<p>No late testing has a completely different test and curve for the test</p>
<p>Here’s what I did (Mechanics only):</p>
<h1>1</h1>
<p>a) F<em>d = F</em>g <a href=“v_t”>b</a>^2 = mg/C**
b)i. 0graph (v<em>t)^2 versus m ii. C = m/slope
c) Can’t remember what I did, I’ll take a look at it again in a few minutes.
d) delta</em>E = E<em>f - E</em>i = K<em>f - U</em>i = <a href=“1/2”>b</a>m(v_t)^2 - mgy**</p>
<h1>2 (These should all be numerical answers - I almost forgot to substitute actual values in the test!)</h1>
<p>a) Draw friction, normal, and gravitational forces, making sure F<em>fr and F</em>n begin at the point of contact between the ball and roof, and F<em>g begins at the center.
b) **Systems of equations.<a href=“F_fr”>/b</a><em>r = I</em>alpha, ma = mgsin(30) - F</em>fr, a=alpha<em>r, solve for F<em>fr
c) Conservation of energy. E</em>i = E<em>f, U</em>i = K<em>f, mgdsin(30) = (1/2)mv^2 +(1/2)Iw^2, v=wr
d) **I said p</em>x was conserved<a href=“not%20really%20sure%20what%20happened%20in%20y-direction%20and%20if%20that%20had%20anything%20to%20do%20with%20x-direction”>/b</a>, so p<em>i,x = p</em>f,x. m</em>v<em>i = (m</em>total)<em>v<em>f, and v</em>i = v</em>cos(30)</p>
<h1>3</h1>
<p>a) dv/dt = a<em>max * sin(pi<em>t/T), integrate to get v= -T/pi</em>a</em>max<em>cos(pi</em>t/T) + T/pi<em>a<em>max
b) W = delta</em>KE = KE<em>f - KE</em>i = KE_f = (1/2)mv^2, substitute T for t in v equation, you get 2</em>m<em>T^2</em>(a<em>max)^2/(pi)^2
c) ma=0, so T=mgsin(theta)
d) J = integral of net force w/ respect to time
e) T = ma + mgsin(theta), a changes for T</em>1 and T_2. Got a sine-looking curve and a negative-exponential-looking curve.</p>
<p>Oh #1c I’m pretty sure I just drew a graph that was concave up, constantly decreasing, and approaching terminal velocity. I was sitting here for a few minutes wondering how to integrate dv/(mg-Cv^2) and just realized that we didn’t have to show any work. Wow, glad I didn’t take this test today!</p>
<p>And then for part ii. I said just take the area under the curve from t=0 to t=T.</p>
<p>I haven’t taken the test but I’m almost positive the light bulb is getting brighter. Does anyone disagree?</p>
<p>I want to agree! I remember saying that the bulb gets brighter b/c v increases in at least one problem…</p>
<p>Why wouldn’t the bulb get dimmer? : /</p>
<p>And ChemE14, did ur sine looking curve (mech3), even out at time T?</p>
<p>Anyone wanna post their work for E&M?</p>
<p>I’m doing all the work now.</p>
<p>Mech 1.
Someone please post these, I’m afraid I did c-e wrong</p>
<p>Mech 2.
(a) Friction at bottom, upslope
Weight straight down
Normal perpendicular to surface
(b) Ff=8.41 N
(c) 5.29 m/s
(d) 2.29 m/s</p>
<p>^Probably way wrong on those</p>
<p>Mech. 3
(a) v=a(max)T/pi * (1 - cos(pi<em>t/T))
(b) 2m(a(max)</em>T/pi)^2
(c) F=mgsin(theta)
(d) 2ma(max)T/pi
(e) LOL</p>
<p>E&M 1.
THIS IS PROBABLY WRONG
(a) Vb>Va=Vc
B is closer to the charge distribution
(b) V = kQ/R
(c) v=sqrt(2kqQ/mR)
(d) to the right (along the x-axis)
(e) E = rad(2)<em>Q / (2pi^2</em>R^2*e0)</p>
<p>E&M 2.
