D.ÈôËù¼Ó´Å³¡´©¹ýÕû¸öÔ²ÅÌ£¬Ô²Å̽«ÔÈËÙת¶¯
16£®ÈçͼËùʾ£¬ÔÚÔÈÇ¿´Å³¡ÖУ¬ABΪ³¤¶ÈΪL´Öϸ¾ùÔȵĽðÊôË¿£¬´¹Ö±´Å³¡·ÅÖã¬Êä³öµçѹºã¶¨µÄµçÔ´½ÓA¡¢BÁ½¶Ëʱ£¬½ðÊôË¿Êܵ½µÄ°²ÅàÁ¦ÎªF£»Èô½«½ðÊô˿ֻ½ØÈ¡Ò»°ëÔÙÍä³ÉÒ»¸ö°ëÔ²ÐΣ¬ÈÔÈ»½ÓÔڸղŵĵçÔ´Á½¶Ë£¬Ôò½ðÊôË¿Êܵ½µÄ°²ÅàÁ¦Îª£¨ £©
FF??F2FA£® B£® C£® D£®
422?17£®ÈçͼËùʾ£¬ÓÃÓëˮƽ·½Ïò³É¦Á½ÇµÄÇáÉþÀСÎïÌ壬СÎïÌåˮƽÏòÓÒ×öÔÈËÙÖ±ÏßÔ˶¯£¬Ô˶¯µÄ¾àÀëΪL£¬ÉþÖÐÀÁ¦´óСºãΪF£¬ÔòÀÁ¦×öµÄ¹¦Îª( )
A.FL B.FLsin a
D.FLtan a C.FLcosa
18£®ÈçͼËùʾ£¬µÈÀë×ÓÆøÁ÷£¨ÓɸßΡ¢¸ßѹµÄµÈµçºÉÁ¿µÄÕý¡¢¸ºÀë×Ó×é³É£©ÓÉ×ó·½Á¬Ðø²»¶ÏµØÒÔËÙ¶Èv0´¹Ö±ÉäÈëP1ºÍP2Á½¼«°å¼äµÄÔÈÇ¿´Å³¡ÖÐ.Á½Æ½Ðг¤Ö±µ¼ÏßabºÍcdµÄÏ໥×÷ÓÃÇé¿öΪ£º0¡«1sÄÚÅų⣬1s¡«3sÄÚÎüÒý£¬3s¡«4sÄÚÅųâ.ÏßȦAÄÚÓÐÍâ¼Ó´Å³¡£¬¹æ¶¨Ïò×óΪÏßȦAÄڴŸÐӦǿ¶ÈBµÄÕý·½Ïò£¬ÔòÏßȦAÄڴŸÐӦǿ¶ÈBËæÊ±¼ät±ä»¯µÄͼÏóÓпÉÄÜÊÇÏÂͼÖеģ¨ £©
19£®Èçͼ¼×Ëùʾ£¬ÀíÏë±äѹÆ÷Ô¡¢¸±ÏßȦµÄÔÑÊý±ÈΪ5 £º1£¬ÔÏßȦ½Ó½»Á÷µçÔ´ºÍ½»Á÷µçѹ±í£¬¸±ÏßȦ½ÓÓС°220V£¬440W¡±µÄÈÈË®Æ÷¡¢¡°220V£¬220W¡±µÄ³éÓÍÑÌ»ú.Èç¹û¸±ÏßȦµçѹ°´Í¼ÒÒËùʾ¹æÂɱ仯£¬ÔòÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨ £©
A£®¸±ÏßȦÁ½¶ËµçѹµÄ˲ʱֵΪu£½2202sin£¨100¦Ðt£©V B£®±äÁ÷µçѹ±íµÄʾÊýΪ11002V C£®1minÄÚ±äѹÆ÷Êä³öµÄµçÄÜΪ3.96¡Á104J
D£®ÈÈË®Æ÷µÄ·¢Èȹ¦ÂÊÊdzéÓÍÑÌ»ú·¢Èȹ¦ÂʵÄ2±¶
¡¤5¡¤
20£®ÈçͼËùʾ£¬Ò»¸öÖÊÁ¿ÎªmµÄÎïÌå(¿ÉÊÓΪÖʵã)£¬ÓÉÐ±Ãæµ×¶ËµÄAµãÒÔijһ³õËٶȳåÉÏÇã½ÇΪ30¡ãµÄ¹Ì¶¨Ð±Ãæ×öÔȼõËÙÖ±ÏßÔ˶¯£¬¼õËٵļÓËÙ¶È´óСΪg£¬ÎïÌåÑØÐ±ÃæÉÏÉýµÄ×î´ó¸ß¶ÈΪh£¬Ôڴ˹ý³ÌÖУ¨ £©
