AAMC FL1 C/P #9, 45, 56
AAMC FL1 C/P #9, 45, 56
#9 Why is this answer the S configuration and not D?
#45 I have no idea how to solve this problem, please help.
#56 I = V/R, plug the numbers in to get the answer, however, what does it mean in the question stem when it says "assume the resistance in the circuit is negligible). I thought that meant to ignore the resistance, so I answered C, setting V = 2 and R=1
Re: AAMC FL1 C/P #9, 45, 56
Hi there!
For #9: remember for SN2 reactions that configuration is always inverted. It is originally configured R with the OH which will be displaced by the Br, turning it into S.
#45: This question requires us to use the thin lens formula and magnification.
Starting with what's given to us, we can see that the object distance is 3 times the focal length, or in other words o=3f.
The thin lens formula is 1/f = 1/o + 1/i. From here we can plug in our numbers, move things around and find out what i is in terms of f.
1/f = 1/3f + 1/i
3/3f -1/3f = 1/i
2/3f = 1/i
i = 3/2f
Now let's go back to question stem, it asks for the ratio of the height of the image to the height of the object, this is essentially magnification. i is 3/2f, o is 1/3f, so our ratio is 1/2.
#56: Wording here is important as always! It tells you to assume the resistance in the circuit is negligible in comparison to the lightbulb. V=IR; I = V/R and then 2.0V/0.5 Ohms = 4.0 A.
Hope this helps!
Raven
For #9: remember for SN2 reactions that configuration is always inverted. It is originally configured R with the OH which will be displaced by the Br, turning it into S.
#45: This question requires us to use the thin lens formula and magnification.
Starting with what's given to us, we can see that the object distance is 3 times the focal length, or in other words o=3f.
The thin lens formula is 1/f = 1/o + 1/i. From here we can plug in our numbers, move things around and find out what i is in terms of f.
1/f = 1/3f + 1/i
3/3f -1/3f = 1/i
2/3f = 1/i
i = 3/2f
Now let's go back to question stem, it asks for the ratio of the height of the image to the height of the object, this is essentially magnification. i is 3/2f, o is 1/3f, so our ratio is 1/2.
#56: Wording here is important as always! It tells you to assume the resistance in the circuit is negligible in comparison to the lightbulb. V=IR; I = V/R and then 2.0V/0.5 Ohms = 4.0 A.
Hope this helps!
Raven
Re: AAMC FL1 C/P #9, 45, 56
thank you!