Convert the acceleration of a free falling body 9.8 m/s^2 to ft/s^2

Question

Jewel Blackwell

Physics

23 September, 20:20

Convert the acceleration of a free falling body 9.8 m/s^2 to ft/s^2

+1

Answers (2)

Know the Answer?

Rebecca · Answer 1

hoping this is the right answer : Use A = log (100/%T) = 2.0 - log %T

a) A = 2.00 - log 53.86 = 2.00 - 1.73 = 0.27

b) A = 2.00 - log 0.56 = 2.00 + 0.25 = 2.25

i feel like ur going to delete this but i tried

Bugsy · Answer 2

i hoping this is the right answer : Use A = log (100/%T) = 2.0 - log %T

a) A = 2.00 - log 53.86 = 2.00 - 1.73 = 0.27

b) A = 2.00 - log 0.56 = 2.00 + 0.25 = 2.25

i feel like ur going to delete this but i tried