Misplaced Pages

User:GabrielVelasquez: Difference between revisions

Article snapshot taken from Wikipedia with creative commons attribution-sharealike license. Give it a read and then ask your questions in the chat. We can research this topic together.
Browse history interactively← Previous editNext edit →Content deleted Content addedVisualWikitext
Revision as of 16:26, 7 September 2009 editGabrielVelasquez (talk | contribs)2,704 editsNo edit summary← Previous edit Revision as of 16:34, 7 September 2009 edit undoGabrielVelasquez (talk | contribs)2,704 editsNo edit summaryNext edit →
Line 43: Line 43:


<br /><br /><br /> <br /><br /><br />
:*:{{User:Richard0612/Userbox Archive/User 911truth}} <br /><br /> :*:{{User:Richard0612/Userbox Archive/User 911truth}}
::<math>T_{\mathrm{p}}= \left (\frac{L (1-A)}{16 \pi \sigma D^2} \right )^{\tfrac{1}{4}}</math>

::<math>T_{\mathrm{p}} = T_{\ast}\left^{\frac{1}{4}}</math>

::<math>\sigma\cdot T_{\mathrm{p}}^4 = \frac{\sigma\cdot T_{\ast}^4}{4\pi\cdot d^2}\cdot\frac{4\pi\cdot R_{\ast}^2}{4\pi\cdot R_{\mathrm{p}}^2}\cdot\pi\cdot R_{\mathrm{p}}^2(1-A_{\mathrm{p}})</math><br /><br />

'''code for Earth at Perihelion:''' <math>f_p= \frac{( ( 6.955 \times 10^8 )^2 ) \times (5.67051 \times 10^{-8}) \times (5778^4)} { ( ( 1- ( 1\times 0.016710219) ) \times 149597876600 )^2 } = 1,412.903 \ W/m^2</math>

'''code for Gliese 581 c at Periastron:''' <math>f_p= \frac{( (0.38 \times 6.955 \times 10^8 )^2 ) \times (5.67051 \times 10^{-8}) \times (3480^4)} { ( ( 0.073 - ( 0.073 \times 0.16 ) ) \times 149597876600 )^2 } = 6.9\times 10^3 \ W/m^2</math>

<math>\frac{Gl \ 581 \ c \ Periastron \ Insolation} { Earth's \ Solar \ Constant } = </math>
<math>\frac{6.9\times 10^3 \ W/m^2} { 1,366.079\ W/m^2 } = 505%</math> <br /><br />

{{Template: User EX-WP}}{{User:Scepia/Burnout Revenge}} {{Template: User EX-WP}} {{User:Scepia/Burnout Revenge}} {{Template: User EX-WP}}{{User:Scepia/Burnout Revenge}} {{Template: User EX-WP}} {{User:Scepia/Burnout Revenge}}
<br /><br /> <br /><br />

Revision as of 16:34, 7 September 2009

This userbox is wrong.
This user is homesick.
This userbox is correct.




This user likes going to the
Seaside
This user likes walking by the
Sea
This user likes walking by the
Coast




This user enjoys camping.
This user plays association football.
7This user has run a 7 minute mile. (Come on! Crank it!)




This user is addicted to travel.
This user is a certified scuba diver.
This user enjoys backpacking.




This user is a beginner archer.
This user is a swimmer.
This user enjoys geocaching.




This user enjoys wakeboarding.
Oxford Encyclopedias This user contributes using Oxford Encyclopedias .
This user enjoys playing golf.




This user loves the Spring.
This user enjoys filmmaking.
This user loves to go camping.




This user does weight training.
This user plays racquetball.
This user's species is a Vulcan.Alien




This user supports the right of all women to go without a top at the beach, the local pool, or in public.
This user is a surfer.
ubx-nThis user communicates exclusively via userboxes.





This user supports Humans United Against Robots.
This user thinks it was Miss Scarlet in the Conservatory with the Candlestick.
This user's computer is infected with malware!




User:Richard0612/Userbox Archive/User 911truth

Oxford EncyclopediasThis user is interested in ].




T p = ( L ( 1 A ) 16 π σ D 2 ) 1 4 {\displaystyle T_{\mathrm {p} }=\left({\frac {L(1-A)}{16\pi \sigma D^{2}}}\right)^{\tfrac {1}{4}}}
T p = T [ ( R d ) 2 1 A p 4 ] 1 4 {\displaystyle T_{\mathrm {p} }=T_{\ast }\left^{\frac {1}{4}}}
σ T p 4 = σ T 4 4 π d 2 4 π R 2 4 π R p 2 π R p 2 ( 1 A p ) {\displaystyle \sigma \cdot T_{\mathrm {p} }^{4}={\frac {\sigma \cdot T_{\ast }^{4}}{4\pi \cdot d^{2}}}\cdot {\frac {4\pi \cdot R_{\ast }^{2}}{4\pi \cdot R_{\mathrm {p} }^{2}}}\cdot \pi \cdot R_{\mathrm {p} }^{2}(1-A_{\mathrm {p} })}

code for Earth at Perihelion: f p = ( ( 6.955 × 10 8 ) 2 ) × ( 5.67051 × 10 8 ) × ( 5778 4 ) ( ( 1 ( 1 × 0.016710219 ) ) × 149597876600 ) 2 = 1 , 412.903   W / m 2 {\displaystyle f_{p}={\frac {((6.955\times 10^{8})^{2})\times (5.67051\times 10^{-8})\times (5778^{4})}{((1-(1\times 0.016710219))\times 149597876600)^{2}}}=1,412.903\ W/m^{2}}

code for Gliese 581 c at Periastron: f p = ( ( 0.38 × 6.955 × 10 8 ) 2 ) × ( 5.67051 × 10 8 ) × ( 3480 4 ) ( ( 0.073 ( 0.073 × 0.16 ) ) × 149597876600 ) 2 = 6.9 × 10 3   W / m 2 {\displaystyle f_{p}={\frac {((0.38\times 6.955\times 10^{8})^{2})\times (5.67051\times 10^{-8})\times (3480^{4})}{((0.073-(0.073\times 0.16))\times 149597876600)^{2}}}=6.9\times 10^{3}\ W/m^{2}}

G l   581   c   P e r i a s t r o n   I n s o l a t i o n E a r t h s   S o l a r   C o n s t a n t = {\displaystyle {\frac {Gl\ 581\ c\ Periastron\ Insolation}{Earth's\ Solar\ Constant}}=} 6.9 × 10 3   W / m 2 1 , 366.079   W / m 2 = 505 % {\displaystyle {\frac {6.9\times 10^{3}\ W/m^{2}}{1,366.079\ W/m^{2}}}=505\%}

This editor has decided to leave Misplaced Pages.
REVENGE TAKEDOWN
This editor has decided to leave Misplaced Pages.
REVENGE TAKEDOWN



Retired This user is no longer active on Misplaced Pages.

Dr.Submillimeter's Humorous Catagories

Categories: