'Vo Ja t = 0 t =35$ vo = (2.50 + 3.5j) m/s vf (-2.58) m/s The velocity of a 3.0-kg object is shown at two times What was the magnitude of the average force exerted on the object between t = 0 and t = 3.5s? 6.26 N 1.74 N 0.845 N 3.23 N 14.7 N'
Added by Heidi L.
Step 1
5s. We can do this by subtracting the initial velocity from the final velocity: Δv = vf - vo Δv = (-2.58 m/s) - (2.50 + 3.5j) m/s Δv = -5.08 - 3.5j m/s Show more…
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