<p>does anyone have the last known grading curve and how to calculate, as of now i see my self getting a 53/80 at worst and at most 5 wrong, and 2 blank on the MC, do u guys think its a 5 status?</p>
<p>@selter01:
I see what you’re saying now about the flux.</p>
<p>@wthalexis1:
That’s a definite 5.</p>
<p>I agree with the others who said that the last part of 7 was D. We need to go down in energy levels so that a photon is emitted, which is what was striking the metal to cause an electron to fly off.</p>
<p>yes, that’s still a 5… the curves i have are like 106-112-118… (not in order)… so i was going for 140… i got like 145 (counting 30 wrong in MC, which it’s a HUGE strech, i answered them all, and i only had trouble with like 3)</p>
<p>btw, to eject an electron, then you it has to be ‘e’. to emit a photon, it has to be ‘d’. I think it’s d, though I missread it…? I gave myself 0/3 in that question, haha… i think i’ll get partial for my explanation</p>
<p>yay, i love the curve, and for the last one i gave my self 4 points max, 3 for getting the first part right, and 1 point for setting up the equation and solving for phi(the work function lolll)and i guessed a on the last one. Hopefully i get points for knowing that an electron is being emitted and noting the scale of the levels xD</p>
<p>^but they’re asking you about what caused the incident photon, so it is d</p>
<h1>5,</h1>
<p>part a- answers no, i put b.c different volumes.
part b-Draw an FBD, then solve for the force of Buoyancy. Fb+T=mg, you can find the mass of the object since you’re given the density of it and volume.
part c-use the force calculate in part b and set it equal to =density<em>volume</em>gravity, and solve for density.Should be a pretty number.
part d-i put increase since not all of the surface area of the object is being covered so the buoyancy force would decrease (complete b.s by the way lol)</p>
<p>wthalexis1, that’s perfect (assuming your numbers were right)</p>
<p>yeayyyyyuhhh, ok i have a pretty good vibe i got a 5 now (watch me get a 4 loll), hopefully all you guys did well too</p>
<p>hey, can someone confirm my solution for frq#1??</p>
<p>a) mgh=1/2kx^2
h=kx^2/2mg</p>
<p>b) quantities i picked to graph are height versus x^2/2mg <–the units for this is m^2/N
so, i filled in the table, but will having too many decimal places lead to point deductions??</p>
<p>c) i graphed it, it was basically a straight line, i picked two points and found the slope between them to calculate the spring constant</p>
<p>d) my procedure for measuring the height is (from the derived equation), 2mgh=kx^2/2, then you repeat this experiment but plot 2mg versus kx^2/2 to find the height via the slope.</p>
<p>awww, alexis you almost got #5 right except for the part you bsed
d asked for if the tension in the string would increase/decrease/stay the same, and the tension would increase because if some of the liquid is drained, there would be less buoyant force F=pVg, because the volume of the object covered would decrease.
but oh well, it’s ok! we all made some silly errors.</p>
<p>@pigs, what did you get for k? </p>
<p>I graphed h and 1/m, since all the other junk was just constants, i don’t think you include them in your plotting…?
part e wanted an actual physical “how?”</p>
<p>yo thats what i put lol, “part d-i put increase since not all of the surface area of the object is being covered so the buoyancy force would decrease (complete b.s by the way lol)”</p>
<p>^lol i don’t remember my k value…=/
i wasn’t quite sure what part e was asking for…i didn’t wanna give a smart ass answer like “you measure it with a ruler” haha</p>
<p>alexis, you prob got 1/2 points for part d then because you checked the correct answer but surface area really has nothing to do with buoyant force…that’s a really logical guess though…so for that prob you most likely got 9/10 unless the graders go crazy</p>
<p>oh i got that one so wrong, i think i put use the same formula in part a, and find the h. i never said how to specifically find it, i guess 1pt there lolll</p>
<p>i hadn’t seen fluids in ages , all i rememeber was the picture of the force of buoyancy and little arrows perpendicular to it all over the surface area so i guessed based on that loll</p>
<p>some of my answers (i think).</p>
<p>[1]
a. Us = Ug > 1/2 kx^2 = mgh > h = kx^2 / 2mg
b. h versus 1/m (fill in 1/m)
d. somewhere in the high 400’s N/m
e. dunno? bs’d smth about a high speed camera.</p>
<p>[2]
b. Ue = k 2Q / LsinTheta
d. LcosTheta = mg; LsinTheta = k Q^2 / (LsinTheta)^2</p>
<p>[3]
a. e = Blv; V = IR > I = Blv / R = 0.25A
b. F = Fb + Ff = BIl + uFn = .535N
c. P = W/t; W = I^2R*t = .375J
d. d = vt = 3.6m; W = Fd = 1.93J
e. not equal: work done against friction</p>
<p>[5]
a. No: Ft = Fg - Fb = mg - pVg. V varies, so does T
b. Fb = Fg - Ft = 0.118N
c. p = Fb / Vg = 1200kg/m^3
d. Increase: Ft = mg - pVg. half submerged in p(1/2 V)g, so more T</p>
<p>4: not sure, think i had 500+K, 900+K, answers like that
but im very confident abt thermal phys. or optics, or atomic
which is why i won’t bother with those :P</p>
<p>yea the temperatures were like that</p>
<p>yea, i got 490 for my k…</p>