ÔòÔÚx=0.4m´¦»¬¿é
A£®ËùÊܵ糡Á¦´óСΪ5.0¡Á10-3N B£®ËùÔÚλÖõĵçÊÆÎª4.0¡Á105V C£®µçÊÆÄÜΪ2.0¡Á10-3J D£®ËÙ¶È´óСΪ0.2m/s
21£®Æû³µÔÚÆ½Ö±¹«Â·ÉÏÐÐÊ»£¬Ä³ÈË´Ó³µ´°Ïà¶ÔÓÚ³µ¾²Ö¹ÊÍ·ÅÒ»¸öСÇò£¬Óù̶¨ÔÚ·±ßµÄÕÕÏà»ú½øÐÐÉÁ¹âÕÕÏàÁ½´Î£¬µÃ
µ½ÈçÏÂÐÅÏ¢£º¢ÙÁ½´ÎÉÁ¹âµÄʱ¼ä¼ä¸ôΪ0.5s£»¢ÚµÚÒ»´ÎÉÁ¹âʱ£¬Ð¡Çò¸ÕÊÍ·Å£¬µÚ¶þ´ÎÉÁ¹âʱ£¬Ð¡Çò¸ÕÂ䵨£»¢ÛÁ½´ÎÉÁ¹âµÄʱ¼ä¼ä¸ôÄÚ£¬Æû³µÇ°½øÁË5m¡¢Ð¡ÇòµÄÎ»ÒÆÎª5m¡£ÖØÁ¦¼ÓËÙ¶ÈÈ¡10m/s2£¬¿ÕÆø×èÁ¦²»¼Æ£¬¸ù¾ÝÒÔÉÏÐÅÏ¢ÄÜÈ·¶¨µÄÊÇ
A£®Æû³µ×öÔÈËÙÖ±ÏßÔ˶¯ B£®Ð¡ÇòÊͷŵãÀëµØµÄ¸ß¶È
C£®µÚ¶þ´ÎÉÁ¹âʱÆû³µµÄËÙ¶È D£®Á½´ÎÉÁ¹âµÄʱ¼ä¼ä¸ôÄÚÆû³µµÄƽ¾ùËÙ¶È
µÚ¢ò¾í
Èý¡¢·ÇÑ¡ÔñÌ⣺¹²174·Ö¡£µÚ22¡«32ÌâΪ±Ø¿¼Ì⣬ÿ¸öÊÔÌ⿼Éú¶¼±ØÐë×÷´ð¡£µÚ33¡«38ÌâΪѡ¿¼Ì⣬¿¼Éú¸ù¾ÝÒªÇó
×÷´ð¡£
£¨Ò»£©±Ø¿¼Ì⣺¹²129·Ö¡£ 22£®£¨5·Ö£©
ÓÃÈçͼËùʾµÄ×°ÖÃÀ´ÑéÖ¤¶¯Á¿Êغ㶨ÂÉ¡£»¬¿éÔÚÆøµæµ¼¹ìÉÏÔ˶¯Ê±×èÁ¦²»¼Æ£¬ÆäÉÏ·½µ²¹âÌõµ½´ï¹âµçÃÅD£¨»òE£©£¬¼ÆÊ±Æ÷¿ªÊ¼¼ÆÊ±£»µ²¹âÌõµ½´ï¹âµçÃÅC£¨»òF£©£¬¼ÆÊ±Æ÷Í£Ö¹¼ÆÊ±¡£ÊµÑéÖ÷Òª²½ÖèÈçÏ£º
a£®ÓÃÌìÆ½·Ö±ð²â³ö»¬¿éA¡¢BµÄÖÊÁ¿mA¡¢mB£» b£®¸øÆøµæµ¼¹ìÍ¨Æø²¢µ÷ÕûʹÆäˮƽ£»
c£®µ÷½Ú¹âµçÃÅ£¬Ê¹ÆäλÖúÏÊÊ£¬²â³ö¹âµçÃÅC¡¢D¼äµÄˮƽ¾àÀëL£»
d£®A¡¢BÖ®¼ä½ôѹһÇᵯ»É£¨ÓëA¡¢B²»Õ³Á¬£©£¬²¢ÓÃϸÏß˩ס£¬Èçͼ¾²ÖÃÓÚÆøµæµ¼¹ìÉÏ£»
e£®ÉÕ¶ÏϸÏߣ¬A¡¢B¸÷×ÔÔ˶¯£¬µ¯»É»Ö¸´Ô³¤Ç°A¡¢B¾ùδµ½´ï¹âµçÃÅ£¬´Ó¼ÆÊ±Æ÷ÉÏ·Ö±ð¶ÁÈ¡A¡¢BÔÚÁ½¹âµçÃÅÖ®¼äÔ˶¯µÄʱ¼ätA¡¢tB¡£
£¨1£©ÊµÑéÖл¹Ó¦²âÁ¿µÄÎïÀíÁ¿xÊÇ £¨ÓÃÎÄ×Ö±í´ï£©¡£ £¨2£©ÀûÓÃÉÏÊö²âÁ¿µÄÊý¾Ý£¬ÑéÖ¤¶¯Á¿Êغ㶨Âɵıí´ïʽÊÇ£º
£¨ÓÃÌâÖÐËù¸øµÄ×Öĸ±íʾ£©¡£
£¨3£©ÀûÓÃÉÏÊöÊý¾Ý»¹Äܲâ³öÉÕ¶ÏϸÏßǰµ¯»ÉµÄµ¯ÐÔÊÆÄÜEp= £¨ÓÃÌâÖÐËù¸øµÄ×Öĸ±íʾ£©¡£ 23£®£¨10·Ö£©
