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2 years ago

(4 points) What field of work generally requires (a) an engineer to have a Professional Engineer

Engineering

1 answer:

Lunna [17]2 years ago

6 0

Answer:

An industrial engineer from UTPL, is a professional who optimizes processes in the company, can work in various areas within the organization such as: occupational safety, logistics, administration, human resources and production systems of goods and services.

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Why will screws never replace nails

cupoosta [38]

Answer:

because people have different opinions on nails and screws

Explanation:

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2 years ago

A six-lane freeway (three lanes in each direction) in rolling terrain has 10-ft lanes and obstructions 5 ft from the right edge.

Anon25 [30]

That is too hard but u got that cuz i believe in you!!!

3 0

2 years ago

Which of the following are some of the problems found in city streets?

storchak [24]

Answer: drugs and rushing cars

Explanation: drug dealers are everywhere on city streets nowadays they have been killing young adults

rushing cars or reckless drivers cut curb fast and potentially someone can get hurt they are speeding and not worrying about other people lives at stake

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2 years ago

During January, at a location in Alaska winds at -20°C can be observed, However, several meters below ground the temperature rem

maw [93]

Answer:

a) $\eta_{th} = 10.910\,\%$ , b) Yes.

Explanation:

a) The maximum thermal efficiency is given by the Carnot's Cycle, whose formula is:

$\eta_{th} =\left(1-\frac{253.15\,K}{284.15\,K} \right) \times 100\,\%$

$\eta_{th} = 10.910\,\%$

b) The claim of the inventor is possible since real efficiency is lower than maximum thermal efficiency.

4 0

2 years ago

Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you

viktelen [127]

Answer:

$\theta_1=15^o\\\theta_2=75^o$

Explanation:

<u>Projectile Motion</u>

In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

$V_x=V_{ox}=V_ocos\theta$

Where $\theta$ is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.

$V_y=V_{oy}-gt=V_osin\theta-gt$

The horizontal and vertical distances are, respectively:

$x=V_{o}cos\theta t$

$\displaystyle y=y_o+V_{o}sin\theta t-\frac{gt^2}{2}$

The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find

$\displaystyle t_f=\frac{2V_osin\theta}{g}$

Using this time in the horizontal distance, we find the Range or maximum horizontal distance:

$\displaystyle R=\frac{V_o^2sin2\theta}{g}$

Let's solve for $\theta$

$\displaystyle sin2\theta=\frac{R.g}{V_o^2}$

This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:

$\displaystyle \theta_1=\frac{asin\left(\frac{R.g}{V_o^2}\right)}{2}$