1mgh2A.ÎïÌå¿Ë·þĦ²ÁÁ¦×ö¹¦
B.ÎïÌåµÄ¶¯ÄÜËðʧÁËmgh
C.ÎïÌåµÄÖØÁ¦ÊÆÄÜÔö¼ÓÁËmgh D.ϵͳ»úеÄÜËðʧÁËmgh
21£®ÖÊÁ¿ÎªmµÄÆû³µÔÚÆ½Ö±Â·ÃæÉÏÆô¶¯£¬Æô¶¯¹ý³ÌµÄËÙ¶ÈͼÏñÈçͼËùʾ£¬ÆäÖÐOAΪ¹ýÔµãµÄÒ»ÌõÖ±Ïß¡£´Ót1ʱ¿ÌÆðÆû³µµÄ¹¦Âʱ£³Ö²»±ä£¬Õû¸öÔ˶¯¹ý³ÌÖÐÆû³µËùÊÜ×èÁ¦ºãΪFf£¬Ôò£¨ £©
A.0¡«t1ʱ¼äÄÚ£¬Æû³µµÄÇ£ÒýÁ¦µÈÓÚ´íÎó£¡Î´ÕÒµ½ÒýÓÃÔ´¡£ B.Æû³µÔÚt1¡«t2ʱ¼äÄڵŦÂʵÈÓÚt2ÒÔºóµÄ¹¦ÂÊ
C.t1¡«t2ʱ¼äÄÚ£¬Æû³µµÄ¹¦ÂʵÈÓÚ´íÎó£¡Î´ÕÒµ½ÒýÓÃÔ´¡£ D.t1¡«t2ʱ¼äÄÚ£¬Æû³µµÄƽ¾ùËٶȵÈÓÚ´íÎó£¡Î´ÕÒµ½ÒýÓÃÔ´¡£
¶þ¡¢ÊµÑéÌâ 22£®£¨6·Ö£©Ä³ÎïÀíС×éµÄͬѧÀûÓÃÈçͼ1ËùʾµÄ×°ÖÃÀ´ÑéÖ¤
»úеÄÜÊØºã¶¨ÂÉ£¬µçԴƵÂÊΪ50 Hz£¬µ±µØµÄÖØÁ¦¼ÓËÙ¶ÈΪ9.80 m/s2£¬ÊÔ·ÖÎöÏÂÁÐÎÊÌ⣺
(1)ËûÃÇÔÚʵÑéÖеõ½Á˼ס¢ÒÒ¡¢±ûÈýÌõʵÑéÖ½´ø£¬Èçͼ2Ëùʾ£¬ÔòӦѡֽ´ø _______ºÃ¡£
(2)ͼ3ÊÇËûÃÇÔÚij´ÎʵÑéÖеõ½µÄÒ»ÌõÖ½´ø£¬Êý¾ÝÈçͼ3Ëùʾ.
¢ÙËûÃÇ×öÁËÈçϵļÆË㣺ÓÉͼÖÐÊý¾Ý¿É¼ÆËã³ö´ò¼ÆÊ±µãBÊ±ÖØ´¸µÄ˲ʱËÙ¶ÈvB=_____ m/s£»ÈôÖØ´¸ÖÊÁ¿Îªm=1 kg£¬ÔòÔÚ´òÏÂBµãÊ±ÖØ´¸µÄ¶¯ÄÜEkB=______J£¬´Ó´òÏÂOµãµ½´òÏÂBµãµÄ¹ý³ÌÖÐÖØ´¸¼õÉÙµÄÖØÁ¦ÊÆÄÜEp=_____ J.±È½ÏEkBºÍEp£¬Èç¹ûÁ½ÕßÔÚÎó²îÔÊÐíµÄ·¶Î§ÄÚÏàµÈ£¬Ôò¿ÉÒÔÑéÖ¤»úеÄÜÊØºã¶¨ÂÉ.
¢ÚÉÏÊöÑéÖ¤ºÏÀíÂð?______.Èô²»ºÏÀí£¬Çë˵Ã÷ÔÒò£º__________________¡£ 23£®£¨10·Ö£©´ÓϱíÖÐÑ¡³öÊʵ±µÄʵÑéÆ÷²Ä,Éè¼ÆÒ»µç·À´²âÁ¿µçÁ÷±íA1µÄÄÚ×èr1.ÒªÇó·½·¨¼ò½Ý,Óо¡¿ÉÄܸߵIJâÁ¿¾«¶È,²¢ÄܲâµÃ¾¡Á¿¶àµÄÊý¾Ý.