¡°²â¶¨Íµ¼Ïߵĵç×èÂÊ¡±ÊµÑéÖпɹ©Ê¹ÓÃµÄÆ÷²ÄÓУº
5
A£®ºá½ØÃæ»ýΪ1.0 mm2¡¢³¤¶ÈΪ100 mµÄÒ»À¦Íµ¼Ïߣ¨µç×èRxÔ¼2¦¸£©£»
G£ºÄÚ×èRg£½100 ¦¸£¬ÂúÆ«µçÁ÷Ig£½3 mA£» B£®µçÁ÷±í¡ð
A£ºÁ¿³Ì0.6 A£¬ÄÚ×èÔ¼1 ¦¸£» C£®µçÁ÷±í¡ð
D£®»¬¶¯±ä×èÆ÷R£º×î´ó×èÖµ5 ¦¸£»
E£®¶¨Öµµç×裺R0£½3 ¦¸£¬R1£½900 ¦¸£¬R2£½1000 ¦¸£» F£®µçÔ´£ºµç¶¯ÊÆ6 V£¬ÄÚ×è²»¼Æ£» G£®¿ª¹Ø¡¢µ¼ÏßÈô¸É¡£ ÇëÍê³ÉÏÂÁÐʵÑéÄÚÈÝ£º
GÓ붨ֵµç×è´®Áª¸Ä×°³ÉÁ¿³ÌΪ3 VµÄµçѹ±í£¬Ôò¶¨Öµµç×èӦѡ_____£¨Ñ¡Ìî¡°R1¡±»ò¡°R2¡±£©¡£ £¨1£©°ÑµçÁ÷±í¡ð
£¨2£©ÎªÁ˾¡¿ÉÄÜ»ñÈ¡¶à×éÊý¾Ý£¬ÊµÑéµç·ͼӦѡÏÂÁÐËÄ·ùÖÐµÄ £¬µç·ÖÐR0µÄ×÷ÓÃÊÇ ¡£
£¨3£©¸ù¾ÝÕýÈ·µÄµç·ͼ£¬Íê³ÉʵÎïͼµÄÁ¬½Ó£¨ÒÑÕýÈ·Á¬½ÓÁ˲¿·Öµ¼Ïߣ©¡£
GµÄ¶ÁÊýΪ2.40mAʱ£¬AµÄ¶ÁÊýΪ0.50 £¨4£©Ä³´Î²âÁ¿ÖУ¬µçÁ÷±í¡ðµçÁ÷±í¡ð
A£¬ÓÉ´ËÇóµÃ͵¼Ïߵĵç×èÂÊΪ______¦¸¡¤m£¨±£Áô2λÓÐЧÊý×Ö£©¡£ 24£®£¨12·Ö£©
µõ´¸´ò×®»úÈçͼa£¬Æä¹¤×÷¹ý³Ì¿ÉÒÔ¼ò»¯ÎªÍ¼b£ºÖÊÁ¿m=2.0¡Á103kgµÄµõ´¸ÔÚÉþ×ӵĺ㶨ÀÁ¦F×÷ÓÃÏ´ÓÓ붤×Ó½Ó´¥´¦Óɾ²Ö¹¿ªÊ¼Ô˶¯£¬ÉÏÉýÒ»¶Î¸ß¶Èºó³·È¥F£¬µ½×î¸ßµãºó×ÔÓÉÂäÏ£¬×²»÷¶¤×Ó½«¶¤×Ó´òÈëÒ»¶¨Éî¶È¡£µõ´¸ÉÏÉý
¹ý³ÌÖУ¬»úеÄÜEÓëÉÏÉý¸ß¶ÈhµÄ¹ØÏµÈçͼc£¬²»¼ÆÄ¦²Á¼°¿ÕÆø×èÁ¦£¬g=10m/s2¡£Çó£º £¨1£©µõ´¸ÉÏÉýh1=1.6mʱµÄËÙ¶È´óС£»
£¨2£©µõ´¸ÉÏÉýh1=1.6mºó£¬ÔÙ¾¹ý¶à³¤Ê±¼äײ»÷¶¤×Ó£» £¨3£©µõ´¸ÉÏÉýh2=0.4mʱ£¬ÀÁ¦FµÄ˲ʱ¹¦ÂÊ¡£
6
25£®£¨20·Ö£©
Èçͼ£¬xÎªÖ½ÃæÄÚµÄÒ»ÌõÖ±Ïߣ¬P¡¢NÊÇxÉϵÄÁ½¸öµã£¬ÔÈÇ¿´Å³¡´¹Ö±Ö½Ãæ¡£Á½¸ö´øµçÁ£×Óa¡¢b·Ö±ð´ÓP¡¢Nͬʱ¿ªÊ¼ÔÚÖ½ÃæÄÚÔ˶¯¡£aµÄ³õËÙ¶È´¹Ö±xÏòÉÏ£¬Ô˶¯¹ì¼£ÈçͼÖÐÐéÏßËùʾ£¬OΪԲÐÄ£¬PCÊÇÖ±¾¶£¬AÊÇÔ²ÖÜÉϵĵ㣻bµÄ³õËÙ¶È·½ÏòÊÇÖ½ÃæÄÚËùÓпÉÄܵķ½Ïò¡£
