One plane departed from New York, and at the same time another plane departed from Mexico City. They flew toward each other at rate of 650 and 550 mph. If New York and Mexico City are 3000 miles apart, in how many hours did the planes pass each other?

30 hours

Explanation:

Average speed = distance/time

Let the time taken for each plane to pass each other be t

Average speed of plane that left from New York = 650 mph

Distance covered = 650t miles

Average speed of plane that left from Mexico = 550 mph

Distance covered = 550t miles

Distance apart = 3000 miles

650t - 550t = 3000

100t = 3000

t = 3000/100 = 30 hours

Explanation:

We also know that, v = S/t which can be rewritten as v * t = S

Where,

v = velocity

S = distance

t = time

i. v1*t = S1

ii. v2*t = S2

iii. s1 + s2 = 3000

Substituting i and ii into equation iii,

St = S1 + S2

v1*t + v2*t = 3000

650*t + 550*t = 3000

1200*t = 3000

t = 3000/1200

t = 2.5 h.

explanation:

wheres the pic

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//