Neat mechanical aptitude test

[OreoGaborio]

50 questions, goes by pretty quick.

Ford Doctors Diesel Technician Society

I scored an 88% but I made a bunch of stupid mistakes... given that it's almost 3am I gotta award myself 8 pity points which puts me at an official 96

Then I went back & "corrected" it to 100%

[Currently]

I got a ninety first shot at it.

[wylee]

94...Why doesn't it tell you which ones you get wrong?

[ZX-12R]

94...Why doesn't it tell you which ones you get wrong?

It does. Click on the Table of Contents icon when you are done.

[Paul_E_D]

Wow, I scored a 68. I think I missed all the questions with the weights and pulleys... What's the formula for that?

[ZX-12R]

Pretty much you count the strands of rope and that is your mechanical advantage. Just make sure that all of the pulleys are in a configuration that contribute to it instead of redirecting force.

[Tricky Mike]

That was pretty cool...
Good find, Pete

[Crash Dummy Denno]

i got a 38





did the first ten and then just clicked answer B

[wylee]

94...Why doesn't it tell you which ones you get wrong?

It does. Click on the Table of Contents icon when you are done.

[Jaynnus]

That was pretty interesting...I am smarter than I thought for a dumb chick!!

I suck at the electrical questions though.

[hammadown]

I got a 98 on my first try! :banana:

And the one I got wrong I actually got right, the test makers are wrong!!

Question: Rapidly compressing air and increasing its pressure does what to the temperature?

They say it increases temperature. That is wrong!

Compressing, ie decreasing volume along with increasing pressure.

According to the ideal gas law: PV=nRT

Thus increasing pressure while decreasing volume will have no effect on temperature (assumption: pressure increase and volume decrease are proportional, which I would think is a logical assumption).

Im writing an email. And that is your science geek lesson for the day

If you got that one wrong and said no effect, add 2 points to your score.

[978chris]

Both boxes weigh 100kg. Which will be more difficult to life with the lever?

# <-box
/
/
^

/
/
# <-box
^

[edit] ok my ascii drawing got fucked... but hopefully you get the idea.

Ok I'm looking at this question, and the answer is obvious, but I cannot tell from the picture if the triangle is the pivot point of the lever or is the triangle where I am lifting from?

Lifted from the left or right side of drawing? It does not say. I could ASSUME that since the left side of the lever is closest to the ground, that I lift from the left. I could also ASSUME that since the triangle is there on the left (and the same shape as the pivot in the seesaw drawings), that IT is the pivot point.

I find way too much ambiguity in things. People hate me for it at work.

[hammadown]

Both boxes weigh 100kg. Which will be more difficult to life with the lever?

# <-box
/
/
^

/
/
# <-box
^

[edit] ok my ascii drawing got fucked... but hopefully you get the idea.

Ok I'm looking at this question, and the answer is obvious, but I cannot tell from the picture if the triangle is the pivot point of the lever or is the triangle where I am lifting from?

Lifted from the left or right side of drawing? It does not say. I could ASSUME that since the left side of the lever is closest to the ground, that I lift from the left. I could also ASSUME that since the triangle is there on the left (and the same shape as the pivot in the seesaw drawings), that IT is the pivot point.

I find way too much ambiguity in things. People hate me for it at work.

That is a perfectly acceptable argument...I was thinking the same thing on that question

[Jaynnus]

94...Why doesn't it tell you which ones you get wrong?

It does.

Click on the magnifying glass icon on the top right after completion and you can walk through to see what you did right or wrong.

[Kurlon]

That was pretty interesting...I am smarter than I thought for a dumb chick!!

I suck at the electrical questions though.

Don't feel too bad, one of the electrical questions is wrong. : )

[978chris]

How come rapidly releasing air from a high pressure state makes it cold then?

I dunno, I think that "rapidly" wasn't just thrown in there for laughs.

I am idiot though, you may be totally correct.

"Rapidly" decreasing volume while also increasing pressure doesn't sound like an even, proportional exhange to me unless they also said "rapidly" before the increasing pressure part.

[edit] I was just thinking of those cans of spray air for dusting off electronics. Shake well, hold the trigger for a good minute, and then have the can surgically removed from your now frozen palm. Maybe that's cause of the other chemicals in there...

I got a 98 on my first try! :banana:

And the one I got wrong I actually got right, the test makers are wrong!!

Question: Rapidly compressing air and increasing its pressure does what to the temperature?

