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1,001 questions about the ASTB (post your scores & ask your questions here!)
- Thread starter edato
- Start date
Ok, I am going to assume those v's are radicals.
Then v50 = v25v2 = 5v2
v72= v36v2 = 6v2
3x(6v2)= 18v2
5v2 + 18v2 = 23v2
I don't think there's a lot of radicals on the astb
djguernsey
Pro-Rec SNA - OCS Class Date 27 May 2012
Ah radicals. the v confused me. I was in algebra mode. Good catch though. I don't remember radicals on the test either.
And as the sprocket wheel, here's my take on how I would solve the problem, though someone correct me if there is another or better way to go about it. I may just get lucky on this:
The sprocket has six spikes to hold six chain links on it but 5 gaps on half of the wheel. Each half of the sprocket has a linear distance of 10". So that means that that each gap is 2 in, so the wheel will pull up 2 in every time it picks up a new chain link.. So it pulls up in one half a revolution 6 chain links x 2in of height the weight moved = 12inches the weight moved. So double that for a full revolution and you get 24inches.
The question is asking how high the weight will be pulled up, not the circumference of the wheel (which is 20). You have to think about the chain and gaps in the sprocket. Sort of combine a gear problem with a pulley problem.
Not sure why I am having a mental block but can't figure out how to arrive at the answer of 35:
There are exactly four times as many V-22s as CH-53Es in a marin expeditionary composite squadron. Which of the following can be the total nbr of V-22s and CH-53Es in the squadron?
Any help?
There are exactly four times as many V-22s as CH-53Es in a marin expeditionary composite squadron. Which of the following can be the total nbr of V-22s and CH-53Es in the squadron?
Any help?
djguernsey
Pro-Rec SNA - OCS Class Date 27 May 2012
I need to know the options available to you. If this is a math problem x+4x = 35 is straight forward (meaning 7 CH-53Es and 28 V-22s) . The other options must not allow X to be whole numbers (as in those without a decimal). If this is a problem about how much you know about the make up of military elements, then I suppose that is just something you got to know.
djguernsey
Pro-Rec SNA - OCS Class Date 27 May 2012
Yeah, this is one of those word problems where if you can recognize that your equation looks like 5x=y, you know that y has to be a multiple of 5 and eliminate all the other options quickly and move on to save you time.
But it looks like the diameter is 10 inches - so if you plug that into the equation for circumference, 2 pieR, you don't get 24. Your answer does seem to make sense, though. The arco book solution simply reads, "the circumference of the wheel is 24 inches. One complete revolution will raise w2 24 inches and lower w1 24 inches, a difference of 48 inches."
Just got word that I will be taking the paper version of the ASTB. Anyone take both by chance (computerized and paper), or just the paper, and know any real nuances or differences between the two (yes, besides the fact that one is on paper and one isn't)? I know the scores are said to be close to the same, but I am looking for more hints I guess than anything.
Alex Wells
New Member
Took the test for the first time this morning.
9/9/9 64
Very excited, it is finally behind me and I can just relax for the rest of my final semester. Thank you to everybody on this site for the study material. I Definitely would not have gotten as high a score without it.
9/9/9 64
Very excited, it is finally behind me and I can just relax for the rest of my final semester. Thank you to everybody on this site for the study material. I Definitely would not have gotten as high a score without it.
Nice work Alex, fantastic scores.
Question that came up from the Marine Mechnical Gouge. I attached the image below, the correct answer A, but I don't understand why.
If Torque=F*D, and A is 2x the distance of B but half the weight (aka force), shouldn't the pull be the same?
Question that came up from the Marine Mechnical Gouge. I attached the image below, the correct answer A, but I don't understand why.
If Torque=F*D, and A is 2x the distance of B but half the weight (aka force), shouldn't the pull be the same?
djguernsey
Pro-Rec SNA - OCS Class Date 27 May 2012
This is one of those that will try and fool you. It's really kind of simple. Weight A weighs 5lbs which means that gravity is pulling down the weight with 5lbs of force. That force is what is being exerted on the bar, because it is resisting that 5lbs of force.
Weight B is 10lbs, so 10lbs of force is being exerted on the bar. Distance in this case doesn't matter. Distance would matter if a fulcrum was involved. There is none.
I hope that helps. If anyone has a more technical answer feel free to offer it.
Weight B is 10lbs, so 10lbs of force is being exerted on the bar. Distance in this case doesn't matter. Distance would matter if a fulcrum was involved. There is none.
I hope that helps. If anyone has a more technical answer feel free to offer it.
Sounds like I was trying to apply the wrong principle. That make's sense since gravity is more or less constant regardless of height. Thanks for the explanation.