(a) I=0A
(b) Q=300uC
(c) 2.25 x 10^-3 J
(d) I = 2.25
(e) LOL (for giggles, I put 0 C)
(f) 600 J</p>
<p>
</p>
<p>Yes. And it was also “half” a sine curve (just had one maximum halfway through, no minimums besides end points). I’ll post my numerical answers for #2 sometime later. I was too lazy to do it earlier.</p>
<p>E&M 3.
(a) Counterclockwise
By the RHR, B through the loop is out of the page but decreasing because I through the wire is decreasing. To compensate, the current will be ccw to oppose this changing flux by generating a B field that is out of the page</p>
<p>(b) Remains the same. The induced emf in the loop is the rate at which flux is changing, which is constant. Therefore, the induced current is constant?</p>
<p>(c) B=u0<em>I0 / (2pi</em>r)</p>
<p>(d) FluxB = u0<em>b</em>(I0 - Kt) / 2pi * ln((d+a)/d)
(e) P = (1/R)<em>(-u0</em>bK/2pi * ln ((d+a)/d))^2</p>
<p>I’m done…you guys can laugh at me, tell me I’m wrong, etc now :)</p>
<p>Let me provide my input as to why the light bulb would get brighter. Since the current due to the long wire is decreasing and it is moving to the right, that means that there must be a decreasing magnetic flux going out of the page, since the rate of change of flux is proportional to the rate of change of current. By Lenz’s law, the magnetic field through the loop must remain constant, so there has to be an induced counterclockwise current that will increase the magnetic flux out of the page. Since the flux due to the LONG wire is decreasing over time, the flux due to the current in the LOOP must be increasing overtime, and thus the current must be increasing over time, which means the power dissipation in the lightbulb must be increasing overtime and thus it will get brighter.</p>
<p>I’ll provide an argument why the bulb should remain the same. The expression for magnetic flux through the loop is FluxB = u0<em>b</em>(I0 - Kt) / 2pi * ln((d+a)/d), which is time dependent. The derivative of this is u0<em>b</em>K / 2pi *ln((d+a)/d), which is no longer time-dependent. The derivative is the emf of the loop wire. Since we know the resistance is a constant R, that means I is constant too. I’m not sure whether current or power absolutely dtermine brightness, but both are constant</p>
<p>JDong, when you took the derivative, it is not time dependent, but there is a “b” in your expression, and if this refers to the magnetic field, then isn’t the magnetic field time dependent?</p>
<p>b is the length of the wire, not the magnetic field I think :x</p>
<p>Sorry lol, I didn’t take the test, but I will work out the later parts of the problem right now</p>
<p>Why would the current in the loop increase over time? Since </p>
<p>EMF = d (flux) / dt
= A<em>dB/dt
= A</em>d/dt(u<em>0 * I / (2 * pi * r)) … using B=u</em>0<em>I/(2</em>pi<em>r)
= A</em>u<em>0 * d/dt(I</em>0 - Kt) / (2 * pi * r) … move constants and plug in current
= -A<em>u_0</em>K / (2 * pi * r) … differentiate</p>
<p>I ignored r because it is irrelevant and you find the flux after this reasoning.</p>
<p>This is a constant emf, constant resistance (not really, but ideally), and therefore constant current.</p>
<p>You assume that because flux decreases, current increases, but no. A CHANGING flux makes a current. If the current were something like I_0-Kt^2 then I’d understand, but it is at a constant change, and therefore constant current.</p>
<p>Bulb stays the same.</p>
<p>Did you guys get these answers for #3 on e&M</p>
<p><a href=“http://img38.imageshack.us/img38/7565/problem3e.jpg[/url]”>http://img38.imageshack.us/img38/7565/problem3e.jpg</a></p>
![http://img38.imageshack.us/img38/7565/problem3e.jpg[/url]](http://img38.imageshack.us/img38/7565/problem3e.jpg%5B/url%5D){kind=link}
<p>lol guys, I posted the answers I got in a separate thread. If you want to, post there and compare :)</p>
<p>But yeah eg, I got those</p>