<p>4:
a) Meterstick, light-intensity meter, large screen, ruler OR tape measure.
b)diagram of the slitted slide projecting an interference portrait onto the large screen, with lengths L, d, X (distance between to maximums) laballed clearly.
c)A sinusoid, with y-values between 0 and Imax, equal to 0 at the origin (because intensity is proportional to sin(theta).
d)Measure distance between light maximums, distance between large screen and slits.
e) wavelength will be abbreviated as "w".
Xm1=(m-1)<em>w</em>L/d, Xm2=m<em>w</em>L/d.
Xm2-Xm1 = mwL/d-(m-1)wL/d = wL/d. Hence, d=wL/(Xm2-Xm1)
where Xm2-Xm1 is the distance between two maximums.</p>
<p>Bsbllallstr8, it depends on whether or not you consider a positive charge to have a field directed towards itself or away from itself. And in the eqn. V=Ed, E(avg) is used, and the E at the origin is taken to be average.</p>
<p>For the last FRQ (modern), did anyone get 8.1eV for KE and then 8.1V for stopping potential?</p>
<p>isnt it standard to use a small positive test charge to determine the direction of the electric field?</p>
<p>i used an equation i found on the equation sheet for Electric Potential.. V=K * SUM (q/r)... </p>
<p>is this the same equation but in a different form? (i think so, just checkin however)</p>
<p>M17, i honestly dont remember, id have to work through it again</p>
<p>Might as well do it out, too.
7.
a)-54.4/4^2=-54.4/16 = -3.4 eV
b) Compton effect states p=h/wavelength = 6.6e-34/121.9e-9 =
5.4e-27 kg*m/s
c) we need to find the frequency, so c/wavelength=2.5e15.
Energy = hf-W0 = 1.65e-18/1.6e-19-4.7 = 5.61 eV
NOTE: hf was divided by 1.6e-19 to convert joules to eV.
d)5.61 eV implies 5.61 V stopping potential.</p>
<p>I got it wrong on the actual test (forgot to convert wavelength into frequency), but this should be the right answer. Please correct me if you find an error.</p>
<p>Bsb, yeah, it's the same equation, cause E=k(SUM:q)/(SUM:r)^2, and if V=Ed, then V=kSUMq/SUMr=kSUM(q/r).</p>
<p>good deal, looks like i got everything as long as i didnt make a careless mistake... phew that was lucky... we didnt do any problems like this in my physics class.</p>
<p>anyone wanna work through the other questions?</p>
<ol>
<li>
a) % submerged = density of raft/density of water = 65%, meaning 35% above water. Height of raft = volume/surface area = 1.8/8.2 = 0.22 m, and 0.22 m<em>0.35= .077 m = 7.7 cm
b) Bouyant force = density of water * Volume displaced * g = 1000 * (.65</em>1.8) *9.8 = 11466 Newtons, upwards.
c) Number of people * m of person *g + m of raft * g = Bouyant force.
Bouyant force = density of water * volume of raft (since raft is fully submerged) * g = 17640
Number of people = (17640 - 11466)/(735) = 8.4, so 8 people max.</li>
</ol>
<p>6
a) we know that PV=nRT, and V=AH, so PAH=nRT.
b) this was drawing a graph...basically a line.
c) from the eqn. in a, we know that PAH=nRT. n=PAH/(RT)=101300<em>.027</em>1.11/(8.31 * 300) = 1.22 mol. (H and T were plugged in from the table.</p>
<p>For #5, I misinterpreted the question and got height of the portion of the raft above water to be .22m! Using this I did all the other questions in the right way, but with the wrong numbers. </p>
<p>Do points carry over? Can they jjust take off points for part 2 and give me credit for all the other parts because i ddid them correctly with with wrong numbers? Help</p>
<p>I think that if you got the wrong answer in one of the first problems that affects the rest of the answers, they give you more or less full credit for the rest of the sections if your method was correct. I think that only one point is awarded for actually having the right number.</p>
<p>yea they do that... i am waiting for my green insert... i wrote my answers in it :) i got my chemistry one back on friday, so perhaps monday?</p>
<p>For number 4, is a light intensity meter really necessary? We did this exact same lab in class, and it was very easy to tell where the maxima were, because they showed up as bright dots on the screen.</p>
<p>For #4, about young's double slit experiment, is the graph supposed to be like a series of hills that get smaller and smaller, with the largest one right in the middle ? It has y-axis symmetry</p>
<p>i didnt say that a light intensity meter was needed.... i said that a ruler was necessary to measure the length between a crest and a trough, or 2 crests, or 2 troughs.. w/e u want</p>
<p>got my green insert back.. did OK i guess... (wrote answers in there just so i could check them)</p>