Alocomotive is pulling 15 freight cars, each of which is loaded with the same amount of weight. the mass of each freight car (with its load) is 37,000 kg. if the train is accelerating at 0.62 m/s2 on a level track, what is the tension in the coupling between the second and third cars? (the car nearest the locomotive is counted as the first car, and friction is negligible.) express your answer using two significant figures.

298,220 N

Explanation:

Let the force on car three is T_23-T_34

Since net force= ma

from newton's second law we have

T_23-T_34 = ma

therefore,

T_23-T_34 = 37000×0.62

T_23= 22940+T_34

now, we need to calculate

T_34

Notice that T_34 is accelerating all 12 cars behind 3rd car by at a rate of 0.62 m/s^2

F= ma

So, F= 12×37000×0.62= 22940×12= 275280 N

T_23 =22940+T_34= 22940+ 275280= 298,220 N

therefore, the tension in the coupling between the second and third cars

= 298,220 N

i think if it were me i whould divide 300 dide 15

step-by-step explanation:

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6