HWCh24Solutions

HWCh24Solutions - PHYSICS 133 HOMEWORK SOLUTIONS CHAPTER 24...

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Unformatted text preview: PHYSICS 133 HOMEWORK SOLUTIONS CHAPTER 24 P24.4 (a) A = 10.0 cm ( 29 30.0 cm ( 29 A = 300 cm 2 = 0.030 0 m 2 E , A = E A cos E , A = 7.80 10 4 ( 29 0.030 0 ( 29 cos 180 E , A = - 2.34 kN m 2 C (b) E , A = EA cos = 7.80 10 4 ( 29 A ( 29 cos 60.0 10.0 cm 30.0 cm 60.0? FIG. P24.4 A = 30.0 cm ( 29 w ( 29 = 30.0 cm ( 29 10.0 cm cos 60.0 = 600 cm 2 = 0.060 0 m 2 E , A = 7.80 10 4 ( 29 0.060 0 ( 29 cos60.0 = + 2.34 kN m 2 C (c) The bottom and the two triangular sides all lie parallel to E , so E = for each of these. Thus, E , total = - 2.34 kN m 2 C + 2.34 kN m 2 C + + + = . P24.11 E = q in Through S 1 E =- 2 Q + Q = - Q Through S 2 E = + Q- Q = Through S 3 E =- 2 Q + Q- Q = - 2 Q Through S 4 E = P24.15 (a) With very small, all points on the hemisphere are...
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This note was uploaded on 06/02/2008 for the course PHY 133 taught by Professor K during the Fall '08 term at Cal Poly Pomona.

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HWCh24Solutions - PHYSICS 133 HOMEWORK SOLUTIONS CHAPTER 24...

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