¡¤6¡¤
(1)»³öÄãÉè¼ÆµÄµç·ͼ,±êÃ÷ËùÓÃÆ÷²ÄµÄ´úºÅ.
(2)ÈôÑ¡²âÁ¿Êý¾ÝÖеÄÒ»×éÀ´¼ÆËãr1,ÔòËùÓõıí´ïʽr1= ,ʽÖи÷·ûºÅָʲô: .
(3)ÈôÌâÖÐûÓиø³öµçÁ÷±íA2,µ«ÌṩÁËÒÔÏÂ3¸öÒÑÖª×èÖµµÄµç×è.ÈÔÈ»ÐèÒª¸ß¾«¶ÈµØ²âÁ¿µçÁ÷±íA1µÄÄÚ×è,ÄãÈÏΪ¿ÉÒÔÔõÑùÉè¼ÆÊµÑé·½°¸,Çë̸̸ÄãµÄÉè¼ÆË¼Â·,²¢»³öʵÑéµç·ͼ.
A.¶¨Öµµç×èR2(×èÖµ5 ¦¸); B.¶¨Öµµç×èR3(×èÖµ25 ¦¸); C.¶¨Öµµç×èR4(×èÖµ1 000 ¦¸).
Èý¡¢¼ÆËãÌâ 24.£¨14·Ö£©Èçͼ¼×Ëùʾ£¬ÖÊÁ¿M=2 kgµÄľ°åÒÔ³õËÙ¶Èv0=5 m/sÔڹ⻬µÄË®Æ½ÃæÉÏÔ˶¯£¬ÖÊÁ¿m=0.5 kgµÄ»¬¿éÂäÔÚľ°åµÄÓÒ¶ËûÓе¯Æð£¬×îÖÕÇ¡ºÃûµôÏÂÀ´£¬´Ó»¬¿éÂ䵽ľ°åÉÏ¿ªÊ¼¼ÆÊ±£¬¶þÕßµÄËÙ¶È¡ªÊ±¼äͼÏñÈçͼÒÒËùʾ£¬gÈ¡10 m/s2£¬Çó£º
(1)»¬¿éÓëľ°å¼äµÄ¶¯Ä¦²ÁÒòÊý¦Ì¡£ (2)ľ°åµÄ³¤¶ÈLºÍϵͳ²úÉúµÄÄÚÄÜQ¡£ 25£®£¨17·Ö£©ÈçͼËùʾ£¬Ë®Æ½ÃæÉϹ̶¨Ò»¸öÇã½ÇΪ¦È=37¡ãµÄ×ã¹»³¤Ð±Ãæ£¬Ð±Ãæ¶¥¶ËÓÐÒ»¹â»¬µÄÇáÖʶ¨»¬ÂÖ£¬¿ç¹ý¶¨»¬ÂÖµÄÇáϸÉþÁ½¶Ë·Ö±ðÁ¬½ÓÎï¿éAºÍB(Á½Îï¿é¾ù¿ÉÊÓΪÖʵã)£¬ÆäÖÐÎï¿éAµÄÖÊÁ¿ÎªmA=0.8 kg£¬Îï¿éBµÄÖÊÁ¿ÎªmB=1.2 kg.Ò»ÇáÖʵ¯»É϶˹̶¨ÔÚÐ±Ãæµ×¶Ë£¬µ¯»É´¦ÓÚÔ³¤Ê±É϶ËλÓÚÐ±ÃæÉϵÄCµã.³õʼʱÎï¿éAµ½CµãµÄ¾àÀëΪL=0.5 m.ÏÖ¸øA¡¢BÒ»´óСΪv0=3 m/sµÄ³õËÙ¶ÈʹA¿ªÊ¼ÑØÐ±ÃæÏòÏÂÔ˶¯£¬BÏòÉÏÔ˶¯.