ÒÑÖª£ºAOÁ¬Ïß´¹Ö±x£¬PO=OC=CN£»aµÄ³õËÙ¶ÈΪv£»a¡¢b´øµÈÁ¿ÒìÖÖµçºÉ£¬aµÄÖÊÁ¿ÎªbµÄÁ½±¶£¬a¡¢b¼äµÄÏ໥×÷ÓÃÁ¦¼°ËùÊÜÖØÁ¦²»¼Æ¡£
£¨1£©Çóa¡¢bµÄÖÜÆÚÖ®±È£»
£¨2£©Èôa¡¢bÔÚAµãÏàÓö£¬ÇóbµÄËÙ¶È´óС£»
£¨3£©bµÄËÙ¶ÈСÓÚij¸öÁÙ½çÖµv0ʱ£¬a¡¢b²»¿ÉÄÜÏàÓö£¬Çóv0µÄ´óС¡£
£¨1£© NH4ClO4 ¸ßηֽâʱ²úÉúË®ÕôÆøºÍÈýÖÖµ¥ÖÊÆøÌ壬 ´Ó¶ø²úÉú¾Þ´óµÄÍÆ¶¯Á¦¡£ ¢ÙÈýÖÖµ¥ÖÊÆøÌåµÄ»¯Ñ§Ê½·Ö±ðΪ________¡£
¢ÚÀûÓÃÏÂͼװÖöԷֽâ²úÉúµÄÈýÖÖµ¥ÖÊÆøÌå½øÐÐÑéÖ¤¡£ ×°Öà A¡¢ B¡¢ C ÖÐËù×°ÊÔ¼ÁÒÀ´ÎΪ________£¨Ìî±êºÅ£©a. ʪÈóµÄµí·Û KI ÊÔÖ½ b. ʪÈóµÄºìÉ«²¼Ìõ c. NaOH ÈÜÒº d. Na2SO3 ÈÜÒº e. CuO f. Cu
£¨2£©NH4ClO4 µÄÖÊÁ¿·ÖÊý¿Éͨ¹ý¼×È©·¨²â¶¨¡£
ÒÑÖª£º NH4ClO4 Óë¼×È©·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º 4NH4++6HCHO=(CH2)6N4H++3H++6H2O (CH2)6N4H+ (CH2)6N4+H+ K=7¡Á10-6
ʵÑé²½Ö裺
¢ñ.È¡ a g NH4ClO4 ²úÆ·ÓÚ×¶ÐÎÆ¿ÖУ¬ ÓÃË®Èܽ⣬ ¼ÓÈë 40mL ¼×ȩˮÈÜÒº£¬ ³ä·Ö·´Ó¦¡£ ¢ò.ÒÔ·Ó̪×÷ָʾ¼Á£¬ Óà c mol¡¤L-1 ±ê×¼ NaOH ÈÜÒºµÎ¶¨¡£ ¼Ç¼Êý¾Ý¡£ ¢ó.ÖØ¸´²½Öè¢ñ¡¢ ¢ò 2~3 ´Î£¬ ´¦ÀíÊý¾Ý¡£ Ôڴ˹ý³ÌÖУº
7
¡£¢ÙʵÑéÊÒÖеõ½±ê×¼ NaOH ÈÜÒºµÄ·½·¨ÊÇ_______¡£ ¢Ú²½Öè¢ò·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ_______¡£
¢ÛÈôʵÑ鯽¾ùÏûºÄ NaOH ÈÜÒºµÄÌå»ýΪ b mL¡£ NH4ClO4 ÖÊÁ¿·ÖÊýµÄ¼ÆËãʽΪ_______¡£
¢ÜʵÑéÖÐʹÓõļ×È©³£»ìÓÐ΢Á¿¼×Ëᣬ »áµ¼Ö²ⶨ½á¹ûÆ« £¨Ìî¡°¸ß¡±»ò¡°µÍ¡±£©¡£ »ìÓм×ËáµÄÔÒòÊÇ_______¡£ 27. £¨14·Ö£©ÒÑÖªÑõ»¯Ð¿Ñ̳¾Ö÷Òª³É·ÖΪ ZnO£¬ »¹º¬ÓÐ PbO¡¢ Ag2O¡¢ CuO¡¢ CdO¡¢ GeO2¡¢ MnO¡¢ FeOµÈ¡£ ´ÓÑ̳¾ÖлØÊÕÕàпµÄ¹¤ÒÕÁ÷³ÌÈçÏ£º
