<p>i have to keep talking so as to kep this site to on th front page..... please posst...... it takes a lot to get a please outta me</p>
<p>smder, your response for the last part of the capacitor problem is incorrect. Charge increases because capacitance increases. Voltage is the same as before because it reaches equilibrium again.</p>
<p>Are you sure d4r7h3v1l??? Because that is actually what I had on the exam...but I thought more about it later and I think voltage decreases since E is affected by dielectric constant (think Gauss's Law)</p>
<p>Wow, i just reread the question and realized that it isn't talking about equilibrium, it's talking about t=4. I'll need to figure this out.</p>
<p>Ok, I did the math and it seems that it is still greater charge.</p>
<p>I just compared the graphs of:</p>
<p>y=1-e^(-t) (the original situation)
and
y=3-3e^(-t/3) (the situation posed in the last part)</p>
<p>Although the second is slower-growing, it is still higher than the first for any value of t.
This can be generalized to the problem.</p>
<p>how about my answers to Eand M #2?</p>
<p>Can anybody help me find the answers for #1 and #2 on the Mechanics test. Thanks!</p>
<p>yeah if anyone have free time please explain to me the entire free response for mechanic</p>