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ASHA 777 [7]
3 years ago
6

Describe the engineering design process in your own words.

Engineering
2 answers:
MrMuchimi3 years ago
8 0

Answer:

Explanation:

asambeis [7]3 years ago
8 0
The engineering design process, by definition, is a series of steps that engineers use to find a solution to a problem. These steps include: defining the problem, brainstorming solutions, designing and building a prototype of the solution, testing the solution, and improving it.
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Technician A says that applying too much braking force can cause tires to lose traction. Technician B says that ABS systems have
Deffense [45]

Answer:

technically both are correct but tech b is more correct because generally the older the vehicle the worse abs is newer vehicles have abs that no matter how hard you push on the brake peddle the wheels shouldnt lock up

Explanation:

7 0
4 years ago
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 cylindrical specimen of some metal alloy having an elastic modulus of 106 GPa and an original cross-sectional diameter of 3.9
kiruha [24]

Answer:

L= 312.75 mm

Explanation:

given data

elastic modulus E = 106 GPa

cross-sectional diameter d = 3.9 mm

tensile load F = 1660 N

maximum allowable elongation ΔL = 0.41 mm

to find out

maximum length of the specimen before deformation

solution

we will apply here allowable elongation equation that is express as

ΔL =     \dfrac{FL}{AE}     ....................1

put here value and we get L

L   =    \dfrac{0.41\times 10^{-3}\times \dfrac{\pi}{4}\times (3.9\times 10^{-3})^2\times 106\times 10^9}{1660}

solve it we get

L = 0.312752 m

L= 312.75 mm

8 0
4 years ago
Two common types of glue
dolphi86 [110]

Answer:

white craft glue

yellow wood glue

Explanation:

the answer is this!

was it correct?sir!?

8 0
2 years ago
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Why are there few effective HCI standards?
Oliga [24]

Answer:

<em>There are few effective Human Computer Interaction (HCI) standards because for one, standards are more suitable for hardware than software because they are relatively unstable. ... Some software product standards have been in place long before any formal standard documents were published.</em>

3 0
3 years ago
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