A freight train has a mass of 1.5 X 10^7 kg. If the
locomotivecan exert a constant pull of 7.5 X1^5 N, how long does it
take toincrease the speed of the train from rest to 80 km/h?

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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