<p>no…my physics teacher showed me how it was “supposed” to be done with h and 1/m</p>
<p>i kept the constants in and just graphed
x^2/2mg versus h, still got the same spring constant, but not the “normal” way to do about doing it</p>
<p>But why would graphing h and 1/m give you the spring constant? If you do the math it doesn’t make any sense.</p>
<p>when you graph h versus 1/m
you get a slope of value kx^2 / 2g
then you multiply by 2g, divide by x^2.
to get k.</p>
<p>btw pigs, i was totally kidding. my way of saying “no big deal”</p>
<p>I hate #1</p>
<p>!!!</p>
<p>Does anybody have a list of all the answers for all the free responses questions?</p>
<p>Thanks.</p>
<p>bump i also want to see the answers</p>
<p>i’ll post it soon, i guess x.x</p>
<p>YAY! PIGGY TO THE RESCUE! =]
<em>answers are from my physics teacher, went over them with him</em></p>
<ol>
<li><p>a) 1/2kx^2=mgh
h=kx^2/2mg
b) i. graph h versus 1/m
ii. (fill in values for h and 1/m)
c) graph should be a pretty straight line with slope
d) k should be somewhere around 490…
e) many ways…design an experiment (i just put graph kx^2 versus 2mg and use slope for h…physics teacher said this was “technically” right, but i’d prob lose a point because they actually wanted me to design an experiment)</p></li>
<li><p>a) google electric field lines
b) Lsin0=r, kQ/Lsin0+kQ/Lsin0= 2kQ/Lsin0
c) tension vector upper right diagonal, mg straight down, and Fe vector to the left
d) Tcos0=mg, Tsin0=Fe, Fe=kQQ/(2Lsin0)^2, so Tsin0=kQQ/(2Lsin0)^2 and Tcos0=mg
<em>yes, you had to expand Fe because it said to express all answers in terms of m, L Q, 0</em></p></li>
<li><p>a) V=BLv=(.8)(.52)(1.8)=.7488V
V=IR, .7488/3=.2496A
b) force=Fb+■■= ILB+ukFn=(.2496<em>.52</em>.8)+(.22.2)=.544 N
c) P=IV= (.2496)(.7488)=.1869J/s—> in 2 seconds= .373 Joules
d) work= Fd= Fvt= (.544)(1.8)(2)=1.958 joules
e) energy is lost to friction</p></li>
<li><p>a) <strong>stupid moles</strong>* 18g=1mole, so 2200 g=122.22 moles
PV=nRT, (3e5)(2)=122.22(8.31)(T) –> T=590.756 K
b) PV=nRT, (4e5)(2)=122.22(8.31)(T) ----> T=787.674 K
c) internal energy decreases, U=Q+W, volume increases so system does work so W is negative
d) work=area under graph= (.5)(4e5)=2e5 joules</p></li>
<li><p>a) no, because mg-pVg=Tension, and mg is the same for all three, but V is not necessarily the same as they have different volumes
b) mg-pVg=Tension, pV=m—> mg=(1300)(1e-5)(10)=.13 N
.13-pVg=.0098N, so pVg=.1202N
c) pVg=.1202, p=.1202/(1e-5)(10)=1202 kg/m^3
d) tension would increase, less buoyant force because less volume is covered by the fluid</p></li>
<li><p>a) c=Fwavelength, F=3e8/(550e-9)=5.454 Hz
b) x=mwavlengthL/d, (m at half numbers) x=.5(550e-9)(2.2)/(1.8e-5)=.0336 m
x=1.5(550e-9)(2.2)/(1.8e-5)=.1008, .1008-.0336=.067 meters
c) frequency is still 5.454 Hz, doesn’t change, trick question
d) distance would decrease because now x=mwavelengthL/nd, so denominator gets bigger</p></li>
<li><p>(■■■!!) a) wavelength=h/p=h/rad(2mK)–> 250e-9=6.63e-34/(2<em>9.11e-31K)—> K=6.87e-4 Joules
*don’t feel like doing b</em>
c) answer is energy transition d, because it has to point down to create a photon, and it needs a greater energy level transition because E=hc/wavelength, and wavelength decreased</p></li>
</ol>
<h1>4 b is wrong because you have to change the volume as well.</h1>
<p>yea, i used p1v1/t1=p2v2/t2 and solved for t2, the answer should be around the 800’s</p>
<p>nope. it’s in the 900s.</p>
<p>4e5<em>2.5=122.22</em>8.31*T</p>
<p>so T=984.59</p>
<p>yea nvm, thats what i got using p1v1/t1=p2v2/t2</p>
<p>=( Epic failed that…think I ended up w/ around a 58-64 out of 80 <em>cries</em></p>
<p>Oh no! You got a 5!!!</p>
<p>sorry :)</p>
<p>but that’s kind of an insult to those who actually DID fail.</p>
<p>How is the internal energy decreasing for 4c? I used U=(3/2)nRT, and it shows the internal energy increases since T increases and n & R are constant.</p>
<p>Man, I think i did a lot of problems wrong because I had no idea. Well, I’ll cross my fingers.</p>
<p>Thank you for the answers, by the way.</p>
<p>Sorry, although I might not have got a 5 as my mc section was cut short by about 30 minutes =(</p>
<p>lol with a generous curve, everyone should at least have a 2, and if you think you failed cause of FRQ #1, don’t worry join the club- “The ones who think they got a 5 after getting raped by #1” lolllll</p>
<p>Oooo you’re right about internal energy increasing.</p>
<p>i didn’t catch that the first time but it increases for sure.</p>
<p>I don’t know why everyone’s complaining about with #1, the only two parts that were challenging were realizing that spring potential equaled gravitational, and that you had to graph 1/m which was weird. I thought it was pretty obvious how they wanted you to do it though. for me the induced current and the thermo questions were the hardest.</p>