<p>There’s no flux through a surface if field lines are parallel to it, right?</p>
<p>common sense + critical thinking :D…
Hmmm what function has a derivative that is equal to itself times a constant???
Hmmm what function has a second derivative equal to itself times a negative constant??? </p>
<p>If not the method known as eigenvalues works extremely well.</p>
<p>Or assume, as you always do in an H or NH linear DFQ, that the answer is of the form e^r*t. </p>
<p>If r is a complex number,p+qi , by Euler’s formula, plus constant analysis it is of the form (e^pt)<em>(c</em>cosqt + C<em>sinqt), where c and C are constants. Plus, using analysis and trig plus the fact that they GAVE an analogy to SHM, the above simplifies in the problem to C</em>sin(qt+phi)</p>
<p>But using commonsense, the above is not even necessary.</p>
<p>@Astroblue - yes, i think, flux = BA cos theta. Theta is considered to be 0 with perpendicular field. So if it is parallel, theta = pi/2. cos (theta) = 0. flux = 0.</p>
<p>I assume the same applies for electrif field?</p>
<p>what did you get for k=-btheta</p>
<p>also on for the force diagram, did you just draw the centripetal and gravitational forces? did you also have to draw the normal force? this was the ball frq…</p>
<p>yes the same applies for an electric field. And were there any diff eqs on the e+m test?</p>
<p>Centripetal force is a combination of two forces not one.
torque = I alpha.
etc. etc.</p>
<p>ologos…i used f=-kx in relation to f=-btheta…therefore b=k in the simple pendulum equation??</p>
<p>A torsional pendulum is different than a spring. Analogous, yes, but not quite.
k—> b… m—> I. (rotational equivalents)</p>
<p>dude…so t= square root of m/b right for the frq???</p>
<p>… no, even for a simple pendulum the period is 2<em>pi</em>sqrt(L/g) or 2<em>pi</em>sqrt(m/k).
And the rotational analog for mass is moment of inertia. Mass does not factor in the answer.</p>
<p>whats the period of a torsion pendulum…2pisquare root of (m/b)?</p>
<p>Not m. I. 10char.</p>
<p>what percentage of the multiple choice questions for a 5? Like 60%??</p>
<p>Ologos, didn’t you need a calculator to do the linear regression on #3? And even if you eyeballed it with a line of best fit, you would still need a calculator to find the value of B, unless you know how to divide a slope by 4pi^2 in your head. Or had the time to do long division.</p>
<p>And you needed to use calculus to solve for the drag force equation, then take its derivative to graph the acceleration function on #2.</p>
<ol>
<li>I eyeballed it, it lined up perfectly. I doubt CB will require an actual linear regression.</li>
<li><p>hmm… i don’t know - i think i might have used it there once, or maybe i just left it in equation format.</p></li>
<li><p>False! Well not integral calculus anyway Drag force equations always are of the form where acceleration/force is directly proportional to velocity where c is the constant (won’t say what it is for security reasons)
hmm… what derivative = itself*constant? OH! e^cx</p></li>
</ol>
<p>Is .2 significant to anyone?</p>
<p>Yes? Think of what it means for whatever the value of the axis is to be 0 (being purposefully vague to dodge College Board censors XD)</p>
<p>^^^
thinking about it… don’t see how .2 could be relevant to THAT
though it may or may not have been relevant somewhere else on the test (how’s that for vague?)</p>