Let the mass of 2500 kg car be
and it's velocity be
and the mass of 1500 kg car be
and it's velocity be
.
After the bumping the mass be M and it's velocity be V.
By law of conservation of momentum we have

2500 * 5 + 1500 * 1=4000 * V
V = 14000/4000 = 7/2 = 3.5 m/s
So the velocity of the two-car train = 3.5 m/s
Answer:
velocity = 472 m/s
velocity = 52.4 m/s
Explanation:
given data
steady rate = 0.750 m³/s
diameter = 4.50 cm
solution
we use here flow rate formula that is
flow rate = Area × velocity .............1
0.750 =
× (4.50×
)² × velocity
solve it we get
velocity = 472 m/s
and
when it 3 time diameter
put valuer in equation 1
0.750 =
× 3 × (4.50×
)² × velocity
velocity = 52.4 m/s
12.5 times 14 and convert to meters its 1.75 meters per second