<p>YB what were the other options besides more steam, was like temp of water and speed of something</p>
<p>Also, with a convex (something) in front of a plane mirror, i just put that a convex (something) would be the image. I though plane mirrors just reflect the same image but on other side of mirror, virtual and upright</p>
<p>Had no clue if this convex object made a difference on that tho</p>
<p>@SheepLionWut I’m pretty sure answer was “concave image”, cuz when a wave is reflected off a plane mirror, the resultant wave’s source appears to be from the other side of the mirror, thus switching the “orientation” of the wave</p>
<p>That actually makes sense. Took the test even though i’ve never taken a physics class haha. But now that i think about it, if i put an ) infront of a mirror, mirror would portray it as (</p>
<p>what were the other questions that other people remember having difficulty with? </p>
<p>what was the answer to the wavecrest and boat question?</p>
<p>the boat one rattled me i had no clue, what was th wave crest one</p>
<p>Does anyone remember a question involving node distance or something? I don’t think I remembered to double node distance for wavelength (or if the problem even required finding wavelength)</p>
<p>And what were the graphs for the maximum energy of photons off a photoelectric surface and the kinetic energy of light as frequency increases?</p>
<p>@recyclingbin was it the question about shortening a guitar’s string and the resultant frequency?</p>
<p>@tamabombXD I believe so…what was the answer?</p>
<p>what did you guys get for the problem asking about distance with the two projectiles. The answers were all in terms of energy.</p>
<p>@recyclingbin If I remember the figures correctly, the q was something along the lines of:
distance between the two nodes of a guitar string (w/standing wave) used to be L, now shortened to 3L/4, what is the resultant frequency (if original frequency was “f”?)</p>
<p>I got 4f/3</p>
<p>kinetic energy of light ends up being extending upwards to infinity at a certain point, almost as if that point were acting as a vertical asymptote if you will. Sorry i’m bad at explaining things</p>
<p>@tamabombXD Hm…I got the same answer. Were there any other questions that involved node distance? I don’t think that one was it because I think L was just the length of the string, not node distance.</p>
<p>I got 2f/3 for that question, I was under the assumption that the distance between the nodes is = to 1/2 lambda, therefor the full wavelength would be 2 x 3/4 = 3/2xlambda, so the frequency would be 2f/3. I think i’m right but I only started learning physics on thursday so idk</p>
<p>@recyclingbin I think there was another one w/node distance too, can’t remember the q though lol</p>
<p>@SheepLionWut was L distance between nodes or the entire length of the string?</p>
<p>Also if each node length is shrunk by 3/4, then the length of the entire string is also shrunk by 3/4, regardless of whether the node length is lambda or 1/2 lambda.</p>
<p>@tamabombXD I’m at -10.5 raw score right now aiming for an 800 haha so I’m stressing out</p>
<p>i believe i got vsqrt(2h/g). My logic was that h=1/2gt^2 horizontal x=vt
rearrange first one to sqrt(2h/g) = T and T times v = displacement in horizontal aka distance between</p>
<p>@recyclingbin I don’t remember the problem well enough tbh, nor am I very confident in my ‘wave knowledge’, i probably spent about an hour total glancing over that chapter. What i do remember is that the fundamental wavelength is = to 2L and since L was changed to 3/4L, i treated it as a completely new problem, one where it’s fundamental wavelength is 2x(3L/4) = 3L/2. If i remember the problem correctly, it asked what the fundamental frequency would be so i just concluded 2f/3 </p>