»Ø´ðÏÂÁÐÎÊÌâ¡£
£¨1£© ÓùýÁ¿½¹Ì¿¸ßλ¹Ô GeO2µÄ»¯Ñ§·½³ÌʽΪ_______¡£ £¨2£© п·ÛµÄ×÷ÓÃÓУº ³ýȥ͡¢ ïÓµÈÔÓÖÊ¡¢______¡£
£¨3£© (NH4)2S2O8ÖÐ SµÄ»¯ºÏ¼ÛΪ+6£¬ Ôò S2O82Öк¬¹ýÑõ¼üÊýĿΪ_____¡£ ÓÃ(NH4)2S2O8³ýÃ̵ÄÀë×Ó·½³ÌʽΪ
£
____________¡£
£¨4£© ¡°ÂËÒº B¡± µÄÖ÷Òª³É·ÝΪ______¡£¡°³Áп¡± Öз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________¡£ £¨5£© Èô¡°ÂËÒº A¡± ÖÐ c(Pb2+)£½2¡Á10-9mol¡¤L-1£¬ ´Ëʱ c(Ag+)£½_______mol¡¤L-1¡£ £¨ÒÑÖª£º Ksp(PbSO4) £½2¡Á10-8mol¡¤L-1£¬ Ksp(Ag2SO4) £½1¡Á10-5mol¡¤L-1£©
28.£¨15 ·Ö£© CO2¼ÓÇâÖÆ±¸¼×ËᣨHCOOH£© ¿ÉÓÃÓÚ»ØÊÕÀûÓà CO2¡£ »Ø´ðÏÂÁÐÎÊÌ⣺ £¨1£© ÒÑÖª£º C(s)+O2(g) £½ CO2(g) ¡÷H1£½£394 kJ¡¤mol-1 C(s)+H2(g) +O2(g) £½ HCOOH(g) ¡÷H2£½£363 kJ¡¤mol-1 ·´Ó¦ CO2(g) +H2(g)
HCOOH(g)µÄ¡÷H£½_____________¡£
£¨2£© ζÈΪ T1ʱ£¬ ½«µÈÎïÖʵÄÁ¿µÄ CO2ºÍ H2³äÈëÌå»ýΪ 1LµÄÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦£º CO2(g)+ H2 (g)
HCOOH(g) K£½1
ʵÑé²âµÃ£º v Õý£½k Õý c(CO2) ¡¤c(H2)£¬ v Äæ£½k Äæ c (HCOOH)£¬ k Õý¡¢ k ÄæÎªËÙÂʳ£Êý¡£ ¢Ùµ± HCOOH µÄÌå»ý·ÖÊýΪ 20%ʱ£¬ CO2µÄת»¯ÂÊΪ___________¡£
¢ÚT1ʱ£¬ kÕý£½_____£¨ÒÔ kÄæ±íʾ£©¡£ µ±Î¶ȸıäΪ T2ʱ£¬ kÕý£½0.9kÄæ£¬ Ôò T2_____T1£¨Ìî¡°>¡±¡¢¡°<¡± »ò¡°£½¡±£©¡£ £¨3£© Óà NaHCO3´úÌæ CO2×÷Ϊ̼ԴÓÐÀûÓÚʵÏÖ CO2¼ÓÇâÖÆ±¸¼×Ëá¡£ Ïò·´Ó¦Æ÷ÖмÓÈëNaHCO3Ë®ÈÜÒº¡¢ Al·Û¡¢Cu·Û£¬ ÔÚ 300¡æÏ·´Ó¦2h£¬ ʵÑé½á¹ûÈçÏÂ±í¡£ ±àºÅ ÎïÖʵÄÁ¿/mmol NaHCO3
̼ת»¯ÂÊ/% Cu 8
Al