<p>Freee Response dicussion: commence...</p>
<p>Starters: What direction was the B-field in FRQ #2?</p>
<p>into the page...3rd right hand rule</p>
<p>Good. And for the speed, it stays the same because the particle is accelerated centripitally by the B-field, so the direction changes, but the magnitude of speed stays constant. And for three you use qvB = mv^2/r right? And for part 4, you use K = qV, right?</p>
<p>that all sounds good to me.</p>
<p>well... maybe i got two parts correct :)</p>
<p>Cool. So how about #6? Just pick a point on the LofBF and use those values? So F is about .29? And how do you do the thing with the tree?</p>
<p>The rays of light coming from the "distant" tree a relatively parallel. So you can treat them as parallel. Where they converge is the focal point.</p>
<p>Oh. I said put the lens at the window, and find where the image appears, then approximate the distance from the building to the tree and the distance from the lens to the image, then use the lensmaker's equation to find the focal length. </p>
<p>For the water, I got the height was like .288. And the water will hit to the left after the height is reduced.</p>
<p>Are the answers to the questions posted anywhere?</p>
<p>I'm not sure, but I wish they were.</p>
<p>Would the rays coming from the tree be parallel to the normal of the lens?</p>
<p>for the last one you had to use e=mc^2 and then the momentum of the two particles was 0</p>
<p>How were you supposed to find the velocity of the fluid</p>
<p>The rays would be relatively parallel. Think about it. If you would use the lensmaker's formula, youll have 1/di + 1/do = 1/f . However, because the tree is so "distant" and we can probably assume the lens has a focal length of less than half a meter... 1/do (distance from object) will be so small that it will practically be zero because D'o is so big.</p>
<p>That makes sense. I just didnt think of it. Any thoughts on the first question, with the plane? It slows down uniformally, right? And we didnt need anyt numbers on tyhe graph, did we?</p>
<p>to find velocity of the fluid, i found volume flow rate for a unit of "m^3/s", then equaled that number to Av (cross-sectional area times velocity). not sure if that's right</p>
<p>^^^ correct.</p>
<p>on number 6, i got around .3 for the focal length. anyone else get that? i also calculated about .29 for the height of the container.</p>
<p>darkruler, yup on the first one, velocity should be flat while on the incline and uniformly decrease once on the flat portion. No we don't need any numbers.</p>
<p>Crap, what was FRQ #6? I don't remember doing anything with trees... well there goes my chance at a 5!</p>