·ÖÎö £¨1£©¸ù¾ÝÁ£×ÓÔڴų¡ÖÐÔ˶¯µÄ°ë¾¶¹«Ê½ºÍÖÜÆÚ¹«Ê½Çó³öÁ£×ÓÔڴų¡ÖÐµÄÆ«×ª½Ç£¬¸ù¾Ý¼¸ºÎ¹ØÏµÇó³öt=t0ʱÁ£×ÓµÄλÖÃ×ø±ê£®
£¨2£©Á£×ÓÔ˶¯¹ý³Ì²»´©Ô½xÖᣬÇÒ×Ý×ø±ê²»³¬¹ýijһֵ£¬Á£×ÓÔ˶¯µ½×î¸ßµãʱËÙ¶ÈÓëxÖáÆ½ÐУ¬×÷³ö¹ì¼£Í¼£¬¸ù¾Ý¼¸ºÎ¹ØÏµ£¬×¥×¡Á£×ÓÔ˶¯µÄÖÜÆÚÐÔÇó³öt¡äµÄ´óС£¬½áºÏ¼¸ºÎ¹ØÏµÇó³öy0µÄ´óС£®
½â´ð ½â£º£¨1£©µ±t=t0ʱ£¬Á£×ÓÔ˶¯µ½Pµã£¬Á£×Ó×öÔ²ÖÜÔ˶¯µÄ¹ìµÀ°ë¾¶Îªr£¬ÖÜÆÚΪT£¬Æ«×ª½ÇΪ¦È£¬Ôò![]()
$q{v}_{0}B=m\frac{{{v}_{0}}^{2}}{r}$£¬
$T=\frac{2¦Ðm}{qB}$£¬$¦È=2¦Ð\frac{{t}_{0}}{T}$£¬
Pµã×ø±êΪ£¨xp£¬yp£©£¬
xp=rsin¦È£¬yp=r£¨1-cos¦È£©£¬
½âµÃP£¨$\frac{\sqrt{3}m{v}_{0}}{2qB}$£¬$\frac{m{v}_{0}}{2qB}$£©
£¨2£©ÓÉÌâÒâ¿ÉÖª£¬Á£×ÓÔ˶¯¹ý³Ì²»´©Ô½xÖᣬÇÒ×Ý×ø±ê²»³¬¹ýijһֵ£¬Á£×ÓÔ˶¯µ½×î¸ßµãʱËÙ¶ÈÓëxÖáÆ½ÐУ¬´øµçÁ£×ÓÔڴų¡ÖеÄÔ˶¯¹ì¼£ÈçͼËùʾ![]()
Óɼ¸ºÎ¹ØÏµ¿ÉµÃ
¦Á=30¡ã
ÓÉÓÚÁ£×ÓÔ˶¯¾ßÓÐÖÜÆÚÐÔ£¬ÓÉͼ¿ÉÖª£¬Á£×Ó½øÈë´Å³¡Ê±µÄʱ¼ä¿ÉÒÔΪ$\frac{1}{2}{t}_{0}$¡¢$\frac{1}{2}{t}_{0}+2{t}_{0}$¡¢$\frac{1}{2}{t}_{0}+4{t}_{0}$¡
¼´$t¡ä=\frac{1}{2}{t}_{0}+2n{t}_{0}$ £¨n=0£¬1£¬2¡£©
Óɼ¸ºÎ¹ØÏµ¿ÉµÃ
y0=2r£¨1-cos¦Á£©£¬
½âµÃ${y}_{0}=\frac{£¨2-\sqrt{3}£©m{v}_{0}}{qB}$£®
´ð£º£¨1£©t=t0ʱÁ£×ÓµÄλÖÃ×ø±êΪP£¨$\frac{\sqrt{3}m{v}_{0}}{2qB}$£¬$\frac{m{v}_{0}}{2qB}$£©
£¨2£©$t¡ä=\frac{1}{2}{t}_{0}+2n{t}_{0}$ £¨n=0£¬1£¬2¡£©£¬${y}_{0}=\frac{£¨2-\sqrt{3}£©m{v}_{0}}{qB}$£®
µãÆÀ ±¾Ì⿼²éÁË´øµçÁ£×ÓÔڴų¡ÖеÄÔ˶¯£¬¹Ø¼ü×÷³öÁ£×ÓÔ˶¯µÄ¹ì¼£Í¼£¬×¥×¡ÁÙ½ç״̬£¬½áºÏ°ë¾¶¹«Ê½ºÍÖÜÆÚ¹«Ê½£¬ÔËÓü¸ºÎ¹ØÏµ½øÐÐÇó½â£¬ÄѶȽϴó£®
| Äê¼¶ | ¸ßÖÐ¿Î³Ì | Äê¼¶ | ³õÖÐ¿Î³Ì |
| ¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£º¶àÑ¡Ìâ
| A£® | B£® | ||||
| C£® | D£® |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | 0.5 m/s2 | B£® | 1 m/s2 | C£® | 1.5 m/s2 | D£® | 2 m/s2 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | ÎïÌåÊܵ½±äÁ¦×÷ÓÃʱ£¬Ò»¶¨×öÇúÏßÔ˶¯ | |
| B£® | ÎïÌåÊܵ½ºãÁ¦×÷ÓÃʱ£¬Ò»¶¨×öÖ±ÏßÔ˶¯ | |
| C£® | µ±ÎïÌåËùÊܺÏÍâÁ¦·½ÏòÓëËÙ¶È·½Ïò²»ÔÚÒ»ÌõÖ±ÏßÉÏʱ£¬Ò»¶¨×öÇúÏßÔ˶¯ | |
| D£® | µ±ÎïÌåËùÊܺÏÍâÁ¦·½Ïò²»¶Ï±ä»¯Ê±£¬ÎïÌå¿ÉÄÜ×öÔȱäËÙÔ˶¯ |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | ·çËÙÔ½´ó£¬ÓêµÎÏÂÂäʱ¼äÔ½³¤ | B£® | ·çËÙÔ½´ó£¬ÓêµÎ×ŵØÊ±ËÙ¶ÈԽС | ||
| C£® | ÓêµÎÏÂÂäʱ¼äÓë·çËÙÎÞ¹Ø | D£® | ÓêµÎ×ŵØËÙ¶ÈÓë·çËÙÎÞ¹Ø |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¹ú¼ÊѧУÓÅÑ¡ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com