kevG Posted May 2, 2016 Report Share Posted May 2, 2016 @IBdoingnothing oh hahaha for 37, I really don't know why its D then and for 17 yeah you can just convert it into series cause remember, the voltage is the same for a parallel circuit Reply Link to post Share on other sites More sharing options...
Guest iblearner Posted May 2, 2016 Report Share Posted May 2, 2016 2 minutes ago, kevG said: @IBdoingnothing oh hahaha for 37, I really don't know why its D then and for 17 yeah you can just convert it into series cause remember, the voltage is the same for a parallel circuit In series the current is same and in parallel the voltage is same, right? Reply Link to post Share on other sites More sharing options...
Guest SNJERIN Posted May 2, 2016 Report Share Posted May 2, 2016 8 minutes ago, kevG said: @IBdoingnothing oh hahaha for 37, I really don't know why its D then The work function Ф is related by the following equation, which is in the data booklet Ekmax= h*f - Ф where f is the frequency of light, h is Planck's constant and Ekmax is the maximum possible kinetic energy of the emitted electron from the metal. By rearranging the formula we get that Ф= h*f - Ekmax and that tells us that the work function is the difference between the energy of the absorbed photon and the maximum kinetic energy of the emitted photon. The answer is clearly not B since the electron can be emitted with an energy ranging from zero to any value. Thus it must be D. Hope this helps. Reply Link to post Share on other sites More sharing options...
Vioh Posted May 2, 2016 Report Share Posted May 2, 2016 @kevG @IBdoingnothing Even though your answer is correct for Question 19, but your reasoning is incorrect. If you look at the picture (from my study guide) attached in my previous post, you can see that: Paverage = Vaverage * Iaverage = Vmax/sqrt(2) * Imax/sqrt(2) = Vmax * Imax / 2 = Pmax / 2 This means that: Pmax = Vaverage * Iaverage * 2 However, note that this is the maximum power for the whole circuit (which consists of 2 resistors connected in series). But the question only asks for one of the resistors, so you need to divided the result above by 2. This is why the answer is B (where P = VI). For question 25, I think the answer is A. The way that I do this question is to first draw the field lines, and then draw the equipotential lines such that the equipotential lines are always perpendicular to the field lines. If you draw all of these correctly, then you should get a vertical equipotential line in the middle of the 2 opposite charges (see the picture below). Thus the picture in the question must be for 2 equal charges of same sign. As for why choice 1 (two equal masses) is also correct, you should think of contour lines. Besides, gravitational field between 2 equal masses is very similar to the electric field between 2 equal charges of the same sign, so the answer must be A. For question 37 & 17, I completely agree with Haitham Wahid and kevG . 1 Reply Link to post Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.