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3 years ago

alguien me ayuda con una tarea? Está en mi perfil de matemática, porfavorrr regalaré corona pero porfavor

Engineering

1 answer:

tensa zangetsu [6.8K]3 years ago

5 0

Answer:

HUH?

Explanation:

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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

What does the DHCP server configures for each host?

lidiya [134]

Answer: IP address

Explanation:

DHCP server automatically assigns an IP address and other information to each host on the network so they can communicate efficiently with other endpoints

8 0

4 years ago

Atmospheric air at 25 °C and 8 m/s flows over both surfaces of an isothermal (179C) flat plate that is 2.75m long. Determine the

vekshin1

Answer:

Re=100,000⇒Q=275.25 $\frac{W}{m^2}$

Re=500,000⇒Q=1,757.77 $\frac{W}{m^2}$

Re=1,000,000⇒Q=3060.36 $\frac{W}{m^2}$

Explanation:

Given:

For air $T_∞$ =25°C ,V=8 m/s

For surface $T_s$ =179°C

L=2.75 m ,b=3 m

We know that for flat plate

$Re$ ⇒Laminar flow

$Re>30\times10^5$ ⇒Turbulent flow

Take Re=100,000:

So this is case of laminar flow

$Nu=0.664Re^{\frac{1}{2}}Pr^{\frac{1}{3}}$

From standard air property table at 25°C

Pr= is 0.71 ,K=26.24 $\times 10^{-3}$

So $Nu=0.664\times 100,000^{\frac{1}{2}}\times 0.71^{\frac{1}{3}}$

Nu=187.32 ( $\dfrac{hL}{K_{air}}$ )

187.32= $\dfrac{h\times2.75}{26.24\times 10^{-3}}$

⇒h=1.78 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=275.25 $\frac{W}{m^2}$

Take Re=500,000:

So this is case of turbulent flow

$Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}$

$Nu=0.037\times 500,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}$

Nu=1196.18 ⇒h=11.14 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=11.14(179-25)

= 1,757.77 $\frac{W}{m^2}$

Take Re=1,000,000:

So this is case of turbulent flow

$Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}$

$Nu=0.037\times 1,000,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}$

Nu=2082.6 ⇒h=19.87 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=19.87(179-25)

= 3060.36 $\frac{W}{m^2}$

7 0

3 years ago

A column in a building is subjected to the following load effects:

vagabundo [1.1K]

Answer:

attached below

Explanation:

6 0

3 years ago

Omar owns a stall selling handmade cosmetics, and wants to launch an e-commerce site. He has lots of experience selling his prod

erica [24]

Answer:

He could put up advertisements all over the town and pay a few websites or search engines to have his company and ads be more common. He should also use some designs and make his company's sign very adequate.

6 0

2 years ago

alguien me ayuda con una tarea? Está en mi perfil de matemática, porfavorrr regalaré corona pero porfavor​

alguien me ayuda con una tarea? Está en mi perfil de matemática, porfavorrr regalaré corona pero porfavor