They say it increases temperature. That is wrong!

Compressing, ie decreasing volume along with increasing pressure.

According to the ideal gas law: PV=nRT

Thus increasing pressure while decreasing volume will have no effect on temperature (assumption: pressure increase and volume decrease are proportional, which I would think is a logical assumption).

Im writing an email. And that is your science geek lesson for the day

If you got that one wrong and said no effect, add 2 points to your score.

[SSearchVT]

I got a 98 on my first try! :banana:

And the one I got wrong I actually got right, the test makers are wrong!!

Question: Rapidly compressing air and increasing its pressure does what to the temperature?

They say it increases temperature. That is wrong!

Compressing, ie decreasing volume along with increasing pressure.

According to the ideal gas law: PV=nRT

Thus increasing pressure while decreasing volume will have no effect on temperature (assumption: pressure increase and volume decrease are proportional, which I would think is a logical assumption).

Im writing an email. And that is your science geek lesson for the day

If you got that one wrong and said no effect, add 2 points to your score.

Except you are rapidly decreasing the volume - the ideal gas law applies to a steady state. The temperature goes up - the best analogy I came up with for this one is how a diesel engine works.

[SSearchVT]

Don't feel too bad, one of the electrical questions is wrong. : )

The one with the switch that goes parallel to the light - I answered that all three lights would go on. The current would still flow through both legs.

[978chris]

The one with the switch that goes parallel to the light - I answered that all three lights would go on. The current would still flow through both legs.

What about resistance (with regards to the path of least)?

[edit] I got 80%... I blame 2 wrong answers on the person who created the test making assumptions.

[Kurlon]

What about resistance?

Add all the resistance you want, current will still flow through both paths. By playing with the resistance (or lack there of) all you're doing is altering the balance of the curent going through each path, but nothing short of infinate resistance is going to keep current from taking both routes.

[Kurlon]

closed switch shorts light. light no workey.

All depends on the light and how much current we've got to play with. : )

[Honclfibr]

All depends on the light and how much current we've got to play with. : )

In cases like these the closed switch should be modelled as a zero resistance path and the lightbulb as a nonzero resistance path. Given those assumptions no current will flow through the bulb which is in parallel with the closed switch.

Edit: Fuck. Beatened.

[hammadown]

How come rapidly releasing air from a high pressure state makes it cold then?

I dunno, I think that "rapidly" wasn't just thrown in there for laughs.

I am idiot though, you may be totally correct.

"Rapidly" decreasing volume while also increasing pressure doesn't sound like an even, proportional exhange to me unless they also said "rapidly" before the increasing pressure part.

[edit] I was just thinking of those cans of spray air for dusting off electronics. Shake well, hold the trigger for a good minute, and then have the can surgically removed from your now frozen palm. Maybe that's cause of the other chemicals in there...

Except you are rapidly decreasing the volume - the ideal gas law applies to a steady state. The temperature goes up - the best analogy I came up with for this one is how a diesel engine works.

Excellent points, this brings up the difference between practice and theory. I agree that in these situations there are no valid steady state approximations. However, the question was a bit vague, and I generally like to keep things simple if I can. More importantly, I like to be difficult and question things, because that is my job (and I'm a PIA )


Also it seems the people that made the test made some assumptions on the electrical questions, but not on this one...bastards!

[KillBill]

NEATO

[ZX-12R]

And the one I got wrong I actually got right, the test makers are wrong!!

Question: Rapidly compressing air and increasing its pressure does what to the temperature?

They say it increases temperature. That is wrong!

Compressing, ie decreasing volume along with increasing pressure.

According to the ideal gas law: PV=nRT

Thus increasing pressure while decreasing volume will have no effect on temperature (assumption: pressure increase and volume decrease are proportional, which I would think is a logical assumption).

Im writing an email. And that is your science geek lesson for the day

If you got that one wrong and said no effect, add 2 points to your score.

The answer to the question is correct and your reasoning is a bit off. PV=nRT is not applicable in this example because V and T refer to the volume and temperature of the container that the gas is contained in and not the gas itself.

The correct formula you need to apply is:

(P1^(1-k))*(T1^k) = (P2^(1-k))*(T2^k)

where:

P1 = initial pressure
T1 = initial temperature
P2 = final pressure
T2 = final temperature
k = heat capacity ratio (Cp/Cv: Air is about 1.4)

This formula assumes that the vessel is not absorbing heat from the process(adiabatic).