<p>Good luck!</p>
<p>What are you guys getting on the MC section?</p>
<p>Here’s a question:</p>
<p>A wire in the plane of the page carries a current
I directed toward the top of the page. If the wire is located in a uniform
magnetic field B directed out of the page, the
force on the wire resulting from the magnetic
field is …</p>
<p>I thought it was to the left, but the answer is to the right.
Pointed my thumb up to the current, palm outwards for the magnetic field. My fingers pointed to the left.</p>
<p>Your fingers are supposed to be the magnetic field and your palm should be the force</p>
<p>Odd, the way I did it used to work for other questions. I guess I’ll do it ^ way from now on.</p>
<p>Next:
Two concentric circular loops of radii b and 2b, made of the same type of wire, lie in the place of the page. The total resistance of the wire loop of radius b is R. What is the resistance of the wire loop of radius 2b?</p>
<p>Doesn’t resistance decrease with a greater area? By resistance = pl/A. So I thought that if the radius was double, the area would be quadruple and thus the resistance would be quartered. Apparently the answer is that it’s double.</p>
<p>could be a typo
whered u get the question from?</p>
<p>Two identical conducting spheres are charged to +2Q and Q. respectively, and are separated by a distance d (much greater than the radii of the spheres) as shown above. The magnitude of the force of attraction on the left sphere is F1 After the two spheres are made to touch and then are reseparated by distance d the magnitude of the force on the left sphere is F2. Which of the following relationships is correct?</p>
<p>A) 2 F1 = F2 (B) F1 = F2 (C) F1 = 2F2 (D) F1 = 4F2 (E) F1 = 8F2</p>
<p>^ B, even though they touch the charges and distance between are stilll the same.</p>
<p>Abrayo-</p>
<p>The equation R= ((rho)L)/A </p>
<p>A= the crossectional area of the wire. the wire is the same thickness. So the only thing that changes is L, the length of the chord.
Inner circle circumference= 2b(pi)
Outer circle circumference= 4b(pi)</p>
<p>So the resistance of the outside circle is 2R</p>
<p>Oh, I misunderstood the circular wire to be a solid circle of some sort. Thanks!</p>
<p>Anyone know about mirror rules?
Like “If an object is in front of a concave mirror, less than the distance of the focal length and then it moves further away, what happens to the image?”
That was poorly worded because I just made it up.
Should it be something like first it’s virtual and upright, and then it becomes real & inverted while also enlarging?</p>
<p>And what do polarizing sheets do? In the Official Book there was a question on them . . . do we need to know about these?</p>
<p>MitDuke your way wrong, the way you read/approached the problem is wrong too…
read it again lol
also answer is E</p>
<p>^ would you mind explaining? I don’t understand.</p>
<p>well, i dont noe how to get the answer but i have some sort of idea…
basically when the charges touch, the charge is evenly distributed over the combined piece of charge which will have magnitude +2Q - Q = +Q.
Now because the net charge is +Q, when you divide the combined piece into two pieces again you get +.5Q and +.5Q on each piece. The resultant force between them is then
K(.5Q)(.5Q)/d^2 … which means its (1/4) of the original k(Q)(Q)/d^2 force.
However answer is not D either but E
and idk why lolol</p>
<p>^ noo idea.</p>
<p>“A child on a swing can greatly increase the amplitude of the swing’s motion by “pumping” at the natural frequency of the swing. This is an example of which of the following?”
A) Conservation of momentum
B) Newton’s First Law
C) Newton’s Second Law
D) Resonance
E) Interference</p>
<p>It’s D. I thought it was a, because the child gains momentum by swinging her arms?
What does Resonance, in this situation, even mean?</p>
<p>Anything with “natural” frequency or wavelength will be resonance. It means it will fully destruct and construct at particular points. So the little kid will be as high as possible at one and as low as possible at another.</p>
<p>@ salzahrah sorry I misread the question as two charges of +Q, however your method only works if the second charge is -Q but the question says a a charge of +2Q and Q, if both charges are positive then the force would be equal to K(1.5Q)^2/(d^2) and F2 would be equal to (9/8)F1</p>
<p>On the part 2 (free response), do you have to use the variables given on the reference sheet or can you use different letters. For example:
Can you use “PE” instead of “U” for potential energy? or
Can you use “d” instead of “x” in the kinematic equations?</p>
<p>Edit: Also, do you have to manipulate the equation to get the only the variable you are solving for on one side or can you simply substitute into the equation. For example
If solving for T using PV=nRT do you have to manipulate the letters to T=PV/nR and substitute into this or can you simply substitute into the original PV=nRT.</p>
<p>@ salzahrah</p>
<p>I took that same practice test. I have no freaking idea how to do it.</p>
<p>@ Abrayo</p>
<p>I tend to think of resonance as a sort of “tuning in” with the frequency. Resonance increases frequency when attained. I suppose that’s all you needed to know for that specific question, but you’re better off taking a brief look at what the interwebs have to say in order to be sure you have it down.</p>
<p>What are all of you bringing to the test?</p>
<p>Calc
pencils
pen
ruler (?)</p>
<p>should we use pens on the FRQ?</p>