Respuesta :
Answer:
t=444.4s
Explanation:
m=1.5*10^7 kg
F=7.5*10^5 N
v=80km/h*(1h/3600s)*(1000m/1km)=22.22m/s
Second Newton's Law:
F=ma
a=F/m=7.5*10^5/(1.5*10^7)=0.05m/s^2
Kinematics equation:
vf=vo+at=at
vo: initial velocity equal zero
t=vf/a=22.22/0.05=444.4s
We have that for the Question "A freight train has a mass of 1.5 X 10^7 kg. If the locomotive can exert a constant pull of 7.5 X1^5 N, how long does it take to increase the speed of the train from rest to 80 km/h?" it can be said that the time it take to increase the speed of the train from rest to 80 km/h is t=44.4sec
- t=44.4sec
From the question we are told
A freight train has a mass of 1.5 X 10^7 kg. If the locomotive can exert a constant pull of 7.5 X1^5 N, how long does it take to increase the speed of the train from rest to 80 km/h?
Generally the equation for the Acceleration is mathematically given as
[tex]a=\frac{F}{m}\\\\Therefore\\\\a=\frac{ 7.5 *10^5}{1.5 X 10^7}\\\\a=0.05m/s^2\\\\[/tex]
Generally the equation for the Motion is mathematically given as
v= u + at
Therefore
22.2 =0+0.05*t
t=44.4sec
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