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2 years ago

PLLLLLEEESSSEEE IIII NEED ASAP

Engineering

2 answers:

Eddi Din [679]2 years ago

6 0

Answer:

It´s A

Explanation:

If I was right please like thank you :)

Please write in the comments if i was wrong and I am sorry if I was :(

pychu [463]2 years ago

6 0

Answer:

Answer is only a: color vision for identifying differences in colored wires

Explanation:

at first I had a bot of truest issues of only it being a so I went ahead and only out a and did some reasrch and in the ned I ended up with a 100% on my unit test review so I am happily to confirm that is is A and only A .

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REVIEW QUESTIONS

babunello [35]

Snap rings, and bearings can be used to keep a gear on a shaft, hope this helps!!

5 0

2 years ago

. In one stroke of a reciprocating compressor, helium is isothermally and reversibly compressed in a piston + cylinder from 298

andriy [413]

Answer:

5.7058kj/mole

Explanation:

Please see attachment for step by step guide

5 0

2 years ago

can someone help me with this engineering mechanics homework, please? I tried to solve it, but I got so confused.

marishachu [46]

Explanation:

Sum of forces in the x direction:

∑Fx = ma

Rx − 250 N = 0

Rx = 250 N

Sum of forces in the y direction:

∑Fy = ma

Ry − 120 N − 300 N = 0

Ry = 420 N

Sum of forces in the z direction:

∑Fz = ma

Rz − 50 N = 0

Rz = 50 N

Sum of moments about the x axis:

∑τx = Iα

Mx + (-50 N)(0.2 m) + (-120 N)(0.1 m) = 0

Mx = 22 Nm

Sum of moments about the y axis:

∑τy = Iα

My = 0 Nm

Sum of moments about the z axis:

∑τz = Iα

Mz + (250 N)(0.2 m) + (-120 N)(0.16 m) = 0

Mz = -30.8 Nm

6 0

3 years ago

AVH

BigorU [14]

Answer: preventive maintenance

Explanation:

3 0

2 years ago

A cylinder with a frictionless piston contains 0.05 m3 of air at 60kPa. The linear spring holding the piston is in tension. The

AleksAgata [21]

Answer:

18 kJ

Explanation:

Given:

Initial volume of air = 0.05 m³

Initial pressure = 60 kPa

Final volume = 0.2 m³

Final pressure = 180 kPa

Now,

the Work done by air will be calculated as:

Work Done = Average pressure × Change in volume

thus,

Average pressure = $\frac{60+180}{2}$ = 120 kPa

and,

Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³

Therefore,

the work done = 120 × 0.15 = 18 kJ

4 0

2 years ago