A ball with a volume of 3 cm3 and a density of 9.0 g/cm3 increases its velocity from 2 m/s to 6 m/s over a 12
1 answer:
Answer:
M = ρ V = 9 gm/cm^ 3 * cm^3 = 27 gm
a = (V2 - V1) / t = (6 - 2) m/s / 12 s = 1/3 m/s^2 the acceleration
F = M a = 27 gm * 1/3 m/s^2 = 9 dynes net force applied
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