ë£¬´Ëʱ»¬¶¯±ä×èÆ÷½ÓÈëµç·µÄ×èÖµ×î´ó£»
¢ÜÓÉͼʾU-IͼÏó¿ÉÖª£¬´ý²âµç×è×èÖµRX=
U
I
=k¡Ö5.0¦¸£»Óɵç·ͼ¿ÉÖª£¬µçÁ÷±í²ÉÓÃÄÚ½Ó·¨£¬ÓÉÓÚµçÁ÷±íµÄ·Öѹ×÷Óã¬Ëù²âµçѹƫ´ó£¬ÓÉÅ·Ä·¶¨ÂÉ¿ÉÖª£¬µç×è²âÁ¿Öµ´óÓÚÕæʵֵ£®ÓÉÓÚ
R
RA
=
5¦¸
1¦¸
=5£¬
RV
R
=
3000¦¸
5¦¸
=600£¬
RV
R
£¾
R
RA
£¬µçÁ÷±íÓ¦²ÉÓÃÍâ½Ó·¨£¬µçÁ÷±í²ÉÓÃÍâ½Ó·¨¿ÉÒÔ¼õСʵÑéÎó²î£®
¹Ê´ð°¸Îª£º£¨1£©¢Ù2.98£»5.680£»¢Ú0.34£»0.39£»£¨2£©¢Ùµç·ͼÈçͼËùʾ£»¢Úa£»¢Ü5.0¦¸£»£¾£»µçÁ÷±í²ÉÓÃÍâ½Ó·¨£®
µãÆÀ£ºÒªÕýÈ·Óα꿨³ßÓëÂÝÐý²â΢Æ÷µÄʹÓü°¶ÁÊý·½·¨£»Óα꿨³ßÖ÷³ßÓëÓαê³ßʾÊýÖ®ºÍÊÇÓα꿨³ßʾÊý£¬ÂÝÐý²â΢Æ÷¹Ì¶¨¿Ì¶ÈÓë¿É¶¯¿Ì¶ÈʾÊýÖ®ºÍÊÇÂÝÐý²â΢Æ÷µÄʾÊý£»Óα꿨³ß²»ÐèÒª¹À¶Á£¬ÂÝÐý²â΢Æ÷ÐèÒª¹À¶Á£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨1£©ÓÃÓαêΪ20·Ö¶ÈµÄ¿¨³ß²âÁ¿Æ䳤¶ÈÈçͼ1£¬ÓÉͼ1¿ÉÖªÆ䳤¶ÈΪ
50.15
50.15
mm£»ÓÃÂÝÐý²â΢Æ÷²âÁ¿ÆäÖ±¾¶Èçͼ£¬ÓÉͼ1¿ÉÖªÆäÖ±¾¶Îª
4.700
4.700
mm£»

