<p>yess. also, was buoyant force the force tension plus the weight of the ball? and did level increase or decrease?
and in the charged sphere question, one of them said to find the masses of the “spheres”. but was there was only one number to be found?</p>
<p>I put buoyant force = T + mg and I put did not change.</p>
<p>Shoot, there’s a formula for single slit diffraction…</p>
<p>Yes, buoyant force was tension plus weight as buoyant force was directed up and weight was directed down. Tension was what kept it from floating up and so was also directed down.</p>
<p>The water level decreased; once the ball floated to the top, it would no longer be displacing the same amount of water.</p>
<p>For the one with the little balls or whatever, to find the mass I plugged in one of the data values from the graph along with the L they specified in the question. Since it was an experiment, you could have chosen any of the data points and they’ll give you full credit as long as the mass you found is within a range of values that they specify.</p>
<p>Did you assume the rubber ball was weightless?</p>
<p>@RawkForGod, it’s still in the water, I’m fairly certain that it still displaces the same amount of water.</p>
<p>Not if it’s floating on top of the water.</p>
<p>Kevin, but since it is not fully submerged any longer, wouldn’t the water level slightly decrease?</p>
<p>Am I the only one who looked at the graphs on number 2 and just randomly drew lines. Ha ha ha :)</p>
<p>No, I just drew a random decreasing line. Number two was so confusing.</p>
<p>Yeah that is what I did too. Number 2 sucked, I did horrible. On the ball in water one, to find the weight of the water was it just density<em>volume</em>gravity?</p>
<p>how many points was the pressure question worth? 10 or 15?</p>
<p>@garfield</p>
<p>CRAP, you’re right. Maximums for single slit diffraction are (m + 1/2)wavelength and not (m)wavelength like in double slit. Well, I guess I screwed up the entirety of that problem, too. </p>
<p>@Zach</p>
<p>You had to find the mass of the water with d = m/v and then multiply the mass by g, so yeah, you’re right.</p>
<p>For the lab question, I set it up so that the beam light from the filament was spread out using a prism then reflected from a plane mirror onto a screen and the distance of light measured (assuming light passed through the entire distance of the slits) so that the measure of the light on the screen would equal the distance b/w the slits.</p>
<p>It’s not like they told us NOT we COULDN’T do something that simple, amirite? :D</p>
<p>I thought the lab question was SO easy. They just wanted the width of the slit.</p>
<p>I just said use the ruler to measure the width of the slit.</p>
<p>@Greed</p>
<p>Did you really? rofl</p>
<p>I was sooo tempted to say use a ruler, but I think they were looking for something else. I also said the single slit, using the prism, and looking at the fringes</p>
<p>The ball was at the surface which means part of it isn’t submerged anymore, so I said the water level went down slightly</p>
<p>Can someone remind me, what was #4?</p>
<p>for number 5 how did you guys start it out? i did potential energy</p>
<p>qV=mgL</p>
<p>q=mgL/V</p>