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Re: Chernaya dyra
13.09.2013 8:40 | A.P. Vasi

Pro slozhenie skorostei
: Vchera v 15:12:05
V principe elementarnaya zadacha: vdol' osi 0h startuet raketa-1 so skorost'yu 4/5 s otnositel'no 0, s nee v tom zhe napravlenii startuet raketa-2 so skorost'yu 4/5 s otnositel'norakety-1; naiti skorost' rakety-2 otnositel'no 0.

Zadacha reshaetsya s pomosh'yu odnoi formuly \[ s=\frac{v+u}{1+vu/{c}^{2}}=\frac{8/5}{1+16/25}=\frac{8 \cdot 25}{41 \cdot 5}=\frac{40}{41}\approx 0.9756 ( c ) \]Lorenc-faktor rakety-2 (otnositel'no 0)\[ \gamma =\frac{1}{\sqrt{1-{s}^{2}/{c}^{2}}}=\frac{1}{\sqrt{1-1600/1681}}=\frac{41}{9} \]

S drugoi storony, Lorenc-faktor rakety-1 (otnositel'no 0)\[ {\gamma }_{(1-0)} =\frac{1}{\sqrt{1-{v}^{2}/{c}^{2}}}=\frac{1}{\sqrt{1-16/25}}=\frac{5}{3} \]To est', summarnaya energiya dvizhusheisya so skorost'yu 4/5 s rakety-1 v 5/3 raza bol'she energii pokoyasheisya rakety-1 otnositel'no 0. Dalee. Raketa-2 dvizhetsya otnositel'no pokoyasheisya dlya nee rakety-1 so skorost'yu 4/5 s.

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V chem oshibka
V tom chto eta zadacha - fizicheskaya,
a fizicheskie zadachi reshayutsya ne formulami, a umstvennymi
sposobnostyami kotorye dolzhny prevyshat' neobhodimyi
minimum nuzhnyi dlya resheniya zadachi.

Eta zadacha reshaetsya v stile - alya chasy da nosim po platforme.

Edet poezd v pervom vagone poezda est' chasolov u nego sachok
s nomerom 1, i on lovit im chasy tozhe s nomerom 1, v konce poezda
est' vtoroi chasolov kotoryi lovit chasy s nomerom 2.

A est' hitrye platformennye chasonosy, - oni smeyutsya nad umstvennymi
sposobnostyami teh kto reshaet podobnye zadachi.

Edet poezd Chasonosy chasy s nomerom odin postavili po sredine platformy
i chasy nomer dva postavili po sredine platformy.
Proezzhayushii poezd mog uznat' svoyu skorost',
a posle togo kak chasonosy pervye chasy otnesli v pered platformy,
a vtorye v konec - to ves' poezd nachal proezzhat' stanciyu
s beskonechnoi skorost'yu.



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