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phy131studiof16:lectures:m2p1sol [2016/10/31 11:29]
mdawber [Question 3 Solution]
phy131studiof16:lectures:m2p1sol [2016/10/31 11:31] (current)
mdawber [Question 3 Solution]
Line 120: Line 120:
 $T\sin40^{o}\times0.7=0.8\sin40^{o}\times g\times 0.35$ $T\sin40^{o}\times0.7=0.8\sin40^{o}\times g\times 0.35$
  
-$T=\frac{0.35\times0.6\times0.8\times g}{0.7}=0.4g=3.92\mathrm{\,​N}$+$T=\frac{0.35\times0.8\times g}{0.7}=0.4g=3.92\mathrm{\,​N}$
  
  
phy131studiof16/lectures/m2p1sol.txt ยท Last modified: 2016/10/31 11:31 by mdawber
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