Asked by Sasha
A 3800 kg open railroad car coasts along with a constant speed of 8.60 m/s on a level track. Snow begins to fall vertically and fills the car at a rate of 3.50 kg/min. ignoring friction with the tracks what is the speed of the car after 90.0 minutes?
Answers
Answered by
bobpursley
momentuminitial=momentumfinal
3800*8.60=(3800+3.50*time)v(t)
solve for v(t) for t=90min
3800*8.60=(3800+3.50*time)v(t)
solve for v(t) for t=90min
Answered by
Anonymous
.088
Answered by
Cy
Correct me if I'm wrong, but...
Momentum before = Momentum after
mv(before)=m'v'(after)
(mass of train before snow)(velocity of train before snow)=m'v'
(3,800 kg)(8.6 m/s)=(3,800 kg + added weight of snow)(v')
32,680=(3,800+315)(v')
32,680=4115(v')
v'= 7.94 m/s
Momentum before = Momentum after
mv(before)=m'v'(after)
(mass of train before snow)(velocity of train before snow)=m'v'
(3,800 kg)(8.6 m/s)=(3,800 kg + added weight of snow)(v')
32,680=(3,800+315)(v')
32,680=4115(v')
v'= 7.94 m/s
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.