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A freight train leaving a yard must exert a force of 2530000 N in order to increase its speed from rest to 17 m/s. During this process, the train must do 1110000000 J of work. How far does the train travel? *

Respuesta :

xero099

Answer:

[tex]\Delta s = 438.735\,m[/tex]

Explanation:

Given that force exerted is constant and parallel to the railroad during acceleration, the equation of work is equal to:

[tex]W = F\cdot \Delta s[/tex]

The travelled distance of the freight train is:

[tex]\Delta s = \frac{W}{F}[/tex]

[tex]\Delta s = \frac{1110000000\,J}{2530000\,N}[/tex]

[tex]\Delta s = 438.735\,m[/tex]

fichoh

Answer: The distance traveled in other to increase its speed from rest to 17m/s is 438.74m

Explanation:

GIVEN the following :

Workdone = 1110000000 J

Force = 2530000 N

Workdone = force × distance

1110000000 = 2530000 × distance

Distance = 1110000000 ÷ 2530000

Distance = 438.735m

Distance = 438.74m

The distance traveled by the freight train in other to increase it's speed from rest to 17m/s is 438.74m

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