Question

19£®Èçͼ1ÀûÓÃÂÖ´¬Éϵĵ綯»úºÍÀÂÉþ´Óºþµ×ÊúÖ±´òÀ̳öÒ»³¤·½ÌåÎïÌ壬ͼ2ÖÐP-tͼÏó±íʾÁ˵綯»úÊä³öµÄ»úе¹¦ÂÊPÓëÎïÌåÉÏÉýʱ¼ätµÄ¹Øϵ£®ÒÑÖª0¡«80sʱ¼äÄÚ£¬ÎïÌåʼÖÕÒÔv=0.1m/sµÄËÙ¶ÈÔÈËÙÉÏÉý£¬µ±t=80sʱ£¬ÎïÌåµ×²¿Ç¡ºÃƽÎȵķÅÔÚÂÖ´¬µÄˮƽ¼×°åÉÏ£®ÒÑÖªµç¶¯»úµÄµçѹÊÇ200V£¬ÎïÌåÉÏÉý¹ý³ÌÖеÄĦ²Á×èÁ¦ºÍË®µÄ×èÁ¦²»¼Æ£¬²»¿¼ÂÇÖØÎï³öˮǰºóÖÊÁ¿µÄ±ä»¯£¬gÈ¡10N/kg£®Çó£º?
£¨1£©ºþµ×Êܵ½Ë®µÄѹǿ¶à´ó£¿?
£¨2£©ÎïÌåÔÚºþË®ÖÐÊܵ½µÄ×î´ó¸¡Á¦ÊǶà´ó£¿?
£¨3£©Èôµç¶¯»úµçÄÜת»»Îª»úеÄܵÄЧÂÊΪ80%£¬ÇóÔÚ0¡«50s ÄÚ£¬µç¶¯»úÏßȦÖеçÁ÷µÄ´óСÊǶàÉÙ£¿

Answer 1

·ÖÎö £¨1£©½áºÏͼÏóµÃ³ö£º¸ù¾Ýs=vt¿ÉÇóµÃºþµ×µÄÉî¶È£¬ÀûÓÃp=¦ÑghÇó³öºþµ×Êܵ½Ë®µÄѹǿ£»
£¨2£©ÒÑÖªÎïÌåÔ˶¯µÄËٶȣ¬ÓÉP=$\frac{W}{t}$½áºÏͼÏó¿ÉÇóµÃµç¶¯»ú¶ÔÀÂÉþµÄÀ­Á¦£»ÒÑÖªÎïÌåÔ˶¯µÄËٶȣ¬ÓÉP=$\frac{W}{t}$½áºÏͼÏó¿ÉÇóµÃÎïÌåµÄÖØÁ¦£¬È»ºó¿ÉÖªÆäÖÊÁ¿£¬ÔÙÀûÓÃG-F_À­¿ÉÇóµÃÆ両Á¦£»
£¨3£©ÓÉͼÏó¿ÉÖªÔÚ0¡«50sÄÚ£¬Æ书ÂÊΪP₁£¬¸ù¾Ýµç¶¯»úµçÄÜת»»Îª»úеÄܵÄЧÂÊΪ80%£¬ÔÙÀûÓÃP=UI¿ÉÇóµÃµç¶¯»úÏßȦÖеçÁ÷µÄ´óС£®

½â´ð ½â£º
£¨1£©t=60sʱ£¬ÎïÌåµ×²¿Ç¡ºÃÀ뿪ˮÃ棬ÔòÓÉv=$\frac{s}{t}$¿ÉµÃ£¬ºþµ×µÄÉî¶È£º
h=vt=0.1m/s¡Á60s=6m£»
ËùÒÔºþµ×Êܵ½Ë®µÄѹǿ£º
p=¦Ñgh=1¡Á10³kg/m³¡Á10N/kg¡Á6m=6¡Á10⁴Pa£»
£¨2£©ÓÉͼÏó¿ÉÖª£¬ÔÚ0¡«50sÄÚ£¬À­Á¦µÄ¹¦ÂÊΪP=1¡Á10³W£¬
ÓÉP=$\frac{W}{t}$=$\frac{Fs}{t}$=FvµÃ£¬
F₁=$\frac{{P}_{1}}{v}$=$\frac{1¡Á1{0}^{3}W}{0.1m/s}$=1¡Á10⁴N£»
ÓÉͼÏó¿ÉÖª£¬ÔÚ60ʱ£¬ÎïÌåÒÑÈ«²¿³öË®£¬¹¦ÂÊΪP₂=2¡Á10³W£¬Ç£ÒýÁ¦£º
F₂=$\frac{{P}_{2}}{v}$=$\frac{2¡Á1{0}^{3}W}{0.1m/s}$=2¡Á10⁴N£¬
ÎïÌåµÄÖØÁ¦£ºG=F₂=2¡Á10⁴N£»
ÎïÌåÔÚÉÏÉý¹ý³ÌÖÐÊܵ½µÄ¸¡Á¦£º
F_¸¡=G-F₁=2¡Á10⁴N-1¡Á10⁴N=1¡Á10⁴N£»
£¨3£©ÔÚ0¡«50s ÄÚ£¬P₁=1¡Á10³W£¬ÓɦÇ=$\frac{{W}_{»úе}}{{W}_{µç}}$=$\frac{{P}_{»úе}}{{P}_{µç}}$¿ÉµÃµç¶¯»úµÄ¹¦ÂÊ£º
P_µç1=$\frac{{P}_{1}}{{¦Ç}_{\;}}$=$\frac{1¡Á1{0}^{3}W}{80%}$=1.25¡Á10³W£¬
ÓÉP=UI¿ÉµÃ£¬µç¶¯»úÏßȦÖеĵçÁ÷£º
I=$\frac{{P}_{µç1}}{U}$=$\frac{1.25¡Á1{0}^{3}W}{200V}$=6.25A£®
´ð£º£¨1£©ºþµ×Êܵ½Ë®µÄѹǿÊÇ6¡Á10⁴Pa£»
£¨2£©ÎïÌåÔÚºþË®ÖÐÊܵ½µÄ×î´ó¸¡Á¦ÊÇ1¡Á10⁴N£»
£¨3£©ÔÚ0¡«50s ÄÚ£¬µç¶¯»úÏßȦÖеçÁ÷µÄ´óСÊÇ6.25A£®

µãÆÀ ±¾Ì⿼²éÁËËٶȹ«Ê½¼°ÆäÓ¦Óᢰ¢»ùÃ×µÂÔ­ÀíºÍµç¹¦ÂÊÓëµçѹ¡¢µçÁ÷µÄ¹Øϵ£¬¹Ø¼üÊÇ»á·ÖÎöµç¶¯»úµÄ»úеÊä³ö¹¦ÂÊËæʱ¼ä±ä»¯µÄͼÏóµÃ³öÏà¹ØµÄÐÅÏ¢£¬Òª×¢Òâ¸ù¾ÝP=FvÇó³öÉþ×ÓµÄÀ­Á¦£¬±¾Ìâ±È½Ï¸´ÔÓ£¬ÊôÓÚÄÑÌ⣬½âÌâ¹ý³ÌÖÐÒ»¶¨Òª×ÐϸÈÏÕ森