Compare the acceleration on stopping with the acceleration of gravity.
2007-09-10 05:45:28 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Physics
9.8m/s^2 * 90m = x * .036m
2007-09-10 05:53:15 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Energy,
m g h = F x distance
(penny) 90m=F 3.6 cm
(penny) 9000 cm = F 3.6cm
(penny)2500= F (force of ground)
2007-09-10 05:58:22 · answer #2 · answered by Anonymous · 0⤊ 0⤋
velocity upon hitting the ground
vi =0m/s
from vf^2 = vi^2 + 2*a*delta(x)
vf = sqrt(2*g*90) = 42.02 m/s
acceleration for the penny in the ground
a = -vi^2/(2*deltax) = -42.02^2/(2*.036) = -24520 m/s^2
g = -9.81 m/s^2
ratio = 24520/9.81 = 2500 g's
2007-09-10 05:57:46 · answer #3 · answered by civil_av8r 7 · 0⤊ 0⤋
90m/3.60cm = 2500:1
2007-09-10 05:50:56 · answer #4 · answered by Alexander 6 · 0⤊ 0⤋
f1d1 = f2d2
f1/f2 = 3.6cm/90m = .0004
2007-09-10 05:51:35 · answer #5 · answered by supastremph 6 · 0⤊ 0⤋