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()
(
)
24
48
2
4
3
env
(5.67 10 W m
K )(0.850)(4 ) 0.500m
350.15K
2.28 10 W.
a
PA
T
σε
π
−
==
×
⋅
=
×
(c) From Eq. 1840, we have
33 3
2.28 10 W 1.23 10 W
1.05 10 W.
nar
PPP
=−=
×
−
×
=
×
55. We use Eqs. 1838 through 1840. Note that the surface area of the sphere is given by
A
= 4
π
r
2
, where
r
= 0.500 m is the radius.
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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