The mass of something is 1.7 X10^7 kg. If it can exert a constant pull of 6.6 X10^5 N, how many minutes does it take to increase the speed of the train from rest to 70 km/h?
2006-09-29 08:14:26 · 5 answers · asked by activegirl 1 in Science & Mathematics ➔ Physics
Start by converting your speed to m/s. (70 km/h)(1000 m / 1 km)(1 h / 3600 s) = 19.4 m/s. Now use the simple relationships F = ma and v = at. You can easily see that a = v/t, so F = mv/t, and t = mv/F. You already have m, the object's mass; v, the final velocity; and F, the exterted force.
2006-09-29 08:19:44 · answer #1 · answered by DavidK93 7 · 0⤊ 0⤋
Use those 2 equations (a million) Accelerating tension (Kg) = Mass (N) x Acceleration (meters/2nd^2) Re manage to acquire the Acceleration (meters/2nd^2) = Kg/N (2) Acceleration = exchange in speed (m/2nd) / time (seconds) Re manage to acquire the time (t seconds) = exchange in speed / acceleration Divide by potential of 60 to get minutes.
2016-10-15 08:35:29 · answer #2 · answered by ? 4 · 0⤊ 0⤋
Remember, F = ma. Just plug in and solve.
6.6e5 N = 1.7e7 kg * a
a = 0.039 m/s^2
Convert 70 km/hr into m/s. You'll get 19.44 m/s.
v = v0 + at
19.44 = 0 + 0.039*t
t = 498 seconds, or 8.3 minutes.
2006-09-29 08:25:21 · answer #3 · answered by Anonymous · 0⤊ 0⤋
First off, 70km/h = 19.44 m/s
F = Ma, or a = F/M
a = 6.6 x 10^5/1.7 x 10^7
a = 0.0388.... m/s^2
v = at
t = v/a
t = 19.44 m/s / 0.388... m/s^2
t = 500.73 s
8 minutes 20 seconds
2006-09-29 08:28:11 · answer #4 · answered by p_rutherford2003 5 · 0⤊ 0⤋
cheese fries duh
2006-09-29 08:16:53 · answer #5 · answered by jenn 2 · 0⤊ 1⤋