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{"ops":[{"insert":"\u00a0Incline 40 degrees, incline-object = 2 kg, hanger mass = 4 kg\nUse 9.8 m\/s\/s for g\nDescribe how mu increased, acceleration down, but tension internal force up\nPlot tension on y-axis, acceleration on x-axis for the 2 kg mass with mu =0.25 using the 2 kg equation of (T, a). Repeat for the 4 kg using the 4 kg equation of (T, a) on the same graph, the intersection is the graphical solution. Repeat for the 2 kg with mu=0.4 using the 2 kg equation of (T, a) on the same graph, the intersection is the graphical solution.\nWriting intensive: explain the graph as a trend justification, the shift of the intersection points to other values.\n\u00a0Incline 40 degrees, incline-object = 2 kg, hanger mass = 4 kg\nUse 9.8 m\/s\/s for g\nDescribe how mu increased, acceleration down, but tension internal force up\nPlot tension on y-axis, acceleration on x-axis for the 2 kg mass with mu =0.25 using the 2 kg equation of (T, a). Repeat for the 4 kg using the 4 kg equation of (T, a) on the same graph, the intersection is the graphical solution. Repeat for the 2 kg with mu=0.4 using the 2 kg equation of (T, a) on the same graph, the intersection is the graphical solution.\nWriting intensive: explain the graph as a trend justification, the shift of the intersection points to other values.\n\n"}]}

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