Physics Question Pack Doubts

[kryptonxenon]

1) Why are the percentages of the change in frequency and wavelength much greater when sound waves are used instead of radio waves in these experiments?
A) Sound waves travel more slowly.
D) Interference in the atmosphere affects sound wave much more.

What makes D incorrect?

2) How will work change if the angle of a ramp to the horizontal is increased? (Question 82 from the question pack)
Here is my reasoning: If angle is increased that means the object is starting at a higher position. Since it is starting at a higher position it has more potential energy and thus gravity has to do more work on the object. However, the answer says that the Work will decrease.

3) The figure above shows an arrow, pointing upward, located a distance p in front of a converging lens of focal length f. If p > f as in the figure, what kind of image of the arrow is formed by the lens? (Question 118 from the question pack)
I made up numbers to figure this out and got the wrong answer. I don't understand the AAMC explanation.

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[theonlytycrane]

1) Why are the percentages of the change in frequency and wavelength much greater when sound waves are used instead of radio waves in these experiments?
A) Sound waves travel more slowly.
D) Interference in the atmosphere affects sound wave much more.

What makes D incorrect?

Which question is this?

2) How will work change if the angle of a ramp to the horizontal is increased? (Question 82 from the question pack)
Here is my reasoning: If angle is increased that means the object is starting at a higher position. Since it is starting at a higher position it has more potential energy and thus gravity has to do more work on the object. However, the answer says that the Work will decrease.

W is the work done by friction as it slides down the ramp, not work done by gravity. The object still slides down the same distance, but the frictional force decreases with a smaller angle. The force of kinetic friction in this case doing the work = coefficient * mg * cos(theta).

3) The figure above shows an arrow, pointing upward, located a distance p in front of a converging lens of focal length f. If p > f as in the figure, what kind of image of the arrow is formed by the lens? (Question 118 from the question pack)
I made up numbers to figure this out and got the wrong answer. I don't understand the AAMC explanation.

Go with the two equations 1 / i + 1 / o = 1 / f and m = -i / o. i is the image distance (what we usually solve for), o is the object distance (p in this case) and f is the focal length. Since the lens is converging, f gets a positive sign. If the lens was diverging, f would get a negative sign.

At this point, plugging in numbers could work if that's easier to use than variables. We just have to make sure that p > f. So lets say p = 4 and f = 2. 1 / i + 1 / 4 = 1 / 2. Solving this gives i = 4. Since i is a positive value, we know the image is real (on the right side of the lens).

Finally, we use the magnification equation m = -i / o. m = -4 / 4 = -1. This tells us the image isn't bigger or smaller (since the value is ~1), but that it's upside down.

The AAMC explanation is talking about using ray diagrams, but that's more confusing in my opinion.

 

[kryptonxenon]

W is the work done by friction as it slides down the ramp, not work done by gravity. The object still slides down the same distance, but the frictional force decreases with a smaller angle. The force of kinetic friction in this case doing the work = coefficient * mg * cos(theta).

Just to confirm you meant that as the angle increases frictional force decreases, right?

Which question is this?

I can't seem to find this passage, but even though the question stem refers to the experiments, I recall that the answer explanation was purely based on knowledge of the Doppler effect. I guess my question is are sound waves and radio waves affected the same amount by interference in the atmosphere?

 

[theonlytycrane]

Just to confirm you meant that as the angle increases frictional force decreases, right?

Yes! Good catch.

I can't seem to find this passage, but even though the question stem refers to the experiments, I recall that the answer explanation was purely based on knowledge of the Doppler effect. I guess my question is are sound waves and radio waves affected the same amount by interference in the atmosphere?

I think I would have to see the passage / q-stem for this one to give a helpful answer.

 

[kryptonxenon]

Just found it! It's question 18 on the Physics Question Pack.

I also have a question on #48 in the question pack. In my MCAT class, my instructor said that in such a case as outlined in #48 the detected frequency is always increasing. So doesn't that mean the frequency shift is also increasing?

 

[theonlytycrane]

For #18, the observed wavelength is slightly shorter than the measured wavelength in the lab. We know that v = lambda * f. If lambda (observed) is a bit smaller, f must be a bit larger. f is a bit larger if the object is coming toward us (doppler shift). f would be a bit smaller if the object was going away from us (choice A).

#48 also deals with a doppler shift. As the object (speaker) moves away the detected frequency is slightly below the actual. Then the object starts coming back down to the ground and the detected frequency increases above the actual. So the shift depends on if the object is going away or toward the detector.

 

[kryptonxenon]

For #18, the observed wavelength is slightly shorter than the measured wavelength in the lab. We know that v = lambda * f. If lambda (observed) is a bit smaller, f must be a bit larger. f is a bit larger if the object is coming toward us (doppler shift). f would be a bit smaller if the object was going away from us (choice A).

Shoot!! My bad, I meant #16 on this one...whoops :/

 

[misterwiggles]

for #1) in relation to radio waves which travel at the speed of light, sound waves travel much slower (340 m/s). Thus changes in frequency and wavelength occur at a greater percentages to speed. Think of it this way: which is a bigger change 1/340 or 1/(3^10) ?