£¨2£©ÔÚ¡°ÑéÖ¤»úеÄÜÊغ㶨ÂÉ¡±µÄʵÑéÖУ¬´òµã¼ÆʱÆ÷ËùÓõçԴƵÂÊΪ50HZ£¬µ±µØÖØÁ¦¼ÓËٶȵÄֵΪ9.80m/s2£¬²âµÃËùÓÃÖØÎïµÄÖÊÁ¿Îª1.00kg£®¼×¡¢ÒÒ¡¢±ûÈýѧÉú·Ö±ðÓÃͬһװÖôò³öÈýÌõÖ½´ø£¬Á¿³ö¸÷Ö½´øÉϵÚ1¡¢2Á½µã¼äµÄ¾àÀë·Ö±ðΪ0.12cm£¬0.19cmºÍ0.25cm£¬¿É¼û²Ù×÷ÉÏÓдíÎóµÄÊÇѧÉú
±û
±û
£¬´íÎó²Ù×÷
ÊÇÏÈ·Å¿ªÖ½´øºó½ÓͨµçÔ´
ÊÇÏÈ·Å¿ªÖ½´øºó½ÓͨµçÔ´
£®
Èô°´ÊµÑéÒªÇóÕýÈ·µØÑ¡³öÖ½´ø½øÐвâÁ¿£¬Á¿µÃÁ¬ÐøÈýµãA£¬B£¬Cµ½µÚÒ»¸öµãµÄ¾àÀëÈçͼ2Ëùʾ£¨ÏàÁÚ¼ÆÊýµãʱ¼ä¼ä¸ôΪ0.02s£©£¬´ÓÆðµãOµ½´òϼÆÊýµãBµÄ¹ý³ÌÖÐÖØÁ¦ÊÆÄܼõÉÙÁ¿ÊÇ¡÷EP=
0.94
0.94
£¬´Ë¹ý³ÌÖÐÎïÌ嶯ÄܵÄÔö¼ÓÁ¿ÊÇ¡÷EK
0.84
0.84
£¨È¡g=9.8m/s2£©£»
£¨3£©Èçͼ3ËùʾÆøµæÊdz£ÓõÄÒ»ÖÖʵÑéÒÇÆ÷£¬ËüÊÇÀûÓÃÆø±Ãʹ´ø¿×µÄµ¼¹ìÓ뻬¿éÖ®¼äÐγÉÆøµæ£¬Ê¹»¬¿éÐü¸¡ÔÚ¹ìµÀÉÏ£¬»¬¿éÔÚ¹ìµÀÉϵÄÔ˶¯¿ÉÊÓΪûÓÐĦ²Á£®ÎÒÃÇ¿ÉÒÔÓôøÊúÖ±µ²°åCºÍDµÄÆøµæ¹ìµÀÒÔ¼°»¬¿éAºÍBÀ´ÑéÖ¤¶¯Á¿Êغ㶨ÂÉ£¬ÊµÑé×°ÖÃÈçͼËùʾ£¨µ¯»ÉµÄ³¤¶ÈºöÂÔ²»¼Æ£©£¬²ÉÓõÄʵÑé²½ÖèÈçÏ£º
a£®µ÷ÕûÆøµæ¹ìµÀ£¬Ê¹µ¼¹ì´¦ÓÚˮƽ£»
b£®ÔÚAºÍB¼ä·ÅÈëÒ»¸ö±»Ñ¹ËõµÄÇᵯ»É£¬Óõ綯¿¨ÏúËø¶¨£¬¾²Ö¹·ÅÖÃÔÚÆøµæµ¼¹ìÉÏ£»
c£®°´ÏµçÅ¥·Å¿ª¿¨Ïú£¬Í¬Ê±Ê¹·Ö±ð¼Ç¼»¬¿éA¡¢BÔ˶¯Ê±¼äµÄ¼ÆÊýÆ÷¿ªÊ¼¹¤×÷£¬µ±A¡¢B»¬¿é·Ö±ðÅöײC¡¢Dµ²°åʱֹͣ¼Æʱ£¬¼ÇÏ»¬¿éA¡¢B·Ö±ðµ½´ïµ²°åC¡¢DµÄÔ˶¯Ê±¼ät1ºÍt2£»
d£®Óÿ̶ȳ߲â³ö»¬¿éAµÄ×ó¶ËÖÁCµ²°åµÄ¾àÀëL1¡¢»¬¿éBµÄÓҶ˵½Dµ²°åµÄ¾àÀëL2£®
¢ÙÊÔÑéÖл¹Ó¦²âÁ¿µÄÎïÀíÁ¿ÊÇ
»¬¿éA¡¢BµÄÖÊÁ¿mA¡¢mB£®
»¬¿éA¡¢BµÄÖÊÁ¿mA¡¢mB£®
£»
¢ÚÀûÓÃÉÏÊö¹ý³Ì²âÁ¿µÄʵÑéÊý¾Ý£¬ÑéÖ¤¶¯Á¿Êغ㶨Âɵıí´ïʽÊÇ
mA
L1
t1
=mB
L2
t2
£¬
mA
L1
t1
=mB
L2
t2
£¬
£»
¢ÛÀûÓÃÉÏÊöʵÑéÊý¾Ýµ¼³öµÄ±»Ñ¹Ëõµ¯»ÉµÄµ¯ÐÔÊÆÄܵıí´ïʽÊÇ
1
2
mA(
L1
t1
)2+
1
2
mB(
L2
t2
)2
£¬
1
2
mA(
L1
t1
)2+
1
2
mB(
L2
t2
)2
£¬
£®
£¨1£©ÓÃÂÝÐý²â΢Æ÷²âÁ¿Ä³Ò»·½¸ÖµÄºñ¶È£¬ÈçͼAËùʾ£¬¸Ö°åµÄºñ¶ÈΪ
 
