Answer:
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Answer:
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Answer:
The distance measure from the wall = 36ft
Explanation:
Given Data:
w = 10
g =32.2ft/s²
x = 2
Using the principle of work and energy,
T₁ +∑U₁-₂ = T₂
0 + 1/2kx² -wh = 1/2 w/g V²
Substituting, we have
0 + 1/2 * 100 * 2² - (10 * 3) = 1/2 * (10/32.2)V²
170 = 0.15528V²
V² = 170/0.15528
V² = 1094.796
V = √1094.796
V = 33.09 ft/s
But tan ∅ = 3/4
∅ = tan⁻¹3/4
= 36.87°
From uniform acceleration,
S = S₀ + ut + 1/2gt²
It can be written as
S = S₀ + Vsin∅*t + 1/2gt²
Substituting, we have
0 = 3 + 33.09 * sin 36.87 * t -(1/2 * 32.2 *t²)
19.85t - 16.1t² + 3 = 0
16.1t² - 19.85t - 3 = 0
Solving it quadratically, we obtain t = 1.36s
The distance measure from the wall is given by the formula
d = VCos∅*t
Substituting, we have
d = 33.09 * cos 36. 87 * 1.36
d = 36ft
Answer:
the volume flow rate per unit depth is:

the ratio is : 
Explanation:
From the question; the equations of the velocities profile in the system are:
----- equation (1)
The above boundary condition can now be written as :
At y= 0; u =0 ----- (a)
At y = h; u =0 -----(b)
At y =
; u =
------(c)
where ;
A,B and C are constant
h = distance between two plates
u = velocity
= maximum velocity
y = measured distance upward from the lower plate
Replacing the boundary condition in (a) into equation (1) ; we have:

Replacing the boundary condition (b) in equation (1); we have:

Replacing the boundary condition (c) in equation (1); we have:

replacing
for A in (d); we get:


replacing the values of A, B and C into the velocity profile expression; we have:

To determine the volume flow rate; we have:

Replacing 


Thus; the volume flow rate per unit depth is:

Consider the discharge ;
Q = VA
where :
A = bh
Q = Vbh

Also; 
Then;

Thus; the ratio is : 