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34_Ch 15 College Physics ProblemCH15 Electric Forces and Electric Fields

34_Ch 15 College Physics ProblemCH15 Electric Forces and Electric Fields

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34 CHAPTER 15 Thus, if R = × , we must have 3 1.27 10 m ( )( ) ( ) 10 2 3 2 2 0 6.90 10 m s 1.27 10 m sin2 0.961 9 550 m s y a R v θ × × = = = 2 73.9 or 2 180 73.9 106.1° giving θ θ = ° = ° − ° = . Hence, 37.0 or 53.0° θ = ° (b) The time of flight for each possible angle of projection is: For 37.0 θ = °
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