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- May 25, 2007
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Feel free to post any general questions about the course and/or the materials.
Hello,
I have signed up for the March 27th MCAT and I have been following SN2's plan and about two weeks into it, however I am noticing that when I get to doing the passages I am missing several more than I would like to be missing. I was wondering if this is typical and if you had any tips on how to work the passages? Do you spend under 45seconds reading the questions and then use P.O.E? Please let me know. Thank you.
why does it take so long to get your books after you mailed the money in?
Is there any discount for the actual class? special promotion or student discount?! This is a silly question since most applicants are also students, but it doesn't hurt to ask!
What is the average time one should spend reading a passage and doing the questions for the passages found in either the chem or physics book?
How is Berkley Prep different from Kaplan? Does the program also come with computerized tests?
If someone signs up for your course do they also get the books or does one have to pay for the course and buy the books?
I'm planning to take Berkeley Review's MCAT course this summer. Fortunately, I have the option to choose between Berkeley, Stanford, Irvine and LA locations. Which location would Todd and Dale be at the most? Which one would you recommend? Thanks!
I'm doing Circuits now and had a question:
In a RC circuit, what is the relationship between the capacitor and the resistor?
And on page 180 in the Physics 2 book (same chapter), eq. 9-35:
I =(V/R) - (q/RC)
what is the q/RC term mean in relation to the example/figure 9-13?
And also, what do we have to understand about eq's 9-36 and 9-37?
Thanks! This part of physics confuses me!
I have been racking my brain over example 6.4a in Physics Part II: Sound, pg 18.
The solution states: ..."which means n and L are linearly related, and more importantly that a bigger sound speed v results in a smaller slope".
I don't understand the latter part of the explanation. I tried manipulating the formula to express it in terms of a straight line, y = mx + b, but I came up empty. So why exactly is the speed of sound lower for pipe A than for pipe B?
ps. I just ordered the TBR exams over USPS. I hope you have the time to help me out there as well! Thanks again!
I'm trying to find the General Chemistry/Physics TBR books, but I cannot get into contact with the Berkeley Staff for shipping charges (I'm from Canada). Has anyone ordered any of the BR books, had them shipped to Canada, and know the actual price of shipping? Or at least had any success in contact the company?
How fast are orders shipping right now?
Hello. Hoping someone might be able to help...
I understand that launching a projectile at 45 degrees maximizes the range of the projectile IF the starting and ending heights are the same. I'm having a little trouble understanding how this changes, however, when the starting and ending heights are different like when you launch a projectile off a cliff.
Particularly, The Berkeley Review (TBR) Physics Part I Chapter 1 (Translational Motion), Practice Passage IV, Question 27 (page 41). I read the explanation (page 55) which says "The best compromise of maximizing the angle to maximize the flight time, while simultaneously minimizing the angle to maximize the x-direction velocity, is achieved at a value of slightly less than 45 degrees." I get why a bigger angle increases flight time and a smaller angle increases x velocity (cos(theta)), but I don't get how we know that an angle less than 45 degrees is the right compromise between these competing factors.
Also, in the problem (page 41), each graph for answers A-D shows the range being the same for 0 degrees and 90 degrees. How is that possible? Isn't zero degrees straight ahead which would definitely have a greater range than 90 degrees which is shooting straight up in the air so the projectile would come right back down???
THANKS!!
It might be easier to think about what the situation allows different than when the launch point and landing point are at the same height. In this case, because you are landing at a lower elevation than you start, the object will be in the air longer than the standard scenario. This means that flight time is increased. If you lower the launch angle slightly, you can convert some of the extra flight time into extra speed, which leads to a balance between flight time and speed in the x-direction.
Another way to think about this involves your instincts when you throw an object onto a roof. You instinctively make the launch angle greater than 45 degrees, because with an elevated landing point, the flight time has been reduced. A greater launch angle will give back some of the lost flight time. Throwing off of a cliff is the opposite situation.
You are right. In the graphs in my picture it looks like the difference in range between 0 and 90 is minimal (I had to use a second piece of paper to see the height difference.