According to the Bernoulli's equation,the pressure difference between the wide and narrow ends of the pipe is given by
Here, 
is the velocity of water through wide ends of cylindrical pipe and 
is the velocity of water through narrow ends of cylindrical pipe.
Given, 
Now from equation continuity,
.
Here, 
and 
are cross- sectional areas of wide and narrow ends of cylindrical pipe.
As pipe is circular, so
.
At the second point, the diameter is halved, which means the radius is also halved. Therefore,
Substituting these values with the density of water is 
in pressure difference formula we get.
The formula for velocity vf = vi + at
First list your given information
2m/s Is your initial velocity (vi)
6m/s is you final velocity (vf)
2 seconds is your time (t)
Since you want the a for acceleration get a by itself
a = (vf-vi)/t
So a= (6-2)/2
a= 4/2
a=2
Now units
the units for acceleration are m/s
2m/s
A material that can easily flow is called a...
A. Fluid.
Answer:
(a) 2.85 m
(b) 16.5 m
(c) 21.7 m
(d) 22.7 m
Explanation:
Given:
v₀ₓ = 19 cos 71° m/s
v₀ᵧ = 19 sin 71° m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
(a) Find Δy when t = 3.5 s.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²
Δy = 2.85 m
(b) Find Δy when vᵧ = 0 m/s.
vᵧ² = v₀ᵧ² + 2 aᵧ Δy
(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy
Δy = 16.5 m
(c) Find Δx when t = 3.5 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²
Δx = 21.7 m
(d) Find Δx when Δy = 0 m.
First, find t when Δy = 0 m.
Δy = v₀ᵧ t + ½ aᵧ t²
(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²
0 = t (18.0 − 4.9 t)
t = 3.67
Next, find Δx when t = 3.67 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²
Δx = 22.7 m