 mm£»ÓÃÓαêΪ20·Ö¶ÈµÄ¿¨³ß²â·½¸ÖµÄ±ß³¤£¬ÈçͼBËùʾ£¬¸Ö°åµÄ³¤¶ÈΪ
 
 cm£®
£¨2£©Ä³µç×èÔª¼þµÄ×èÖµRËæζÈt±ä»¯µÄͼÏóÈçͼ¼×Ëùʾ£®Ò»¸öͬѧ½øÐÐÁËÈçÏÂÉè¼Æ£º½«Ò»µç¶¯ÊÆE=1.5V£¨ÄÚ×è²»¼Æ£©µÄµçÔ´¡¢Á¿³Ì5mAÄÚ×èΪ100¦¸µÄµçÁ÷±í¡¢µç×èÏäR¡äÒÔ¼°Óøõç×èÔª¼þR£¬´®Áª³ÉÈçͼÒÒËùʾµÄµç·£®Èç¹û°ÑµçÁ÷±íµÄµçÁ÷¿Ì¶È¸ÄΪÏàÓ¦µÄζȿ̶ȣ¬¾ÍµÃµ½ÁËÒ»¸ö¼òµ¥µÄ¡°½ðÊôµç×èζȼơ±£®
¢ÙµçÁ÷¿Ì¶È½ÏС´¦¶ÔÓ¦µÄζȿ̶È
 
£»£¨Ñ¡Ìî¡°½Ï¸ß¡±»ò¡°½ÏµÍ¡±£©
¢ÚÈôµç×èÏä×èÖµR¡ä=250¦¸£¬Í¼±ûÖÐ3mA¿Ì¶È´¦¶ÔÓ¦µÄζÈÊýֵΪ
 
¡æ£®
£¨3£©Ä³Í¬Ñ§ÎªÁ˲âÁ¿µçÁ÷±íA1µÄÄÚ×辫ȷֵ£¬ÓÐÈçÏÂÆ÷²Ä£º
µçÁ÷±íA1£¨Á¿³Ì300mA£¬ÄÚ×èԼΪ5¦¸£©£»µçÁ÷±íA2£¨Á¿³Ì600mA£¬ÄÚ×èԼΪ1¦¸£©£»µçѹ±íV£¨Á¿³Ì15V£¬ÄÚ×èԼΪ3k¦¸£©£»
»¬¶¯±ä×èÆ÷R1£¨0¡«5¦¸£¬¶î¶¨µçÁ÷Ϊ1A£©£»»¬¶¯±ä×èÆ÷R2£¨0¡«50¦¸£¬¶î¶¨µçÁ÷Ϊ0.01A£©£»µçÔ´E£¨µç¶¯ÊÆ3V£¬ÄÚ×è½ÏС£©£»
¶¨Öµµç×èR0 £¨5¦¸£©£»µ¥µ¶µ¥ÖÀ¿ª¹ØÒ»¸ö¡¢µ¼ÏßÈô¸É£®
¢ÙÒªÇó´ý²âµçÁ÷±íA1µÄʾÊý´ÓÁ㿪ʼ±ä»¯£¬ÇÒ¶à²â¼¸×éÊý¾Ý£¬¾¡¿ÉÄܵļõÉÙÎó²î£¬ÒÔÉϸø¶¨µÄÆ÷²ÄÖ묶¯±ä×èÆ÷Ӧѡ
 
£®ÔÚ´ðÌ⿨µÄ·½¿òÄÚ»­³ö²âÁ¿Óõĵç·ԭÀíͼ£¬²¢ÔÚͼÖбê³öËùÓÃÒÇÆ÷µÄ´úºÅ£®
¢ÚÈôÑ¡²âÁ¿Êý¾ÝÖеÄÒ»×éÀ´¼ÆËãµçÁ÷±íA1µÄÄÚ×èr1£¬Ôòr1µÄ±í´ïʽΪr1=
 
£»ÉÏʽÖи÷·ûºÅµÄÎïÀíÒâÒåÊÇ
 
£®
¾«Ó¢¼Ò½ÌÍø
£¨1£©ÓÃÓαêΪ20·Ö¶ÈµÄ¿¨³ß²âÁ¿Æ䳤¶ÈÈçͼ1£¬ÓÉͼ1¿ÉÖªÆ䳤¶ÈΪ______mm£»ÓÃÂÝÐý²â΢Æ÷²âÁ¿ÆäÖ±¾¶Èçͼ£¬ÓÉͼ1¿ÉÖªÆäÖ±¾¶Îª______mm£»

