Posted by saru on Friday, October 18, 2013 at 1:25am.
When the sun is in the zenith and the sky is perfectly clear, the solar power we receive on the surface of the earth is roughly 1 kW per square meter (on a horizontal surface that is normal to the direction of the sun).
Calculate the average solar power per square meter (in Watts) on a horizontal surface for a day when the sun goes through the zenith at noon.
The sun goes through the zenith exactly twice a year on latitudes that are close to the equator. Take the angle to the sun into account.

Physics pls anyone can help me? its urgent  bobpursley, Friday, October 18, 2013 at 3:56am
The incident power is given by
power=Powermax*sinTheta
where theta is the angle measured from horizontal East to the sun. At noon, Theta=90 degrees This takes 12 hours, so theta=time*PI/12, where time is in hours, ie, noon is time 6
avg=total energy /time
lets measure energy in kwhr
total energy= INTEGRal power*dtime from time=0 to 12
total energy= INT 1kw*sin(PI*t/12 )dt
= 1kw* (cosPIt/12)*12/PI from zero to 12
= 12/PI * (cos (PI) cos0)
= 12/PI ( 1 1)=24/PI kwhr
average power= total energy/(12)
= 2/PI kwhrs
check this.

Physics pls anyone can help me? its urgent  saru, Friday, October 18, 2013 at 4:15am
It should be in watts. so 2/PI kwhr should be 2*1000/(PI*3600)watts or ? please

Physics pls anyone can help me? its urgent  Steve, Friday, October 18, 2013 at 4:28am
something got mangled in the units
kwhr is energy
kw is power
Better check the units in the math above.
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