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deff fn [24]
2 years ago
8

Compute the mean and maximum velocities for a liquid with a flow rate of 20 L/min in a 1.5-in nominal diameter sanitary pipeline

. The liquid has a density of 1030 kg/m3 and viscosity of 50 cP. Is the flow laminar or turbulent?
Physics
1 answer:
vlada-n [284]2 years ago
7 0

Answer:

Mean velocity = 0.292 m/s

Maximum velocity = 0.584 m/s

The flow is laminar as Re = 229.2

Explanation:

D = 1.5 inches = 0.0381 m

Q = volumetric flow rate = 20 L/min = 0.000333 m³/s

Q = A × v

A = Cross sectional Area = πD²/4 = π(0.0381)²/4 = 0.00114 m²

v = average velocity

v = Q/A = 0.000333/0.00114 = 0.292 m/s

For flow in circular pipes, maximum velocity = 2 × average velocity = 2 × 0.292 = 0.584 m/s

To check if the flow is laminar or turbulent, we need its Reynolds number

Re = (ρvD)/μ

ρ = 1030 kg/m

v = 0.292 m/s

D = 1.5 inches = 0.0381 m

μ = 50 cP = 0.5 poise = 0.05 Pa.s

Re = (1030 × 0.292 × 0.0381)/0.05 = 229.2

For laminar flow, Re < 2100

For turbulent flow, Re > 4000

And 229.2 < 2100, hence, this flow is laminar.

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A stationary 1.67-kg object is struck by a stick. The object experiences a horizontal force given by F = at - bt2, where t is th
Usimov [2.4K]

Answer:

v_{f}  = 3289.8 m / s

Explanation:

This exercise can be solved using the definition of momentum

     I = ∫ F dt

Let's replace and calculate

     I = ∫ (at - bt²) dt

We integrate

      I = a t² / 2 - b t³ / 3

We evaluate between the lower limits I=0  for t = 0 s and higher I=I for t = 2.74 ms

      I = a (2,74² / 2- 0) - b (2,74³ / 3 -0)

      I = a 3,754 - b 6,857

We substitute the values ​​of a and b

      I = 1500 3,754 - 20 6,857

      I = 5,631 - 137.14

      I = 5493.9 N s

Now let's use the relationship between momentum and momentum

      I = Δp = m v_{f} - m v₀o

      I = m v_{f}  - 0

     v_{f}  = I / m

    v_{f}  = 5493.9 /1.67

    v_{f}  = 3289.8 m / s

5 0
2 years ago
If you add 700 kJ of heat to 700 g of water at 70 degrees C, how much water is left in the container? The latent heat of vaporiz
makkiz [27]

Answer:A

Explanation:Find attached picture file for details

3 0
2 years ago
A target in a shooting gallery consists of a vertical square wooden board, 0.250 m on a side and with mass 0.750 kg, that pivots
Alenkasestr [34]

Here in this case since there is no torque about the hinge axis for the system of bullet and block then we can say that angular momentum of this system will remain conserved

L_i = L_f

mv \frac{L}{2} = (I_1 + I_2)\omega

here we will have

L = 0.250 m

v = 385 m/s

m = 1.90 gram

now moment of inertia of the plate will be

I_1 = \frac{ML^2}{3}

I_1 = \frac{0.750 (0.250)^2}{3} = 0.0156 kg m^2

I_2 = m(\frac{L}{2})^2 = 0.0019(0.125)^2 = 2.97 \times 10^{-5} kg m^2

now from above equation

0.0019 (385)(0.125) = (0.0156 + 2.97 \times 10^{-5})\omega

\omega = 5.85 rad/s

8 0
2 years ago
(Another tomato/skyscraper problem.) You are looking out your window in a skyscraper, and again your window is at a height of 45
Ivan

Answer:

1027.2 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.2 ft/s

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{450-\frac{1}{2}\times 32.2\times 2^2}{2}\\\Rightarrow u=192.8\ ft/s

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{192.8^2-0^2}{2\times 32.2}\\\Rightarrow s=577.20\ m

The height the tomato would fall is 450+577.2 = 1027.2 m

6 0
3 years ago
The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t , where the time tis in seconds. The
Reika [66]

Complete question:

The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t³ , where the time tis in seconds. The

particle is momentarily at rest at t is:

Select one:

a. 9.3s

b. 1.3s

C. 0.75s

d.5.3s

e. 7.3s

​

Answer:

b. 1.3 s

Explanation:

Given;

position of the particle, x(t)=1 6t- 3.0t³

when the particle is at rest, the velocity is zero.

velocity = dx/dt

dx /dt = 16 - 9t²

16 - 9t² = 0

9t² = 16

t² = 16 /9

t = √(16 / 9)

t = 4/3

t = 1.3 s

Therefore, the particle is momentarily at rest at t = 1.3 s

6 0
2 years ago
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