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3 years ago

What kinds of problems or projects would a mechanical engineer work on?

Engineering

1 answer:

Mademuasel [1]3 years ago

3 0

Answer:

Design power producing machines like electric generators, internal combustion engines, turbines and power using machines like refrigiration or air conditioning. And alot like elevators or such.

Explanation:

If that is what your asking for? Sorry if its not what you wanted. In genral all hard work science and maths

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What is an example of a product made of textile?

Otrada [13]

$beach \: towel \\ \\ hope \: it \: helps$

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2 years ago

A nail that is rectangular in<br> cross section and has a blunt<br> point is called a

Travka [436]

Answer:

A nail that is rectangular in cross section and has a blunt point is called a Cut nail.

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3 years ago

An ice hockey player is skating on an ice rink. The rink has a coefficient of kinetic friction of roughly 0.1. If the normal for

AlekseyPX

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3 years ago

Engineer Smith, who is licensed in several jurisdictions, recently had his license acted upon by a licensing authority in one of

Nimfa-mama [501]

Answer:

Assuming the infraction in the other jurisdiction is an infraction in Florida, having approximately the same penalty imposed by the FBPE as was imposed in the other jurisdiction

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The idea of 8 hours continuation in the options is irrelevant to this case. Rather, Engineer Smith should expect about the same penalty imposition.

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3 years ago

A missile flying at high speed has a stagnation pressure and temperature of 5 atm and 598.59 °R respectively. What is the densit

alexdok [17]

Answer:

$5.31\frac{kg}{m^3}$

Explanation:

Approximately, we can use the ideal gas law, below we see how we can deduce the density from general gas equation. To do this, remember that the number of moles n is equal to $\frac{m}{M}$ , where m is the mass and M the molar mass of the gas, and the density is $\frac{m}{V}$ .

For air $M=28.66*10^{-3}\frac{kg}{mol}$ and $\frac{5}{9}R=K$

So, $598.59 R*\frac{5}{9}=332.55K$

$pV=nRT\\pV=\frac{m}{M}RT\\\frac{m}{V}=\frac{pM}{RT}\\\rho=\frac{pM}{RT}\\\rho=\frac{(5atm)28.66*10^{-3}\frac{kg}{mol}}{(8.20*10^{-5}\frac{m^3*atm}{K*mol})332.55K}=5.31\frac{kg}{m^3}$

7 0

4 years ago