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Tanya [424]
2 years ago
13

A horizontal tube consists of a 7.0-cm-diameter pipe that narrows to a 2.0-cm-diameter throat. In the pipe, the water pressure i

s twice atmospheric pressure and the water flows with a speed of 0.40 m/s. What is the pressure in the throat, assuming that the water behaves like an ideal fluid? The density of water is 1000 kg/m3 , and atmospheric pressure is 1.01 × 105 Pa.
Physics
1 answer:
cestrela7 [59]2 years ago
4 0

Answer:

1.9\times10^{5} Pa

Explanation:

d_{p} = diameter of the pipe = 7 cm

v_{p} = speed of water in the pipe = 0.40 m/s

A_{p} = Area of cross-section of pipe = (0.25)\pi d_{p}^{2}

d_{t} = diameter of the throat = 2 cm

v_{t} = speed of water in the throat

A_{t} = Area of cross-section of throat = (0.25)\pi d_{t}^{2}

Using equation of continuity

A_{p} v_{p} = A_{t} v_{t} \\(0.25)\pi d_{p}^{2} v_{p} = (0.25)\pi d_{t}^{2} v_{t} \\(7)^{2} (0.40) = (2)^{2} v_{t}\\v_{t} = 4.9 ms^{-1}

P_{o} = atmospheric pressure = 1.01 x 10⁵ Pa

P_{p} = Pressure in the pipe = 2 P_{o} = 2.02 x 10⁵ Pa

P_{t} = Pressure in the throat

Using Bernoulli's theorem

P_{t} + (0.5)\rho v_{t}^{2} = P_{p} + (0.5)\rho v_{p}^{2}\\P_{t} + (0.5)(1000) (4.9)^{2} = 2.02\times10^{5} + (0.5)(1000) (0.40)^{2}\\P_{t} + 12005 = 202080\\P_{t} = 190075 Pa\\P_{t} = 1.9\times10^{5} Pa

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Answer:

9.98 m/s

Explanation:

The force acting on the particle is defined by the equation:

F=(0.850) sin (\frac{x}{2.00}) [N]

where x is the position in metres.

The acceleration can be found by using Newton's second law:

a=\frac{F}{m}

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m = 150 g = 0.150 kg is the mass of the particle. Substituting into the equation,

a=\frac{0.850}{0.150}sin (\frac{x}{2.00})=5.67 sin(\frac{x}{2.00}) [m/s^2]

When x = 3.14 m, the acceleration is:

a=5.67 sin(\frac{3.14}{2.00})=5.67 m/s^2

Now we can find the final speed of the particle by using the suvat equation:

v^2-u^2=2ax

where

u = 8.00 m/s is the initial velocity

v is the final velocity

a=5.67 m/s^2

x = 3.14 m is the displacement

Solving for v,

v=\sqrt{u^2+2ax}=\sqrt{8.00^2+2(5.67)(3.14)}=9.98 m/s

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A uniform cylindrical steel wire (density: 7.8 x 103 kg/m3), 58.0 cm long and 1.34 mm in diameter, is fixed at both ends. To wha
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Answer:

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Explanation:

Given that

ρ= 7800 kg/m³

L= 58 cm

d=1.34 mm

f= 311 Hz

T= Tension

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V = f λ      

V = f L

V= 311 x 0.58 m/s

V=180.38 m/s

Area of cross sectional

A= πr² mm²

A= 3.14 x 0.67² mm²

A=1.41 mm²

Mass = Density x Volume

m=7800\times 1.41\times 10^{-6}\ kg/m

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Answer:

W = -510.98J

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Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The power created  is  P_{avg} =  F  *  v_{avg}

Explanation:

From the question we are told that

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        So  

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Now this  displacement can be represented mathematically as  

            s =  v_{avg} *  \Delta  t

Where v_{avg } is the average  velocity and \Delta  t is the time  taken  

So  

            P_{avg} =  \frac{F *v_{avg} *  \Delta  t    }{\Delta  t  }

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