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Nat2105 [25]
3 years ago
8

I think it’s a wedge. Am I right?

Engineering
2 answers:
Hitman42 [59]3 years ago
4 0

Answer:

yeah its A > wedge

Explanation:

just search it on the internet of something easy

german3 years ago
3 0

Answer:

i believe you correct  

Explanation:

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A 1.8-m3 rigid tank contains water steam at 220oC. One-third of the volume is in the liquid phase and the rest is in the vapor f
svetoff [14.1K]

Answer:

a) 2319.6 kPa

b) 0.027

c) 287.86 kg/m^3

Explanation:

The pressure is determined from table  in the appendix for the given temperature:  

 P_220=2319.6 kPa

to calculate the quality we need to determine the masses of the vapor and the liquid and for that we need the respective specific volumes which can also be found in table.  

m_liq  = V_liq/α_liq

           = 504.2 kg

m_vap = V_vap/α_vap

            = 13.94 kg

        q = m_vap/m_liq

          = 0.027

Finally, the density is simply calculated as follows:  

        p = m_tot/V_tot

           = 518.14 kg/1.8 m^3

           = 287.86 kg/m^3

8 0
3 years ago
Which of the following identifies three advantages of PLM software?
DerKrebs [107]

Answer:

reduced cost, faster marketing, and process transparency

Explanation:

The correct answer is reduced cost, faster marketing, and process transparency. PLM software helps coordinate tasks during implementation.

5 0
3 years ago
Read 2 more answers
A small submarine has a triangular stabilizing fin on its stern. The fin is 1 ft tall and 2 ft long. The water temperature where
Arturiano [62]

Answer:

\mathbf{F_D \approx 1.071 \ lbf}

Explanation:

Given that:

The height of a  triangular stabilizing fin on its stern is 1 ft tall

and it length is 2 ft long.

Temperature = 60 °F

The objective is to determine the drag on the fin when the submarine is traveling at a speed of 2.5 ft/s.

From these information given; we can have a diagrammatic representation describing how the  triangular stabilizing fin looks like as we resolve them into horizontal and vertical component.

The diagram can be found in the attached file below.

If we recall ,we know that;

Kinematic viscosity v = 1.2075 \times 10^{-5} \ ft^2/s

the density of water ρ = 62.36 lb /ft³

Re_{max} = \dfrac{Ux}{v}

Re_{max} = \dfrac{2.5 \ ft/s \times 2  \  ft }{1.2075 \times 10 ^{-5} \ ft^2/s}

Re_{max} = 414078.6749

Re_{max} = 4.14 \times 10^5 which is less than < 5.0 × 10⁵

Now; For laminar flow;  the drag on  the fin when the submarine is traveling at 2.5 ft/s can be determined by using the expression:

dF_D = (\dfrac{0.664 \times \rho  \times U^2 (2-x) dy}{\sqrt{Re_x}})^2

where;

(2-x) dy = strip area

Re_x = \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}

Therefore;

dF_D = (\dfrac{0.664 \times 62.36  \times 2.5^2 (2-x) dy}{\sqrt{ \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}}})

dF_D = 1.136 \times(2-x)^{1/2} \ dy

Let note that y = 0.5x from what we have in the diagram,

so , x = y/0.5

By applying the rule of integration on both sides, we have:

\int\limits \  dF_D =  \int\limits^1_0 \  1.136 \times(2-\dfrac{y}{0.5})^{1/2} \ dy

\int\limits \  dF_D =  \int\limits^1_0 \  1.136 \times(2-2y)^{1/2} \ dy

Let U = (2-2y)

-2dy = du

dy = -du/2

F_D =  \int\limits^0_2 \  1.136 \times(U)^{1/2} \ \dfrac{du}{-2}

F_D = - \dfrac{1.136}{2} \int\limits^0_2 \ U^{1/2} \ du

F_D = -0.568 [ \dfrac{\frac{1}{2}U^{ \frac{1}{2}+1 }  }{\frac{1}{2}+1}]^0__2

F_D = -0.568 [ \dfrac{2}{3}U^{\frac{3}{2} }   ] ^0__2

F_D = -0.568 [0 -  \dfrac{2}{3}(2)^{\frac{3}{2} }   ]

F_D = -0.568 [- \dfrac{2}{3} (2.828427125)}   ]

F_D = 1.071031071 \ lbf

\mathbf{F_D \approx 1.071 \ lbf}

8 0
3 years ago
A) For Well A, provide a cross-section sketch that shows (i) ground elevation, (ii) casing height, (iii) depth to
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Don’t go on that file will give a virus! Sorry just looking out and I don’t know how to comment!
7 0
3 years ago
Calculate the scale and speed of the pattern in order to gain useful results for a turbine operate at 150 rev/min at height diff
Rzqust [24]

Answer:

first mark me as a brainleast

6 0
3 years ago
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