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Semmy [17]
3 years ago
14

A fluid flows through a pipe whose cross-sectional area changes from 2.00 m2 to 0.50 m2 . If the fluid’s speed in the wide part

of the pipe is 3.5 m/s, what is its speed when it moves through the narrow part of the pipe?
Physics
1 answer:
borishaifa [10]3 years ago
4 0

Answer:

v₂ = 7/ (0.5)= 14 m/s

Explanation:

Flow rate of the fluid

Flow rate is the amount of fluid that circulates through a section of the pipeline (pipe, pipeline, river, canal, ...) per unit of time.

The formula for calculated the flow rate is:

Q= v*A Formula (1)

Where :

Q is the Flow rate (m³/s)

A is the cross sectional area of a section of the pipe (m²)

v is the speed of the fluid in that section (m/s)

Equation of continuity

The volume flow rate Q for an incompressible fluid at any point along a pipe is the same as the volume flow rate at any other point along a pipe:

Q₁= Q₂

Data

A₁ = 2m² : cross sectional area 1

v₁ = 3.5 m/s : fluid speed through A₁

A₂ = 0.5 m² : cross sectional area 2

Calculation of the fluid speed through A₂

We aply the equation of continuity:

Q₁= Q₂

We aply the equation of Formula (1):

v₁*A₁= v₂*A₂

We replace data

(3.5)*(2)= v₂*(0.5)

7 = v₂*(0.5)

v₂ = 7/ (0.5)

v₂ =  14 m/s

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Mashutka [201]

Answer: The power required by the pump to produce a discharge of 9m³/s is 5990joules/secs.

Explanation: The given parameters from the questions are:

Flow rate Q = 9m³/s, Diameter D = 0.98m, acceleration a = 1.0, head loss(Pressure P) is given by the function 10v²/2g.

STEP 1. Find the velocity of water in the pipe from the equation:

Diameter D = (√4.Q/π.v), where v is the velocity, and Q is flow rate

Making v subject of the formula gives:

v = 4Q/π.√D =[ 4 × 9m³/s / 3.142 × (√0.98m)] = 11.69m/s.

STEP 2. Find the pressure from the relationship, P = 10v²/2g, NB. g = a

P = 10 × (11.69m/s)² / 2× 1.0m/s²

P = 683.25N/m² or Pascal.

STEP 3. Find force exerted by the pump;

Recall that Pressure P = Force/Area

But Area A = π.r², where r = D/2

Therefore, A = π.(D/2)²

A = 3.142 × [0.98m/2]² = 0.75m²

Therefore, Force = Pressure × Area

Force F = 683.25N/m² × 0.75m²

F = 512.44N.

STEP 4. Find work done

Work done W by the pump is = Force × distance d moved by the water

W = F . d

Also recall that flow rate Q = Velocity/time.

Q = v/t, we can write t = v/Q.

Time t = 11.69m/s / 9m³/s = 1.298s

Also recall that velocity v = distance d/time t, v = d/t, making d subject of formula gives v × t

Distance d = v × t = 11.69m/s × 1.298s = 15.17m.

Hence,

Work Done W = Force × distance

W = 512.44N × 15.17m = 7775.56Nm or joules.

Lastly, Power P = Work done/ time

P = 7775.56joules/1.298s

P = 5990.4joules/s.

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Answer:

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Explanation:

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To find the new volume of the gas we will use the combined gas law.

The combined gas law combines Boyle's law, Charles law and Gay-Lussac's law. Its states that the ratio of the product of pressure and volume to temperature is equal to a constant. its expressed mathematically below.

\frac{V1P1}{T1} = \frac{V2P2}{T2}

V2 =\frac{V1P1T2}{P2T1}

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4 0
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valentinak56 [21]

Answer:

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S = ut + ½at²

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S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

S = 4.66*2.35 + ½*5.66*2.35²

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If you increase the voltage you increase the current
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