<p>Oh **** my bad, yeah I see it now.
The flux is the change in the magnetic field within the coil. Because more (arrows) of the field are being added that point upward, a current will be generated that tries to counteract that. So therefore the current generated will try to produce a downward magnetic field, using the right hand rule its clockwise.</p>
<p>I understand everything except one thing you said.
Why are there more field lines upwards if you bring a bar magnet, south pole first, close to a loop of wire? How do you know there can’t be more field lines downwards?</p>
<p>Well,
Consider a bar magnet like this
Lines come out of the north pole and into the south pule. Therefore if the magnet is lowered into a coil that is vertical more magnetic field lines that are pointing up into the south pole are being put into the coil. I don’t know if that makes sense.</p>
<p>Oh I got it. The field lines actually are the field lines that go through the magnet from S to N, which point upwards. So B increases upwards. </p>
<p>So B needs to decrease, so a clockwise current is induced. Thanks.</p>
<p>Magnetic field lines exit out the north and enter in the south. Not sure if thats what you meant to say, just want to clear that out.</p>
<p>Wow, you guys really helped! Thanks for all the explanations! Keep the practice questions coming please.</p>
<p>For the question above, dealing with countering the upward magnetic field lines, you guys are seeing (using the right hand rule) that clockwise is the correct choice; however, using the right hand rule as I think I remember it, I am getting the opposite answer. How does one exactly position his hand to execute the right hand rule correctly…?</p>
<p>@ Realitybugll When magnetic fields induce currents, the currents always try to induce a magnetic fields that induced its current in the first place, so if the magnetic field is downward the magnetic field induced by the current would be upward and vice versa.</p>
<p>I still don’t get that questiom… If the magnet’s south pole has field lines pointing upward, then the change in the magnetic field would be upward. Using the right hand rule, I point my fingers downards and wrapped them around the coil… and got counte-clockwise. What am I doing wrong?</p>
<p>Scourpower, if your really using your right hand, and you wrap them around the coil so that your fingers are facing down when INSIDE the coil, your thumb should be pointing clockwise.</p>
<p>seeing as em, mech, thermo have all been posted
here’s an optics question</p>
<p>an object placed 60cm in front of a spherical mirror forms a real image at a distance of 30cm from the mirror
1.Is the mirror concave or convex?
2.What’s the focal length?
3.Is the image shorter or taller?</p>
<ol>
<li>Notice: Real Image, that tells us that it is convex.</li>
<li>We can find this 1/60+1/30=1/20, 20 cm is focal length</li>
<li>Magnification is -30/60 so it is inverted and half the height.</li>
</ol>
<p>alright, someone explain this one please. It is number 15 in the collegeboard book if you have it. Assume that every projecctile fired by the toy cannon shown above experiences a constant net force F along the entire length of the barrel. If a projectile of mass m leaves the barrel of the cannon with a speed v, at what speed will a projectile of mass 2m leave the barrel?</p>
<p>A. v/2
B. v / (sqrt 2)
C. v
d. 2v
E. 4v</p>
<p>The answer is B, but I put A, and I cannot figure out why.</p>
<p>^^ I think you mean concave? For concave mirrors, real images are on the same side as the object. (For lenses, it’s the opposite; bi-convex/converging lenses create real images on the OPPOSITE side of the object.)</p>
<p>Correct me if I’m wrong.</p>
<p>The kinetic energy of a projectile with mass m is 1/2mv^2. So, if you have a projectile with mass 2m, the kinetic energy would have to be 1/2(2m)(v/sqrt 2)^2 because (after squaring the velocity and canceling out the 2’s) you will be left with 1/2mv^2.</p>
<p>I’m not sure if that explains it properly, but it gets you the right answer.</p>
<p>@ Miamirulz29: start with F=ma. since it is 2m, you know have f=2ma. thus, a=f/2m (acceleration is half as much). Plug this into a kinematic equation. originally: Vf^2=Vi^2 +2a(delta x).
Vi is zero, so Vf = sq. rt. of 2a delta x. Now plug in half acceleration: Vf^2=Vi^2 +2(1a/2) (delta x)[vi is still zero] so Vf^2 = a(deltax), thus Vf = sq rt of a(delta x). compare this with the first velocity: Vf(2) = (Vf(1)/sq rt 2)</p>
<p>can anyone explain why II is true? (answer is D)
An experimenter adds thermal energy to a block of some solid substance, but the temperature of the substance does not change. Which of the following explains this effect?</p>
<p>I. The thermal energy is used in a phase transformation.
II. The thermal energy increases the entropy of the substance.
III. The thermal energy is lost because heat engines can never be perfectly efficient.</p>
<p>(A) I only<br>
(B) II only<br>
(C) III only<br>
(D) I and II only<br>
(E) I and III only</p>
<p>^ I just took that practice test and I was confused as well. I’m pretty sure Sparknotes made a typo because in the explanation, it only gave reasoning for why I is correct.</p>
<p>@Mandu my bad yeah I meant concave mispoke.</p>
<p>@Miamirulez
Just use the kinematic equation v^2=2ax
The cannonball is accelerated over the same distance, it just has half the acceleration. So for v^2 to account for the halved acceleration it has to be lowered by a factor of sqrt(2).
@Born4soccer
You have to know that entropy increases whenever heat is added to a system phase change or not.</p>
<p>I have another question too. It is number 34. Two blocks of identical mass are connected by a light string as shown above. The surface is frictionless and the pulley is massless and frictionless The acceleration of the two-block system is most nearly</p>
<p>A. 20 m/s^2
B. 15 m/s^2
C. 10 m/s^2
D. 5 m/s^2
E. 2.5 m/s^2</p>
<p>The answer is D. Whatever I do, I seem to keep getting C.</p>