A supertanker of mass 1.0 x 10^8 kg travels 3.5 km, reaching a speed of 4.1 km/h from rest. What was the magnitude of the unbalanced force acting on it?
2007-06-20 10:50:00 · 5 answers · asked by Chess 2 in Science & Mathematics ➔ Physics
time=(distance traveled*2)/velocity
time=(3500*2)/1.138888
time=6,146.34146 seconds (1:42:26.34146)
acceleration=velocity/time
acceleration=1.138888/6,146.34146
acceleration=0.0001852954 m/s/s
force=acceleration*mass
force=0.0001852954 * 10^8
force=18,529.54 newtons
2007-06-20 11:38:28 · answer #1 · answered by farwallronny 6 · 0⤊ 0⤋
Use the motion equation that has all but one of the variables you are given:
vf^2 = vi^2 + 2*a*x
(1.1388 m/s)^2 = (0)^2 + 2*a*(3500 m)
(1.1388 m/s)^2 / (2 * (3500 m)) = a
a = 0.0001853 m/s^2
Now we know the acceleration, so we use Newton's second law to find the force:
F = m*a
F = (1.0*10^8 kg)*(0.0001853 m/s^2)
F = 19000 newtons, to 2 significant digits.
2007-06-20 18:06:42 · answer #2 · answered by lithiumdeuteride 7 · 0⤊ 0⤋
first, find acelleration using
d=1/2at^2 and v=at
d = 1/2 a (v/a)^2
d = 1/2 v^2/a
a = 1/2 v^2/d
a = 1/2 4.1^2/3.5
once you've got a, use f=ma where m = 1.0x!0^8
2007-06-20 18:05:19 · answer #3 · answered by Piglet O 6 · 0⤊ 0⤋
v^2=2*a*s
v=41/36 m/s
s=3500 m
(41/36)^2=2*a*3500
a=(41/36)^2/2/3500 m/s^2
F=ma=1.0*10^8*(41/36)^2/2/3500 newtons
=18529.5414 N
2007-06-20 17:58:15 · answer #4 · answered by Anonymous · 0⤊ 0⤋
help you
2007-06-20 17:55:12 · answer #5 · answered by Anonymous · 0⤊ 0⤋