ÌâÄ¿ÄÚÈÝ

3£®ÔÚ×ö¡°ÑéÖ¤»úеÄÜÊغ㶨ÂÉ¡±µÄʵÑéÖУ¬Ä³ÊµÑéС×éÓÃÌìƽ³ÆÁ¿µÃµ½ÖØ´¸ÖÊÁ¿m=0.3kg£¬´òµã¼ÆʱÆ÷¹¤×÷ƵÂÊΪ50Hz£¬ËûÃÇÔÚʵÑéÖлñµÃÒ»ÌõÖ½´øÈçͼËùʾ£¬A¡¢B¡¢C¡¢D¡¢E¡¢FΪÒÀ´Î´òϵĵ㣬¸ù¾ÝÖ½´øÉϵÄÊý¾Ý£¬¿É²âµÃ´òÏÂDµãÖØ´¸µÄ¶¯ÄÜEkD=0.198J£®Ä³Í¬Ñ§¼ÆËã·¢ÏÖAµ½DµãÖØÁ¦ÊÆÄܵļõÉÙÁ¿¡÷EpСÓÚDµãʱÖØ´¸µÄ¶¯ÄÜEkD£¬ÓÉ´ËËûµÃ³ö½áÂÛ£ºÔÚÖØ´¸ÏÂÂä¹ý³ÌÖУ¬»úеÄܲ»Êغ㣮ÄãÈÏΪËûµÄ¹ÛµãÊÇ·ñÕýÈ·£¿ÊÔ˵Ã÷Ô­Òò£®²»ÕýÈ·£¬´òAµãʱÖØ´¸µÄ¶¯Äܲ»ÎªÁ㣨»ò¡°Î´¼ÆÈëAµãʱÖØ´¸µÄ¶¯ÄÜ¡±»ò¡°±È½ÏµÄӦΪÊÆÄܱ仯Á¿Ó붯Äܱ仯Á¿¡±£©£®

·ÖÎö DµãÖØ´¸µÄ¶¯ÄÜ¿ÉÒÔÏÈËã³ö¸ÃµãËٶȣ¬¼´CE¶ÎµÄƽ¾ùËٶȣ¬ÔÙ¼ÆËãDµã¶¯ÄÜ£®´ÓAµ½Dµã¡°ÑéÖ¤»úеÄÜÊغ㶨ÂÉ¡±Òª×¢ÒâA¡¢DµãµÄ¶¯ÄÜ£®

½â´ð ½â£ºÓÉÖ½´øÉÏËٶȵļÆËã·½·¨¿ÉÖª£º
${v}_{D}=\frac{{x}_{E}-{x}_{C}}{2T}=\frac{£¨8.54-3.95£©¡Á1{0}^{-2}}{2¡Á0.02}m/s¡Ö1.15m/s$£¬
DµãÖØ´¸µÄ¶¯ÄÜΪ£º
${E}_{KD}=\frac{1}{2}m{v}_{D}^{2}=\frac{1}{2}¡Á0.3¡Á1.1{5}^{2}J=0.198J$
¹Ê´ð°¸Îª£º0.198£¨0.190¡«0.207¾ùÕýÈ·£©£»²»ÕýÈ·£¬´òAµãʱÖØ´¸µÄ¶¯Äܲ»ÎªÁ㣨»ò¡°Î´¼ÆÈëAµãʱÖØ´¸µÄ¶¯ÄÜ¡±»ò¡°±È½ÏµÄӦΪÊÆÄܱ仯Á¿Ó붯Äܱ仯Á¿¡±£©

µãÆÀ ¡°ÑéÖ¤»úеÄÜÊغ㶨ÂÉ¡±ÊµÑéҪעÒâʼĩÁ½¸öλÖõĶ¯ÄÜ£¬²»ÊÇËùÓгõλÖõĶ¯Äܶ¼Îª0£¬±È½ÏµÄӦΪÊÆÄܱ仯Á¿Ó붯Äܱ仯Á¿£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø