EK 1001 PHYSICS Questions/Answers Thread

[Bernoull]

I started this Thread because a lot of MCATers use the EK 1001 series and frequently ask questions, however the questions are all over the place as opposed to one specific thread. There's something similar for EK VR 101 and it's great..
I think a One-Stop-Shop for those using EK 1001 PHYSICS will make asking/answering questions more user-friendly and reduce the number of redundant questions.

It will be great if other EK 1001 Physics threads could be migrated over here.

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[Verse]

I am having trouble with number 737 in physics 1001:
How far from the vertical will the pendulum swing to the left while the boxcar accelerates down the incline?

I know it's hard without diagrams, but could someone please try to explain it to me? I understand that the restoring force is reduced to mgcosBeta and so the angle distended will be more than theta, but how do you find out by how much it increases (i.e., beta)?

 

[Verse]

Another question, on #978. The problem is about a two-lens system. The object is to the left of the left lens. It asks, "If the image created by Lens 1 is formed to the right of Lens 2, the final image must be:

a) virtual upright
b) virtual inverted
c) real upright
d) real inverted

I am confused...I know the object distance for the second lens is negative, and the first lens must have produced a real, inverted image because it is on the opposite side from the object. If, for the second lens, f is positive and do is negative, di must be positive, and a positive di indicates a real, inverted image. Since it was already inverted, wouldn't that lead to a real, upright image?

Or is f negative because it's on the opposite site to the initial object? But in that case, wouldn't the final image depend on whether the second object is within the focal distance of the second lens or not?

 

[G1SG2]

I am confused...I know the object distance for the second lens is negative, and the first lens must have produced a real, inverted image because it is on the opposite side from the object. If, for the second lens, f is positive and do is negative, di must be positive, and a positive di indicates a real, inverted image. Since it was already inverted, wouldn't that lead to a real, upright image?

You mean a real, inverted image, right? So yeah, if do is negative, and f is positive, then from 1/f=1/do+1/di, we get 1/f=-1/do+1/di, then 1/f-(-1/do)=1/di), then 1/di=1/f+1/do. Seems to give us a positive di, which would mean a real and inverted image. Anyone else want to take a stab at it?