645 * 590 = 380,550 W or 381 kW
Ke = (1/2)m v^2 = (1/2)(4.5*10^4)(900)
= 2025*10^4 = 2.025*10^7 Joules
energy needed = (2.025/.91)10^7
= 2.225 * 10^7 Joules
3.8055*10^5 t = 2.225*10^7
t = .585*10^2 = 58.5 seconds, about a minute
a = delta v/delta t = 30/58.5 = .513 m/s^2
approx 60* (50 km/h/30 mi/h) = 100 km/h
100/3.6 = 28 m/s
28 m/s / 10 s = 2.8 m/s^2
car accelerates 2.8/.513 = 5.5 times as fast
A light–rail commuter train draws 645 A of 590–V DC electricity when accelerating. What is its power consumption rate in kilowatts?
How long does it take to reach 30.0 m/s starting from rest if its loaded mass is 4.50×104 kg, assuming 91.0% efficiency and constant power?
Find its average acceleration.
What is the ratio of the acceleration you found for the light–rail train compares to what might be typical for an automobile? (A typical automobile can accelerate from 0 to 60 mph in 10 s.)
1 answer
0 answers