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Ad libitum [116K]
3 years ago
12

Water at 20oC flows through a long elliptical duct 30 cm wide and 22 cm high. What average velocity, in m/s, would cause the wei

ght flow to be 500 lbf/s?
Physics
2 answers:
mars1129 [50]3 years ago
8 0

Explanation:

The given data is as follows.

   Fluid is water so, density \rho = 1000 kg/m^{3}

  Weight flow rate = 500 lbf/s = 2224.11 N/sec

  Cross-sectional area (A) = \pi \times \frac{30}{2} \times \frac{22}{2}

                                         = 0.05184 m^{2}

Hence, weight flow rate will be given as follows.

        w = \rho \times g \times A \times V

    2224.11 N/sec = \rho \times g \times A \times V

        V = \frac{2224.11}{1000 \times 9.81 \times 0.05184} m/s

            = 4.373 m/s

Thus, we can conclude that average velocity in the given case is 4.373 m/s.

gizmo_the_mogwai [7]3 years ago
8 0

Answer:

Explanation:

length of major axis, 2a = 30 cm

a = 15 cm = 0.15 m

length of minor axis, 2b = 22 cm

b = 11 cm = 0.11 m

Weight flow, mg = 500 lbf/s = 2224.1 N/s

Area of duct, A = π ab = 3.14 x 0.15 x 0.11 = 0.0518 m²

Let v be the velocity

Volume per second = mass per second / density

Area x velocity = mass per second / density

0.0518 x v = 2224.1 / (9.8 x 1000)

v = 4.4 m/s

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nikitadnepr [17]

Answer:

The extension of the second wire is   e_2 = 0.0024 \  m =  2.4 mm

Explanation:

From the question we are told that

    The length of the wire is L  = 3 \ m

     The elongation of the wire is  e =  1.2mm =  \frac{1.2}{1000} =  0.0012 m

        The tension is F  =  200 \ N

       The length of the second wire is  L_2   =  6 \ m

     

Generally the Young's modulus(Y) of this material is  

        Y  = \frac{stress}{strain }

Where stress =  \frac{F}{A}

    Where A is the area which is evaluated as  

           A = \pi r^2

  and   strain = \frac{extention}{length} =  \frac{e}{L}

   So

        Y  = \frac{\frac{F}{\pi r^2 } }{ \frac{e}{L}  }

Since the wire are of the same material Young's modulus(Y)  is constant

So we have  

              \frac{F * L }{r^2 e}  =  \pi * Y = constant

              F * L   =  constant   * r^2 e

Now the ration between the first and the second wire is

         \frac{F_1}{F_2}  * \frac{L_1}{L_2} =  \frac{r*2_1}{r^2}  *  \frac{e_1}{e_2}

Since tension , radius are constant

   We have

           \frac{L_1}{L_2} =   \frac{e_1}{e_2}

substituting values

          \frac{3}{6} =   \frac{0.0012}{e_2}

          0.5 e_2 =  0.0012

         e_2 = \frac{ 0.0012  }{0.5}

          e_2 = 0.0024 \  m =  2.4 mm

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3 years ago
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<span>1 cal = 4,185 J
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Answer:

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Answer:

social cognitive theory

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If a planet has a radius 20% greater than that of the Earth but has the same mass as the Earth, what is the acceleration due to
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Answer:

g_2=6.8125 m^2/sec

Explanation:

We know that

Acceleration due to gravity g is given by the formula

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