Physics doubts!!

[kevG]

@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

[Guest iblearner]

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? 

[Guest SNJERIN]

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 E_kmax= h*f - Ф where f is the frequency of light, h is Planck's constant and E_kmax is the maximum possible kinetic energy of the emitted electron from the metal. By rearranging the formula we get that Ф= h*f - E_kmax  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. 

[Vioh]

@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:

P_average = V_average * I_average = V_max/sqrt(2) * I_max/sqrt(2) = V_max * I_max / 2 = P_max / 2

This means that:

P_max = V_average * I_average * 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

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