3 c^6
----------------------------------------------=?
32π (6.67259 × 10^ -11)^2 G^3
[ π是圓周率( 3.1415926 或 355/133 )，G是萬有引力常數( 6.67259×10-11)，c是光速( 299,792,458 ) ]
thx！
一條數學問題
2007-02-13 3:37 am
回答 (3)
2007-02-13 7:33 am
✔ 最佳答案
你想問什麼，你將所有資料代入，便可以計得結果3c6
----------------------------------------------
32π(6.67259 × 10-11)2 G3
3(299792458)6
=------------------------------------------------------------------
32π(6.67259 × 10-11)2 (6.67259x10-11)3
= 1.64x10100
2007-02-14 7:58 pm
3c6
----------------------------------------------
32π(6.67259 × 10-11)2 G3
3(299792458)6
=------------------------------------------------------------------
32π(6.67259 × 10-11)2 (6.67259x10-11)3
= 1.64x10100
----------------------------------------------
32π(6.67259 × 10-11)2 G3
3(299792458)6
=------------------------------------------------------------------
32π(6.67259 × 10-11)2 (6.67259x10-11)3
= 1.64x10100
參考: book
2007-02-14 3:46 am
secθ = -17/8 where π/2 < θ < π
cosθ = -8/17
sinθ = 15/17 (as π/2 < θ < π)
tanθ = sinθ/cosθ = -15/8
QUESTION 2
sin a + sec a = k/2
sin a sec a = 3/2
sin a + sec a = k/2
tan a = 3/2
As π < θ < 3π/2
tan a = 3/2 gives
sin a = 3/√13
cos a = 2/√13
sin a + sec a
= sin a + (1/cos a)
= (3/√13) + [(√13)/2]
= (3√13/13) + [(√13)/2]
= [(6√13)/13]/2 + [(√13)/2]
= [(6√13)/13 + (√13)]/2
= k/2
Hence, k = (6√13)/13 + (√13)
cosθ = -8/17
sinθ = 15/17 (as π/2 < θ < π)
tanθ = sinθ/cosθ = -15/8
QUESTION 2
sin a + sec a = k/2
sin a sec a = 3/2
sin a + sec a = k/2
tan a = 3/2
As π < θ < 3π/2
tan a = 3/2 gives
sin a = 3/√13
cos a = 2/√13
sin a + sec a
= sin a + (1/cos a)
= (3/√13) + [(√13)/2]
= (3√13/13) + [(√13)/2]
= [(6√13)/13]/2 + [(√13)/2]
= [(6√13)/13 + (√13)]/2
= k/2
Hence, k = (6√13)/13 + (√13)
收錄日期: 2021-04-23 16:39:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070212000051KK03620