¡¤7¡¤
ÈôÎï¿éBʼÖÕδµ½´ïÐ±Ãæ¶¥¶Ë£¬Îï¿éAÓëÐ±Ãæ¼äµÄ¶¯Ä¦²ÁÒòÊýΪ¦Ì=0.5£¬sin37¡ã=0.6£¬cos37¡ã=0.8£¬È¡g=10 m/s2£¬¿ÕÆø×èÁ¦²»¼Æ¡£
(1)ÊÔÇóÎï¿éAÏòϸÕÔ˶¯µ½CµãʱµÄËÙ¶È´óС£»
(2)Èôµ¯»ÉµÄ×î´óѹËõÁ¿Îª¦¤x=0.2 m£¬ÊÔÇ󵯻ɵÄ×î´óµ¯ÐÔÊÆÄÜEpm£» (3)ÔÚ(2)µÄ»ù´¡ÉÏ£¬ÈôÎï¿éB¸Õ¿ªÊ¼Ô˶¯Ê±ÀëË®Æ½ÃæµÄ¸ß¶Èh=0.6 m£¬Çóµ±Îï¿éBÂ䵨ºó£¬Îï¿éAÄܹ»ÉÏÉýµÄ×î´ó¸ß¶È¡£
26£®(14·Ö)ÏÂͼÖУ¬A¡«LΪ³£¼ûÎïÖÊ»ò¸ÃÎïÖʵÄË®ÈÜÒº£¬BÔÚAÆøÌåÖÐȼÉÕ²úÉúר»Æ
É«ÑÌ£¬B¡¢GΪÖÐѧ»¯Ñ§Öг£¼ûµÄ½ðÊôµ¥ÖÊ£¬EµÄÏ¡ÈÜҺΪÀ¶É«£¬IµÄÑæÉ«·´Ó¦Îª»ÆÉ«£¬×é³ÉJµÄÔªËØÔ×ÓºËÄÚÖ»ÓÐÒ»¸öÖÊ×Ó£¬FΪÎÞÉ«¡¢Óд̼¤ÐÔÆøÎ¶µÄÆøÌ壬ÇÒÄÜʹƷºìÈÜÒºÍÊÉ«¡£ Çë»Ø´ðÏÂÁÐÎÊÌ⣺ £¨1£©¿òͼÖÐËùÁÐÎïÖÊÖÐÊôÓڷǵç½âÖʵÄ
ÎïÖÊÊÇ_______£»
£¨2£©½«DµÄË®ÈÜÒºÕô¸É²¢×ÆÉյõ½µÄ
¹ÌÌåÎïÖʵĻ¯Ñ§Ê½Îª___ _£» £¨3£©ÔÚÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÏõËáï§ÈÜ
ÒºÖеμÓÊÊÁ¿µÄKÈÜÒº£¬Ê¹ÈÜÒºµÄpH=7£¬ÔòÈÜÒºÖÐ c£¨Na+£©____c£¨NO3¨D£©£¨Ñ¡Ìî¡°£¾¡±¡°£½¡±»ò¡°£¼¡±£©¡£
£¨4£©¢Ù4 g JÔÚ´¿ÑõÖÐÍêȫȼÉÕÉú³ÉҺ̬»¯ºÏ
Î·Å³öÈÈÁ¿ÎªQkJ£¬Ð´³ö±íʾJȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ ¡£
¢ÚAÊÇÖØÒªµÄ»¯¹¤ÔÁÏ£¬¹¤ÒµÉÏÖÆÈ¡AµÄ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________¡£
£¨5£©ÓëF×é³ÉÔªËØÏàͬµÄÒ»ÖÖ-2¼ÛËá¸ùÀë×ÓM£¬MÖÐÁ½ÖÖÔªËØµÄÖÊÁ¿±ÈΪ4¡Ã3£¬ÒÑÖª
1 mol Aµ¥ÖÊÓ뺬1 mol MµÄÈÜÒºÄÜÇ¡ºÃÍêÈ«·´Ó¦£¬·´Ó¦Ê±½ö¹Û²ìµ½ÓÐdz»ÆÉ«³Áµí²úÉú¡£È¡·´Ó¦ºóµÄÉϲãÇåÒº¼ÓÈëÑÎËáËữµÄÂÈ»¯±µÈÜÒº£¬Óа×É«³Áµí²úÉú¡£ÔòAµ¥ÖÊÓ뺬MµÄÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º____ ¡£ £¨6£©25¡æÊ±£¬ÈôKsp£¨H£©=2.2¡Á10-20£¬Ïò0.022mol/LµÄEÈÜÒºÖÐÖðµÎµÎÈëÈÜÒºK£¬µ±¿ª
ʼ³öÏÖ³Áµíʱ£¬ÈÜÒºÖеÄc(OH-)= ¡£
27.(15·Ö). ¼×±½ÊÇÓлú»¯¹¤Éú²úµÄ»ù±¾ÔÁÏÖ®Ò»¡£ÀûÓÃÒÒ´¼ºÍ¼×±½ÎªÔÁÏ£¬¿É°´ÏÂÁз
ÏߺϳɷÖ×Óʽ¾ùΪC9H10O2µÄÓлú»¯¹¤²úÆ·EºÍJ¡£
¡¤8¡¤
Ïà¹ØÍÆ¼ö£º