To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.
By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,

Where,
Velocity in each state
g= Gravity
h = Height
Our values are given as,



Replacing at the kinetic equation to find
we have,



Applying the concepts of continuity,

We need to find A_2 then,

So the cross sectional area of the water stream at a point 0.11 m below the faucet is



Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is 
Answer: A. The total displacement divided by the time and C. The slope of the ant's displacement vs. time graph.
Explanation:
Hi! The question seems incomplete, but I found the options on the internt:
A. The total displacement divided by the time.
B. The slope of the ant's acceleration vs. time graph.
C. The slope of the ant's displacement vs. time graph.
D. The average acceleration divided by the time.
Now, since we know the ant is travelling at a constant speed, its average velocity
will be expressed by the following equation:

Where:
is the ant's total displacement
is the time it took to the ant to travel to the kitchen
Hence one of the correct options is: A. The total displacement divided by the time
On the other hand, this can be expressed by a displacement vs. time graph graph, where the slope of that line leads to the equation written above. So, the other correct option is:
C. The slope of the ant's displacement vs. time graph.
The answer is B high pressure.
F = 750 N (Force)
d = 10 m (displacement
)
t = 25 s (time)
L = ? (Mechanical work
) = (Energy)
P = ? (Power)
Solve:
L = F × d = 750 × 10 = 7500 Joules
P = L / t = 7500 / 25 = 300 Watts
The law of reflection states that the angle of incidence is equal to the angle of reflection. Furthermore, the law of reflection states that the incident ray, the reflected ray and the normal all lie in the same plane.
hope this helps :)