The momentum of both the identical balls would eventually be transferred to one another when it comes to a point wherein they will collide. In addition, the phenomenon is called an elastic collision wherein both the momentum and energy of the system would considered to be conserved.
(a) 5.66 m/s
The flow rate of the water in the pipe is given by

where
Q is the flow rate
A is the cross-sectional area of the pipe
v is the speed of the water
Here we have

the radius of the pipe is
r = 0.260 m
So the cross-sectional area is

So we can re-arrange the equation to find the speed of the water:

(b) 0.326 m
The flow rate along the pipe is conserved, so we can write:

where we have

and where
is the cross-sectional area of the pipe at the second point.
Solving for A2,

And finally we can find the radius of the pipe at that point:

Answer:
Explanation:
If the object is falling straight down it is in free fall. The difference between that and two-dimensional motion is that 2D motion is parabolic (projectile)
Answer:
a) 17.8 m/s
b) 28.3 m
Explanation:
Given:
angle A = 53.0°
sinA = 0.8
cosA = 0.6
width of the river,d = 40.0 m,
the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,
The river itself was 100 m below the ramp H = 100 m,
(a) find speed v
vertical displacement

putting values h=15 m, v=0.8
............. (1)
horizontal displacement d = vcosA×t = 0.6×v ×t
so v×t = d/0.6 = 40/0.6
plug it into (1) and get

solving for t we get
t = 3.734 s
also, v = (40/0.6)/t = 40/(0.6×3.734) = 17.8 m/s
(b) If his speed was only half the value found in (a), where did he land?
v = 17.8/2 = 8.9 m/s
vertical displacement = 
⇒ 
t = 5.30 s
then
d =v×cosA×t = 8.9×0.6×5.30= 28.3 m
Answer:
D. −F
Explanation:
the rest of the answers are
2/3F
The force is represented as a positive quantity and is repulsive.
Electrostatic force is inversely proportional to the square of the distance.
The direction of the force changes, and the magnitude of the force quadruples.
hope this helps sorry if i was too late! :)