The problem -

http://i244.photobucket.com/albums/gg16 ... j/prob.jpg

3) I thought answer was 4 because of mg = weight.

17) i thought 1 A/2 * 2L = > R

18) i thought 3) because need parallel circuit to make effective I smaller.

19) not sure how to do

32) not sure

33) I believed it was 2 but it might be 1 because of way + or - are defined

I thought it was C.

## Quick Questions about Regents exam for preparing for PGRE

### Re: Quick Questions about Regents exam for preparing for PGRE

3) the answer is 2 because the question asks for the object MASS, not its weight. It's weight depends on gravity according to W = mg, but it's mass is constant, even if it is floating in space where gravity is zero.kebel23 wrote:The problem -

http://i244.photobucket.com/albums/gg16 ... j/prob.jpg

3) I thought answer was 4 because of mg = weight.

17) i thought 1 A/2 * 2L = > R

18) i thought 3) because need parallel circuit to make effective I smaller.

19) not sure how to do

32) not sure

33) I believed it was 2 but it might be 1 because of way + or - are defined

I thought it was C.

17) R is proportional to L/A so if you double the length and half the area your new resistance is four times greater. (4).

Think about it. Resistance measures how tough it is for an electron to travel through a metal. If the metal is longer, it is more difficult to travel through, so INCREASING the wire's length should INCREASE the resistance. If the wire is thicker, the electron has more paths to take through the metal--or think of water flowing through a pipe. It is easier in a bigger pipe, right? So increasing the area DECREASES the resistance, and decreasing the area INCREASES the resistance. 2 divided by 1/2 is 4.

Sorry I'm getting bored. Someone else might help you with the others.

### Re: Quick Questions about Regents exam for preparing for PGRE

OK well now I'm at work and I've nothing better to do so:

18) why would you need to make the current smaller when choice 4 would have a current of 0.25A with the switch closed? The best choice is 4.

19) Since they are in parallel the voltage across each lamp is the same-120V. P=IV, so I = (10/12)A for each lamp. There are six of them, so I(total) = 60/12 = 5 Amps.

32) Charge comes in discrete units of 1.602E-19, which is the charge of a proton or electron. You can't have half a proton's worth of charge. You can only have a unit number of elementary charge, so your total net charge must be evenly divisible by 1.602E-19. Only choice 2 works.

33) Only A will have a net gain of electrons. Sphere A has more positive charge. When they touch, sphere A will become less positively charged until the two have equal net charge. In this case, the electrons are moving so sphere A becomes less positively charged by taking some of sphere B's electrons.

18) why would you need to make the current smaller when choice 4 would have a current of 0.25A with the switch closed? The best choice is 4.

19) Since they are in parallel the voltage across each lamp is the same-120V. P=IV, so I = (10/12)A for each lamp. There are six of them, so I(total) = 60/12 = 5 Amps.

32) Charge comes in discrete units of 1.602E-19, which is the charge of a proton or electron. You can't have half a proton's worth of charge. You can only have a unit number of elementary charge, so your total net charge must be evenly divisible by 1.602E-19. Only choice 2 works.

33) Only A will have a net gain of electrons. Sphere A has more positive charge. When they touch, sphere A will become less positively charged until the two have equal net charge. In this case, the electrons are moving so sphere A becomes less positively charged by taking some of sphere B's electrons.

### Re: Quick Questions about Regents exam for preparing for PGRE

are you really preparing for the physics GRE? Isn't this a high school test?

### Re: Quick Questions about Regents exam for preparing for PGRE

oh, ok. cool. what year are you planning on taking the exam?