ch26-p026 - 26. Let r = 2.00 mm be the radius of the kite...

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26. Let 2.00 mm r = be the radius of the kite string and 0.50 mm t = be the thickness of the water layer. The cross-sectional area of the layer of water is 22 32 6 2 ( ) [(2.50 10 m) (2.00 10 m) ] 7.07 10 m Ar t r ππ −−− ⎡⎤ =+ = × −× = × ⎣⎦ . Using Eq. 26-16, the resistance of the wet string is
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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.

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