<p>I’m solving many practice tests for the AP Physics exam,
and I have few questions that I cannot figure out myself…
Pls help if you know the logic behind the answer!</p>
<ol>
<li><p>A lens is used to produce a sharp image on a screen. When the right half of the lens is covered with an opaque material, how will the image be affected?
A: The image brightness will become approximately half of the original brightness. </p></li>
<li><p>It is certain that a rod is electrically charged if it…
A: repels a pith ball
(however, if the pith and the rod are oppositely charged, can’t they also attract?)</p></li>
<li><p>The rate of heat production of a wire immersed in ice water and carrying an electric current is proportional to…
A: the square of the current
(why is it not the current? I thought P=IV so it’s proportional to the current)</p></li>
</ol>
<p>1) Image brightness depends on diameter of lens and focal length</p>
<p>2) You are misinterpreting the question. Doesn’t matter whether they attract or repel. If it repels, you know it’s charged.</p>
<p>3) Remember that V and I are functions of each other through Ohm’s law. Thus, you need to look at the power equation such that P = I^2*R. R is always constant, thus you know that P is proportional to I^2.</p>
<p>@WiseGuy
Thanks for answering. </p>
<p>However, in question 2, there was another option which says ‘attracts a pith ball.’
I don’t know why ‘attracts a pith ball’ cannot be the answer…</p>
<p>Also regarding question 3, why is it wrong to use the equation P=IV? I know that P=I^2R is also correct, but why not P=IV in this question?</p>
<p>I think number two is a poor question. Either that, or the question writers wanted you to think that there is some attractive force between the two due to gravity, but that is extremely negligible and you’d never even consider it.</p>
<p>For question 3, you can’t use P=IV because V itself is a function of the current. V=I<em>R. Sub that into the P=IV equation, and you get that P=I</em>(I<em>R), or P=I^2</em>R. You know that R is a constant that depends on the physical characteristics of the resistor, so you can’t simplify any further.</p>
<p>I get it now, thanks a lot for your explanations 
BTW, are you really a student in MIT…?</p>
<p>well wlsnehf there is a logical reasoning behind question two. You have to actually try out a lab to understand this. When you have a neutral pith ball and you put a negative charge by it, it will be attracted. If you have a positive pith ball it will also be attracted to the negative charge. Suppose you did not know the charge of the pith ball, you wouldn’t be able to identiffy the charge through attraction. But if you were to put a positive charge, it would only react if the ball was positve in which case it would repel.</p>
<p>I think of it as this. Repulsion is a team effort with both charges involved while attraction is a one man thing.</p>
<p>eswara20</p>
<p>but don’t positive and neutral charges also attract?</p>
<p>yes i’m sorry. I was mistaken above. BUT that doubles the point. ONly repulsion is an indicator of charge. Any charges object will attract a neutral object. But only like charges will repel.</p>