£¨2£©ÔÚ¡°ÑéÖ¤»úеÄÜÊغ㶨ÂÉ¡±µÄʵÑéÖУ¬´òµã¼ÆʱÆ÷ËùÓõçԴƵÂÊΪ50HZ£¬µ±µØÖØÁ¦¼ÓËٶȵÄֵΪ9.80m/s2£¬²âµÃËùÓÃÖØÎïµÄÖÊÁ¿Îª1.00kg£®¼×¡¢ÒÒ¡¢±ûÈýѧÉú·Ö±ðÓÃͬһװÖôò³öÈýÌõÖ½´ø£¬Á¿³ö¸÷Ö½´øÉϵÚ1¡¢2Á½µã¼äµÄ¾àÀë·Ö±ðΪ0.12cm£¬0.19cmºÍ0.25cm£¬¿É¼û²Ù×÷ÉÏÓдíÎóµÄÊÇѧÉú______£¬´íÎó²Ù×÷______£®
Èô°´ÊµÑéÒªÇóÕýÈ·µØÑ¡³öÖ½´ø½øÐвâÁ¿£¬Á¿µÃÁ¬ÐøÈýµãA£¬B£¬Cµ½µÚÒ»¸öµãµÄ¾àÀëÈçͼ2Ëùʾ£¨ÏàÁÚ¼ÆÊýµãʱ¼ä¼ä¸ôΪ0.02s£©£¬´ÓÆðµãOµ½´òϼÆÊýµãBµÄ¹ý³ÌÖÐÖØÁ¦ÊÆÄܼõÉÙÁ¿ÊÇ¡÷EP=______£¬´Ë¹ý³ÌÖÐÎïÌ嶯ÄܵÄÔö¼ÓÁ¿ÊÇ¡÷EK______£¨È¡g=9.8m/s2£©£»
£¨3£©Èçͼ3ËùʾÆøµæÊdz£ÓõÄÒ»ÖÖʵÑéÒÇÆ÷£¬ËüÊÇÀûÓÃÆø±Ãʹ´ø¿×µÄµ¼¹ìÓ뻬¿éÖ®¼äÐγÉÆøµæ£¬Ê¹»¬¿éÐü¸¡ÔÚ¹ìµÀÉÏ£¬»¬¿éÔÚ¹ìµÀÉϵÄÔ˶¯¿ÉÊÓΪûÓÐĦ²Á£®ÎÒÃÇ¿ÉÒÔÓôøÊúÖ±µ²°åCºÍDµÄÆøµæ¹ìµÀÒÔ¼°»¬¿éAºÍBÀ´ÑéÖ¤¶¯Á¿Êغ㶨ÂÉ£¬ÊµÑé×°ÖÃÈçͼËùʾ£¨µ¯»ÉµÄ³¤¶ÈºöÂÔ²»¼Æ£©£¬²ÉÓõÄʵÑé²½ÖèÈçÏ£º
a£®µ÷ÕûÆøµæ¹ìµÀ£¬Ê¹µ¼¹ì´¦ÓÚˮƽ£»
b£®ÔÚAºÍB¼ä·ÅÈëÒ»¸ö±»Ñ¹ËõµÄÇᵯ»É£¬Óõ綯¿¨ÏúËø¶¨£¬¾²Ö¹·ÅÖÃÔÚÆøµæµ¼¹ìÉÏ£»
c£®°´ÏµçÅ¥·Å¿ª¿¨Ïú£¬Í¬Ê±Ê¹·Ö±ð¼Ç¼»¬¿éA¡¢BÔ˶¯Ê±¼äµÄ¼ÆÊýÆ÷¿ªÊ¼¹¤×÷£¬µ±A¡¢B»¬¿é·Ö±ðÅöײC¡¢Dµ²°åʱֹͣ¼Æʱ£¬¼ÇÏ»¬¿éA¡¢B·Ö±ðµ½´ïµ²°åC¡¢DµÄÔ˶¯Ê±¼ät1ºÍt2£»
d£®Óÿ̶ȳ߲â³ö»¬¿éAµÄ×ó¶ËÖÁCµ²°åµÄ¾àÀëL1¡¢»¬¿éBµÄÓҶ˵½Dµ²°åµÄ¾àÀëL2£®
¢ÙÊÔÑéÖл¹Ó¦²âÁ¿µÄÎïÀíÁ¿ÊÇ______£»
¢ÚÀûÓÃÉÏÊö¹ý³Ì²âÁ¿µÄʵÑéÊý¾Ý£¬ÑéÖ¤¶¯Á¿Êغ㶨Âɵıí´ïʽÊÇ______£»
¢ÛÀûÓÃÉÏÊöʵÑéÊý¾Ýµ¼³öµÄ±»Ñ¹Ëõµ¯»ÉµÄµ¯ÐÔÊÆÄܵıí´ïʽÊÇ______£®

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

¾«Ó¢¼Ò½ÌÍø