# T1a8s - Assignment 8 solutions(Fall term Physics 102 8.8 If...

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8.8 If the mass of a hydrogen atom is 1.00 u (i.e., 1 unit), then the mass of the oxygen atom is 16.0 u. () ( ) ( ) 16.0 u 0 2 1.00 u 0.100 nm cos 53.0 16.0 1.00 1.00 u ii cg i mx x m Σ == = 3 6.69 10 nm × Σ+ + i my y m Σ = Σ ( ) ( ) ( ) ( ) + ° ⎡⎤ ⎣⎦ ++ 16.0 0 1.00 0.100 sin 53.0 1.00 0.100 sin 53.0 u nm 16.0 1.00 0 Assignment 8 solutions (Fall term Physics 102) 8.12 Requiring that 0 i x gives m Σ Σ ( ) () ( )( ) ( )( ) ( ) 5.0 kg 0 3.0 kg 0 4.0 kg 3.0 m 8.0 kg 0 5.0 3.0 4.0 8.0 kg x + = +++ , or 8. which yields 0 12 m 0 x += 1.5 m x = − . Also, requiring that 0 i y m Σ = Σ = gives ( ) 5.0 kg 0 3.0 kg 4.0 m 4.0 kg 0 8.0 kg 0 5.0 3.0 4.0 y + = , or 8. yielding 0 12 m 0 y 1.5 m y = − . Thus, the 8.0-kg mass should be placed at coordinates ( ) 1.5 m, 1.5 m −− . 8.18 Consider the torques about an axis perpendicular to the page through the left end of the scaffold. τ Σ=⇒ + = 12 0 0 700 N 1.00 m 200 N 1.50 m 3.00 m 0 TT . From which, 2 T = 333 N . 239

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CHAPTER 8 Then, from Σ , we have , or = 0 y F +− 12 700 N-200 N=0 TT 900 N- 900 N-333 N= == 567 N T 1 700 N 200 N T 2 1.50 m 1.00 m 2.00 m 8.20 (a) See the diagram below: 200 N V H 700 N 80.0 N T 60.0° x 3.00 m 3.00 m (b) If x = 1.00 m, then )( ) ( ) ( )( ) () 0 700 N 1.00 m 200 N 3.00 m 80.0 N 6.00 m sin 60.0 6.00 m 0, left end T τ Σ= + °= giving T = 343 N Then, , or cos 60.0 0 x FH T Σ=⇒− ° = ( ) 343 N cos 60.0 H = 171 N and ( ) 980 N+ 343 N sin 60.0 0 y FV ° = , or V = 683 N (c) When the wire is on the verge of breaking, T = 900 N and ) ( )( ) () () 700 N 200 N 3.00 m 80.0 N 6.00 m 900 N
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T1a8s - Assignment 8 solutions(Fall term Physics 102 8.8 If...

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