Two barges, each with a displacement (mass) of 570 Mg, are loosely moored in calm water. A stunt driver starts his 1870-kg car from rest at A, drives along the deck, and leaves the end of the 15° ramp at a speed of 30 km/h relative to the barge and ramp. The driver successfully jumps the gap and brings his car to rest relative to barge 2 at B. Calculate the velocity v2 imparted to barge 2 just after the car has come to rest on the barge.

The velocity imparted to barge 2 just after the car has come to rest on the barge = 0.096 km/h

Explanation:

Mass of the two barges, = 570000 kg

Mass of the car,

The horizontal component of car velocity:

The driver brings his car to rest relative to barge 2 at B:

Applying the principle of conservation of momentum:

d) the number of neutrons plus the number of protons

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14