Comments on: What total solar cell area would you need to provide energy needs? Mastering Physics problem? http://www.eblips.net/solar-energy/what-total-solar-cell-area-would-you-need-to-provide-energy-needs-mastering-physics-problem Helping Relieve Poverty Though The Provision of Solar Energy. Join Us! Thu, 15 Jul 2010 16:45:59 -0400 http://wordpress.org/?v=2.8.4 hourly 1 By: ericnutsch http://www.eblips.net/solar-energy/what-total-solar-cell-area-would-you-need-to-provide-energy-needs-mastering-physics-problem/comment-page-1#comment-1990 ericnutsch Tue, 09 Feb 2010 09:13:59 +0000 http://www.eblips.net/solar-energy/what-total-solar-cell-area-would-you-need-to-provide-energy-needs-mastering-physics-problem#comment-1990 The correct rate is 11kWhr per day. 11kw would be a horrendous amount of power.(maybe Al Gore uses it at this rate) I will calc it at your rate for problem sake. (this is not real life by the way, there are lots more variables, length of day, etc) 200W/m^2 * 15% = 30w/meter 11kw / 30w/meter = 366m^2 On a side note I think mastering physics is a crappy program, made for lazy physics teacher that dont ever care about application of knowledge.<br><b>References : </b><br>If you want to know the real calculation visit: http://www.aurorapower.net/alternative-energy/solar-electric.aspx The correct rate is 11kWhr per day. 11kw would be a horrendous amount of power.(maybe Al Gore uses it at this rate)

I will calc it at your rate for problem sake. (this is not real life by the way, there are lots more variables, length of day, etc)

200W/m^2 * 15% = 30w/meter

11kw / 30w/meter = 366m^2

On a side note I think mastering physics is a crappy program, made for lazy physics teacher that dont ever care about application of knowledge.
References :
If you want to know the real calculation visit: http://www.aurorapower.net/alternative-energy/solar-electric.aspx

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