All categories

2 years ago

Which of the following is an example of an observation?

Engineering

2 answers:

Novay_Z [31]2 years ago

6 0

The monkey is brown!

Lorico [155]2 years ago

3 0

The monkey is brown, because when you look at it without doing anything you can clearly notice that hope this helps

You might be interested in

The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp

Aliun [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

$Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1$

The Lower Surface Cp is given as

$Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1$

The difference of the Cp over the airfoil is given as

$\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32$

Now the Lift Coefficient is given as

$C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192$

Now the coefficient of moment about the leading edge is given as

$C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306$

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

5 0

2 years ago

Exercises

Feliz [49]

Answer:

Rocket

Gas

Explanation:

5 0

2 years ago

How did Atlantis benefit from lessons learned in construction of earlier orbiters?

Alenkasestr [34]

Answer:

Atlantis benefited from lessons learned in the construction and testing of Enterprise, Columbia and Challenger. ... The Experience gained during the Orbiter assembly process also enabled Atlantis to be completed with a 49.5 percent reduction in man hours (compared to Columbia).

Explanation:

8 0

3 years ago

What motivated software engineers to move from the waterfall model to the incremental or spiral model

scoray [572]

Answer:

1. They needed to develop multiple components in software programs.

2. The ability to overlap the development to be more evolutionary in nature.

3. The need to be more risk-averse or the unwillingness to take risks led to the use of a spiral model.

Explanation:

Software development life cycle (SDLC) can be defined as a strategic process or methodology that defines the key steps or stages for creating and implementing high quality software applications.

In SDLC, a waterfall model can be defined as a process which involves sequentially breaking the software development into linear phases. Thus, the development phase takes a downward flow like a waterfall and as such each phase must be completed before starting another without any overlap in the process.

An incremental model refers to the process in which the requirements or criteria of the software development is divided into many standalone modules until the program is completed.

Also, a spiral model can be defined as an evolutionary SDLC that is risk-driven in nature and typically comprises of both an iterative and a waterfall model. Spiral model of SDLC consist of these phases; planning, risk analysis, engineering and evaluation.

<em>What motivated software engineers to move from the waterfall model to the incremental or spiral model is actually due to the following fact;</em>

They needed to develop multiple components in software programs.

The ability to overlap the development to be more evolutionary in nature.

The need to be more risk-averse or the unwillingness to take risks led to the use of a spiral model.

6 0

2 years ago

(SI units) Molten metal is poured into the pouring cup of a sand mold at a steady rate of 400 cm3/s. The molten metal overflows

maxonik [38]

Answer:

diameter of the sprue at the bottom is 1.603 cm

Explanation:

Given data;

Flow rate, Q = 400 cm³/s

cross section of sprue: Round

Diameter of sprue at the top $d_{top}$ = 3.4 cm

Height of sprue, h = 20 cm = 0.2 m

acceleration due to gravity g = 9.81 m/s²

Calculate the velocity at the sprue base

$V_{base}$ = √2gh

we substitute

$V_{base}$ = √(2 × 9.81 m/s² × 0.2 m )

$V_{base}$ = 1.98091 m/s

$V_{base}$ = 198.091 cm/s

diameter of the sprue at the bottom will be;

Q = AV = (π $d_{bottom}^2$ /4) × $V_{base}$

$d_{bottom}$ = √(4Q/π $V_{base}$ )

we substitute our values into the equation;

$d_{bottom}$ = √(4(400 cm³/s) / (π×198.091 cm/s))

$d_{bottom}$ = 1.603 cm

Therefore, diameter of the sprue at the bottom is 1.603 cm

6 0

2 years ago