Asked by mssailormouth
A water balloon with a mass of 1.20 kg is falling from a roof. If the average force of friction is 0.500 N, what is the acceleration on the balloon in m/s2?
I am not sure how to get started
I am not sure how to get started
Answers
Answered by
mssailormouth
1.20kg x 9.80 m/s^2= 11.76-0.500=11.26
11.26=1.20a
a=9.38 m/s^2??
11.26=1.20a
a=9.38 m/s^2??
Answered by
oobleck
looks good to me
Answered by
Damon
Yes, but not sure I would say it that way
Fdown = m a down
mg - .5 = ma
a = (mg -.5)/m = (1.2*9.8 - .5) / 1.2 = 11.26/1.2 = 9.38 down
Fdown = m a down
mg - .5 = ma
a = (mg -.5)/m = (1.2*9.8 - .5) / 1.2 = 11.26/1.2 = 9.38 down
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