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 
<span> In </span>transverse waves<span>, </span>particles<span> of the</span>medium<span> vibrate </span>to<span> and from in a direction perpendicular </span>to<span> the direction of energy transport. </span>
Answer:
Electrical force, F = 90 N
Explanation:
It is given that,
Charge on sphere 1, 
Charge on sphere 2, 
Distance between two spheres, d = 6 cm = 0.06 m
Let F is the electrical force between them. It is given by the formula of electric force which is directly proportional to the product of charges and inversely proportional to the square of distance between them such that,


F = 90 N
So, the electrical force between them is 90 N. Hence, this is the required solution.
Line up in a direction parallel to the magnetic field lines<span />
<span>R = rate of flow = 0.370 L/s
H = height = 2.9 m
T= time = 3.9 s
V = velocity of water when it hits the bucket = sqrt(2gh) = sqrt(2 x 9.8 x 2.9) =7.539 m/s2
G value = 9.8 m/s2
Wb = weight of bucket = 0.690 kg x 9.8 m/s2 = 6.762 N
Wa = weight of accumulated water after 3.9 s
Fi = force of impact of water on the bucket
S = reading on the scale = Wa + Wb + Fi
mass of water accumulated after 3.9 s = R x T = 0.370 x 3.9 = 1.443 L = 1.443 kg
Therefore, Wa = 1.443 x 9.8 = 14.1414 N
Fi = rate of change of momentum at the impact point = R x V (because R = dm/dt)
= 0.37 x 7.539 = 2.78943 N
S = 14.1414 + 6.762 + 2.78943 = 23.692 N</span>