# 7r confusions

Discussion in 'MCAT: Medical College Admissions Test' started by Omyss, Aug 10, 2006.

1. ### Omyss Member 2+ Year Member

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i'm having some problems with Q's on AAMC 7... was wondering if some of guys could help me.

15. Is this a typo? the answer is a but clearly disobeys hunds rule?

63. i was thinking that since the frictional forces are less for the hard rubber, more tire air pressure would be required since this would icnrease the force on the ground to compensate for the frictional losses... thus i put my answer as b: the fuel efficieny, i thought would change because more energy would be needed to make up for lack of frictional forces, and more fuel would be used

71. the anser was c i put d... and the answers don;t give a good explanation at all... can some clear this is up for me?

thanks alot

2. ### vicinihil Member Moderator Emeritus 10+ Year Member

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Having same problems...HELP

3. ### 87138 Guest

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I don't think the fact that A is the answer is a typo, as much as the actual diagram given in A is done slightly incorrectly. The pictures in the answer choices are a bit misleading, but you still have to remember the general rule for this situation. First of all, you can immediately eliminate answer D since it has too many electrons. Then you're left with A, B, and C. You know that each electron has to be added parallel to each other before "doubling them up." You also know that the 3p orbital will contain 6 electrons, which means the first three would theoretically be pointing the same direction. You only have two, although they still need to be pointing in the same direction.

But, I do admit I didn't initially notice the dashed lines beneath the arrows when I first took the test. I do believe those are incorrect. Each arrow should be on its own line.

Still, the things I outlined above were what screamed out at me that A was the correct answer, based on the other two possible answers having the first two 3p electrons anti-parallel. I assume the lines beneath them are incorrectly placed.

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4. ### WilliamsF1 2+ Year Member

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I didn't like the answer for the tire question. Harder tires wear less than soft tires because they have less friction. They also get better gas mileage because there's less friction between it and the road. Their logic for the air pressure was lacking.

5. ### surag kobayashi 7+ Year Member

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I TOTALLY F*KING AGREE. I had that same reasoning for mileage. I also thought it was off topic compared to the rest. so i figured, thats the right answer....because of that, im leaving 1-3 questions as probably getting wrong on the actual test because AAMC people's minds are from planet x.

6. ### 87138 Guest

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Wouldn't greater friction increase gas mileage? It's static friction (not kinetic) that determines the car's ability to "grip" the road. With lower static friction, there's essentially a chance for more "slipping" and less efficient use of the car's rotating axel.

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7. ### phorensic SDN Lifetime Donor Lifetime DonorVerified Account 10+ Year Member

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no that doesn't make sense at all. if all the cars on the road were slipping half the time, then i guess you would be right.

8. ### 87138 Guest

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I'm not saying they go slip-sliding all over the place. I'm talking at a very miniscule level (probably smaller than is really practical).

I don't see how less friction would help gas mileage. Cars don't skid to get where they need to go. They "grip" or "paw" the ground (for lack of a good term). A better "grip" would lead to a more efficient utilization of resources.

At least that's how I'm picturing it.

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9. ### phorensic SDN Lifetime Donor Lifetime DonorVerified Account 10+ Year Member

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i know, but after a certain point, more grip = less efficiency. all tires will give you enough grip so that you don't go slipping all over the place. but putting any more grip than that will slow you down. picture it at extremes...what if you just put some of the stickiest substances all over your tires in order to grip the road better...your car would have to work much harder than usual to overcome the unusual "grippiness" of your tires.

10. ### WilliamsF1 2+ Year Member

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Friction also helps to slow you down, not just speed up. If there's say 100% friction (a lot of normal force), you won't go anywhere and your mpg is 0 since you're giving the car throttle, but it won't move. If friction is 0, then you won't go anywhere either since your wheels freely spin and your mpg is also 0. So there are different levels on friction and how it plays into mpg.

If you have very soft (sticky) tires, the road will be working to slow you down more. If you get off the throttle, your car will slow down quicker by itself without brakes because there's a lot of friction. For you to overcome this added resistive force to maintain the same same speed, you have to give the car more throttle (more gas) to overcome the extra friction. If you have a harder (less sticky) tire, there's not as much friction pushing the car to slow down so you don't have to give the car as much gas to overcome the amount of friction.

11. ### estairella Senior Member 10+ Year Member

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15 is definitely a bad image; it was intended for there to be 3 dashes, if you thought they were together, you would have picked b) (the least energetic) because a) would be an impossible configuration (two positive electron spins)

63 man I had no idea... I assumed that hard rubber requires more pressure to pump it up as opposed to soft rubber (like blowing up a balloon versus blowing up a basketball), so yea, I picked the wrong-but-common answer too

71 okay this one I can explain. It says the rupture is going south to north, key here is to realize that the SOURCE is the rupture. if you go about thinking that the rupture is a wave (what I thought at first), then the question itself makes no sense (how can a wave propagate more waves?). So you have a source propagating waves going south to north, the waves going northward are gonna be compressed (frequency goes up) and those going southward are gonna be lengthened (frequency goes down).

12. ### Knickerbocker 2+ Year Member

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15. The PDF you have is screwed up

63. A, B, and C are directly related to the friction between the car and the road. I chose D because it required more justification than A, B, and C. I also know that there are airless tires out there, but that's probably not what they had in mind.

71. This problem was difficult to understand, IMO. Think about the center of the earthquake as if it were a train blowing a whistle while moving to the north. To the north, waves will have a higher frequency and thus a shorter wavelength. Opposite for waves in the southern direction.

13. ### 87138 Guest

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Ok I've got a question actually. Question number 40, talking about circuits.

Now, in the picture, the way the battery is set up, CURRENT will flow clockwise when the switch is to the left, correct? (positive side of the battery is on the "top"). Of course I know that current is the opposite direction as electron flow . . .

Anyway, technically (assuming zero resistance wires) the current encounters no resistance before reaching the capacitor, right? I'm not 100% clear on capacitors in circuits.

The answer to the question is B, the electrical potential energy that the capactor gains as it's charged is less than the work done by the battery throughout the charging process. The explanation is that the current encounters resistance and some heat is dissipated. But again, it looks like the current reaches the capitor first. Which then begs the question (and I'm embarassed for not knowing this), does a current actually complete a full loop in a circuit when a capacitor is present? Do you have to consider that since electrons are flowing counter-clockwise they are going through the resistor?

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