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Question
9 fill in the blank 1 point a 10 n force and a 40 n force are acting in opposite direction on the same 7 kg ball. what is the magnitude of the acceleration of the ball? friction 10 n 7 kg applied force 40 n answer type your answer... choose your answer...
Step1: Calculate net force
Net force \(F_{net}=F_{applied}-F_{friction}\). Given \(F_{applied} = 40N\) and \(F_{friction}=10N\), so \(F_{net}=40 - 10=30N\).
Step2: Use Newton's second law \(F = ma\)
We know \(F = F_{net}\), \(m = 7kg\), and we need to find \(a\). Rearranging the formula for \(a\), we get \(a=\frac{F_{net}}{m}\). Substituting \(F_{net}=30N\) and \(m = 7kg\), we have \(a=\frac{30}{7}\approx4.29m/s^{2}\).
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\(4.29m/s^{2}\)