ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Ð¡Ã÷ÓÃÌìƽ¡¢ÉÕ±­¡¢ÓÍÐԱʼ°×ãÁ¿µÄË®²âÁ¿Ò»¿é¶ìÂÑʯµÄÃܶȣ¬ÊµÑé²½ÖèÈçÏ¡£

(1)½«Ììƽ·ÅÔÚˮƽ×ÀÃæÉÏ£¬°ÑÓÎÂ벦ÖÁ±ê³ß×ó¶ËµÄ¡°0¡±¿ÌÏß´¦£¬·¢ÏÖºáÁºÎȶ¨Ê±Ö¸ÕëÆ«Ïò·Ö¶ÈÅ̵ÄÓҲ࣬ҪʹºáÁºÔÚˮƽλÖÃƽºâ£¬Ó¦½«Æ½ºâÂÝĸÍù_______£¨×ó/ÓÒ£©µ÷¡£ÓÃÌìƽ·Ö±ð²â³ö¶ìÂÑʯµÄÖÊÁ¿31.8gºÍ¿ÕÉÕ±­µÄÖÊÁ¿ÊÇ90g£»

(2)Èçͼ¼×£¬°Ñ¶ìÂÑʯ·ÅÈëÉÕ±­ÖУ¬ÍùÉÕ±­µ¹ÈëÊÊÁ¿µÄË®£¬ÓñÊÔÚÉÕ±­±Ú¼ÇÏ´ËʱˮÃæλÖÃΪM£¬È»ºó·ÅÔÚÌìƽ×óÅÌ£¬Èçͼ±û£¬±­¡¢Ë®ºÍ¶ìÂÑʯµÄ×ÜÖÊÁ¿Îª_______g£»

(3)½«¶ìÂÑʯ´ÓË®ÖÐÈ¡³öºó£¬ÔÙÍùÉÕ±­ÖлºÂý¼ÓË®£¬Ê¹Ë®ÃæÉÏÉýÖÁ¼ÇºÅM£¬ÈçͼÒÒËùʾ£¬ÓÃÌìƽ²â³ö±­ºÍË®µÄ×ÜÖÊÁ¿Îª142g£»

(4)¸ù¾ÝËù²âÊý¾Ý¼ÆËã³ö¶ìÂÑʯµÄÃܶÈΪ_______g/cm3£»£¨¼ÆËã½á¹û±£ÁôһλСÊý£©

(5)°´ÕÕÕâÑùµÄ·½·¨²â³öÀ´µÄÃܶÈÖµ_______£¨Ñ¡Ìî¡°Æ«´ó¡±¡°Æ«Ð¡¡±»ò¡°ÈÔȻ׼ȷ¡±£©£»

(6)ÕûÀíʵÑéÆ÷²Äʱ·¢ÏÖ£¬ÌìƽµÄ×óÅÌÓÐÒ»¸öȱ½Ç£¬ÔòÖÊÁ¿µÄ²âÁ¿½á¹û_______¡££¨Ñ¡Ìî¡°Æ«´ó¡±¡°Æ«Ð¡¡±»ò¡°ÈÔȻ׼ȷ¡±£©

¡¾´ð°¸¡¿×ó 161.6 2.6 ÈÔȻ׼ȷ ÈÔȻ׼ȷ

¡¾½âÎö¡¿

(1)[1]°ÑÌìƽ·ÅÔÚˮƽ̨ÉÏ£¬ÏÈÒª°ÑÓÎÂ벦µ½±ê³ß×ó¶ËµÄÁã¿Ì¶ÈÏß´¦£¬Í¨¹ýµ÷½ÚƽºâÂÝĸʹÌìƽƽºâ£»Ö¸ÕëÆ«Ïò·Ö¶ÈÅ̵ÄÓҲ࣬ҪʹºáÁºÔÚˮƽλÖÃƽºâ£¬Ó¦½«Æ½ºâÂÝĸÍù×óµ÷¡£

(2)[2]ÓÉͼ±û¿ÉÖª£¬±­¡¢Ë®ºÍ¶ìÂÑʯµÄ×ÜÖÊÁ¿

m=100g+50g+10g+1.6g=161.6g

Ë®µÄÖÊÁ¿Îª

mˮ1=161.6g-31.8g-90g=39.8g

´ËʱˮµÄÌå»ýΪ

Vˮ1==39.8cm3

(4)[4]½«¶ìÂÑʯ´ÓË®ÖÐÈ¡³öºó£¬ÔÙÍùÉÕ±­ÖлºÂý¼ÓË®£¬Ê¹Ë®ÃæÉÏÉýÖÁ¼ÇºÅM£¬ÓÃÌìƽ²â³ö±­ºÍË®µÄ×ÜÖÊÁ¿Îª142g£¬´Ëʱ£¬±­ÖÐË®µÄÌå»ý

