Ask a New Question

Question

A 17000 kg railroad car travels along on a level frictionless track with a constant speed of 23.0 m/s. A 5100 kg additional load is dropped onto the car. What then will be its speed in meters/second?
12 years ago

Answers

Elena
m₁v=(m₁+m₂)u
u= m₁v/(m₁+m₂)=
=17000•23/(17000+5100)=17.7 m/s
12 years ago

Related Questions

An individual has $17000 income in period 0 and $60,000 income in period 1. If the individual des... Shawn Bixby borrowed 17000.00 on a 120-day, 12% note. After 65 days, Shawn paid 2000.00 on the note... suppose you invest 17000 at an annual interest rate of 3.9% compounded continuously? how much will u... A 14000 kg railroad car travels alone on a level frictionless track with a constant speed of 24.0 m/... 1.) A 10,000 kg railroad car traveling at a speed of 24.0 m/s strikes an identical car at rest. If t... 17000 at 2.5% for 5 year show much inter trust would you have ? What happens to the railroad tracks as they get farther away? 17000 customers last years. They gain a net of 1800 customers this year. They wish to reach at least... y=17000(0.1225)^14 round to the nearest cent. y=17000(0.03)^10
Ask a New Question
Archives Contact Us Privacy Policy Terms of Use