p1+(1/2)

*p(*v1^2)+

*p*gy1=p2+(1/2)

*p(*v2^2)+

*p*gy2

Since we have no change in height and we are looking at the change in air pressure the equation we use becomes:

Δp=(1/2)

*p(*v1^2)-(1/2)

*p(*v2^2)

*p*(air)= 1.225kg/m^3 v(air)= 500m/s v(average subway)= 24m/s

Δp= ((.5)( 1.225kg/m^3)(500m/s)^2) - ((.5)( 1.225kg/m^3)(524m/s)^2)=-15,052 N

Rearranging P=F/A to F=PA and assuming the average area of person is 1.7m^2 we get:

F=(15,052N)(.85m^2)= 12,794N *since the air pushes only on half a person's body the area used is multiplied by a factor of 0.5.

12,974 newtons would certainly be enough to move a small child though bear in mind that to fully experience this force an individual would need to be a the entrance to the tunnel and extremely close to the train. The force would dramatically decrease as the distance between an individual and a train increased. Even so it is not advisable to stand near a moving train, and I guess one should sometimes listen to their parents.

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