VË®==52cm3

¶ìÂÑʯµÄÌå»ýµÈÓÚ¼ÓÈëË®µÄÌå»ý£¬Ôò

V=Vˮ1-Vˮ=52cm3-39.8cm3=12.2cm3

¶ìÂÑʯµÄÃܶÈ

¦Ñ=¡Ö2.6g/cm3

(5)[5]¼ÆËã¶ìÂÑʯÌå»ýʱºòµÄʱºò£¬¼ÓÈëË®µ½±ê¼ÇλÖ㬲»¹ÜÊDz»ÊÇ´ø³öË®£¬¶¼»á¼Óµ½±ê¼ÇλÖã¬×îºó²âÁ¿³öÀ´µÄ×ÜÖÊÁ¿²»±ä£¬ËüµÄÌå»ý²»±ä£¬ÓɦÑ=¿ÉÖªËüµÄÃܶȲ»±ä¡£

(6)[6]ÌìƽµÄ×óÅÌÓÐÒ»¸öȱ½Ç£¬µ÷½ÚƽºâÂÝĸºó£¬×óÓÒÁ½ÅÌÖÊÁ¿ÏàµÈ£¬²âÁ¿ÎïÌåÖÊÁ¿Ê±£¬×óÓÒÁ½ÅÌÖÊÁ¿Ò²ÏàµÈ£¬²»Ó°Ïì²âÁ¿½á¹û£»¼´Ëù²âµÄÃܶÈÖµÈÔȻ׼ȷ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Èçͼ¼×ËùʾΪһ×ÔÖÆÎÕÁ¦²âÊÔÆ÷£¬ÐéÏß¿òÄÚ²¿·ÖΪÎÕÁ¦²âÊԶˣ¬B¹Ì¶¨²»¶¯Á¬½ÓÓÚÇᵯ»ÉÒ»¶Ë£¬A¶ËÓë½ðÊôƬPÏàÁ¬²¢Á¬½Óµ¯»ÉÁíÒ»¶Ë£¬µ¯»Éµç×è²»¼Æ£¬A¡¢BÓëÊÖ½Ó´¥´¦¾øÔµ¡£

(1)µç·Éè¼Æ£ºÒªÊ¹ÎÕÁ¦¼ÆÄÜÕý³£¹¤×÷£¬ÏÂÁеç·Á¬½ÓÕýÈ·µÄÊÇ£¨______£©

A.µã¡°1¡±Óëµã¡°2¡±Á¬½Ó¡¡¡¡

B.µã¡°1¡±Óëµã¡°3¡±Á¬½Ó

C.µã¡°2¡±Óëµã¡°3¡±Á¬½Ó¡¡¡¡

(2)ÕýÈ·Íê³Éµç·Á¬½Óºó£¬±ÕºÏ¿ª¹ØS£¬¸Ä±äÊ©¼ÓÔÚA¡¢BÁ½¶ËÎÕÁ¦´óС£¬·¢ÏÖµçѹ±íʾÊýʼÖÕ²»±äÇÒ²»ÎªÁ㣬·¢Éú¹ÊÕϵÄÔ­Òò¿ÉÄÜÊÇ£¨______£©

A.R0¶Ï· B.R1¶Ì·

C.µçѹ±í¶Ì· D.½ðÊô»¬Æ¬P´¦¶Ï·

(3)¿Ì¶Èµ÷ÊÔ£ºµ±A¡¢B¶Ë²»ÊÜÎÕÁ¦Ê±£¬Ê¹½ðÊô»¬Æ¬P´¦ÓÚR1µÄa¶Ë£¬´Ëʱµçѹ±íµÄʾÊýΪÁã¡£µ±A¡¢B¶ËËùÊÜÎÕÁ¦Îª10Nʱ£¬½ðÊô»¬Æ¬P¸ÕºÃÖ¸ÔÚR1ÉϾàÀëa¶Ë³¤¶È´¦£¬Çë¼ÆËã»Ø´ð¡°10N¡±Ó¦±êÔÚµçѹ±í_____·üµÄλÖÃÉÏ£¿£¨µç×èR0=5¦¸£¬µç×èR1=20¦¸£¬µçÔ´µçѹU=3.0V£¬µçѹ±íµÄÁ¿³ÌΪ 0~3.0V£©

(4)Èô½ðÊô»¬Æ¬PÔÚa¶Ëʱ£¬µ¯»É¸ÕºÃ´¦ÓÚÔ­³¤£¬µ¯»ÉËùÊÜѹÁ¦ºÍµ¯»ÉѹËõÁ¿Ö®¼äµÄ±ä»¯¹ØϵÈçͼÒÒËùʾ¡£µ±µçѹ±íʾÊýÈçͼ±ûËùʾ£¬´Ëʱ¶ÔÓ¦µÄÎÕÁ¦Îª _____N¡£

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

¾«Ó¢¼Ò½ÌÍø