<p>I think everyone is worried about Physics B. Everyone on CC seems to be feeling worried and I know for a fact that everyone in my physics class is expecting 2s. Oh boy!</p>
<p>this question may have been answered b4, but wat % do u have get correct on each MC and FR, inorder 2 get a 5?</p>
<p>Over 75% correct is about a 5.</p>
<p>I’m literally expecting a 1 or a 2.</p>
<p>I don’t understand this question: </p>
<p>Planet X does not have an atmosphere or any air resistance. Say an object is released from rest, and it falls 3 meters in 1 second. What is the acceleration due to gravity on planet X?</p>
<p>It says the answer is 6 m/s^2. How is that possible?</p>
<p>From what I’ve read 65% of everything gives you a good shot at a 5.</p>
<p>Question: Can anyone explain the equations like f = nv/2L as it applies to waves? This has never really clicked for me.</p>
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<p>This is easier than it sounds. Just use the old standby, y(f) = y(i) + y(vi)t + 1/2(ay)t^2.</p>
<p>3 = 0 + 0 + 0.5a(1)
3 = 0.5a
a = 6m/s^2</p>
<p>(Answer omitted because it is WRONG.)</p>
<p>^Final position after 1s =/= final velocity after 1s though</p>
<p>Yes, that’s what I got. I did v = vo + at, and solved from there. The answer key says it’s 6 m/s^2.</p>
<p>No,</p>
<p>x=.5at^2
6 m/s/s=a</p>
<p>And the curve is about 64% normally…the PR book says 75%, but that’s just to encourage you to work even harder and make it seem like their book actually helped you…pretty shady imo =P</p>
<p>You aren’t given the final velocity…so you can’t solve it that way</p>
<p>Ohh, right. Thanks for clarifying. I feel silly lol.</p>
<p>Electrons that have been accelerated from rest through a potential difference of 150 V have a de Broglie wavelength of approximately 1 x 10-10 m. In order to obtain electrons whose de Broglie wavelength is 5 X 10-11 m, what accelerating potential is required? </p>
<p>& plz show work…ty!</p>
<p>Can someone clarify waves in open/closed pipes for me first >_>…what f = nv/2L means?</p>
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<p>Don’t be fooled by the complex language; just look at the bare-bones components, and you’ll note that it’s a straight line motion problem.</p>
<p>The formula we need is S = Vit + 1/2at^2.</p>
<p>Since Vi is equal to zero, the equation simplifies to:</p>
<p>S = 1/2at^2</p>
<p>Plug in:</p>
<p>3 = 1/2a(1^2)</p>
<p>Simplify:</p>
<p>3 = 1/2a</p>
<p>Solve:</p>
<p>a = 6 m/s^2</p>
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<p>600 V? not sure about this one…</p>
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<p>That’s correct. Can you post your work?</p>
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<p>First time I’m encountering this so bare with me…dBw = h/p, and p = mv. Thus h/dBw = mv = 1.325 x 10^-23. So that divided by the mass of an electron [9.11 x 10^-31 kg], v = 14544457 m/s. So to link that up with voltage, we’d use that U = qV = (1.6 x 10^-19)V = 0.5mv^2. So V = 0.5(9.11 x 10^-31)(14544457^2)/(1.6 x 10^-19) = 600V</p>
<p>wewt</p>
<p>Now someone please answer my waves question ;(</p>
<p>Well, my work was purely conceptual so I don’t know if it’s correct, but here we go:</p>
<p>Assume y= lambda </p>
<p>y= h/(mv)
thus, if we decrease y by a factor of 1/2, it yields a two fold increase in velocity</p>
<p>next, we have</p>
<p>qV=.5mv^2</p>
<p>plugging in 2 v, we see a 4x increase in V, yielding a voltage of 600 V</p>
<p>f = nv/2L essentially means that the f in a string or open pipe is equal to nv/2L</p>
<p>the equation is used for instruments and open pipes</p>
<p>conversely, f= nv/4L is used for closed pipes and n can only be odd…</p>
<p>But how conceptually does frequency relate to those values? That’s what’s messing me up…</p>