# Giancoli Solutions on Video

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## Giancoli 7th Edition, Chapter 20, Problem 10

By livanessa98 on Tue, 07/12/2016 - 13:26Why would the magnetic field come to an before the electron reaches the bottom of its circular path?

## Giancoli 6th Edition, Chapter 6, Problem 9

By trandzuyhieu on Fri, 07/08/2016 - 21:01Can you explain how M(0.10g) is equal to Force up - Force mg = ma?

I get the equation: F=ma and sum of F =Fup-Fmg=ma. I don't get how M(0.10g) is equal to all that.

Thanks

Hi trandzuyhieu, thanks for the question. $M(0.10g)$ replaces the $ma$ in $F_{up} - mg = ma$. Data from the question say the helicopter mass is $M$, and it's acceleration is $(0.10g)$, so we substitute for each of these in $ma$: $M$ replaces $m$, and $(0.10g)$ replaces $a$.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 19, Problem 33

By livanessa98 on Tue, 07/05/2016 - 09:52Two quick questions:

1) Why is the answer for part b different from what you would find by doing Req calculations based on the rules for parallel and in-series resistors? (I got 5R/8 that way).

2) By considering all three resistors that are arranged in series as having the same current by definition, you could solve for I with 3 equations and 3 unknowns. Would this be a correct approach?

Hi livanessa98, good questions.

1) The series/parallel formulas don't apply here. Looking at the circuit diagram, the junction with $I_2$ going in and $I_3$ and $I_4$ coming out makes all the difference. None of the resistors in this picture are in series. Surprising, I know! When resistors are in series it means that

allthe current going through one resistormustalso go through the next, but with these junctions interfering, some of the current through one is being diverted among multiple resistors downstream, and for this reason they're not in series. None of the resistors are in parallel either because of these pesky junctions, which makes it a clever circuit diagram.2) See (1) :)

Best wishes,

Mr. Dychko

## Giancoli 6th Edition, Chapter 12, Problem 28

By kcecchini on Mon, 07/04/2016 - 16:23how do I do better and pass my last exam and not have to repeat physics?

thank you for your solutions, they are terrific.

Yay! I'm glad you don't have to repeat your course! :) I'm glad I could help, and thank you very much for the feedback.

Best wishes with your future studies,

Mr. Dychko

## Giancoli 7th Edition, Chapter 18, Problem 16

By kniffin.1 on Wed, 06/15/2016 - 07:06how would you find the length of the short wire?

Hi kniffin.1, thanks for spotting that I missed that part. I'll try to fill the gap with a typed answer. Since $R = \rho \dfrac{l}{A}$ and we're told that $R = R_1 + R_2$ and $R = R_1 + 4R_1$. We can re-write that in terms of length by substituting for each resistance: $\rho \dfrac{L}{A} = \rho \dfrac{L_1}{A} + 4\rho \dfrac{L_1}{A}$. All the common factors cancel leaving $L = L_1 + 4L_1 = 5L_1$. So the total length of the wire is five times that of the first segment, or written another way after dividing both sides by 5: $L_1 = \dfrac{L}{5}$ or $L_1 = 20\%L$. The short wire is 20% of the total length of the wire.

Hope that helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 16, Problem 25

By williams.dpw on Mon, 06/13/2016 - 09:54Should there be 3x's as many field lines going into the -3Q charge compared to the +1Q charge?

Hi williams.dpw, thanks for the question. Yes, I suppose that's an interesting point to consider, although things are not quite as simple as they might seem. Taking the two points in isolation (considering a case that is simple first), it's true that $E = \dfrac{kQ}{r^2}$ which means the field strength is directly proportional to the charge. This means that a point with three times the charge of another point will have an electric field strength 3 times as strong, and yes, in that case the number of field lines should be three times as well. When the two charges are near each other, as in this question, and their fields are interacting, then this interaction changes the number of field lines you would expect compared to the case when they're in isolation. Nevertheless I would agree that the picture could use more field lines going into the $-3Q$ charge. When considering a position very close to a point charge, the effect of the other point charge becomes negligible since their field strengths are inversely proportional to distance from the point charge. What this means is that, close to one point charge, the other point charge doesn't matter, and if you zoom in on any point charge it will be possible to find a "zoom level" at which the field lines look the same as a point charge in isolation.

I suppose I'm arguing both for and against your excellent point. The only way to get a quantitatively correct picture is to use a computer to create the vector field by calculating the resultant electric field (which is not a trivial calculation since these fields add as vectors) at every point. The hand drawn picture is meant just to show an approximation. I think an important feature of the drawing is to show the field emanating left of the left point charge bending around to the right hand charge, which would happen only if the right hand charge has a greater magnitude of charge. Also, yes, I would agree for the most part that the density of field lines around the right hand charge should be nearly 3 times that of the left hand charge, but not exactly so since the left hand charge reduces the field strength around the right hand charge.

Hope that helps!

Mr. Dychko

## Giancoli 7th Edition, Chapter 6, Problem 39

By poojamukund10 on Sat, 06/11/2016 - 12:02For part b why can't you use the initial kinetic energy of the ball using the velocity you found from part to find the potential energy (mgh) and then solve for h? Why do you need to use the potential energy of the spring?

Hi poojamukund10, thanks for the question. The tricky part is in the meaning of "h". You could do what you're suggesting, and calculate "h" to be the height that gives an amount of potential energy that is equivalent to the kinetic energy that the ball has when it is at the tip of the extended spring, but in this approach "h" will be the height

above the extended spring, not the height above the initial starting point. As an interesting exercise (I haven't tried this, but it should work), why not add the answer for "h" in your approach to the amount by which the spring was compressed ($0.160 \textrm{ m}$) and see if you get the same answer? Let me know how it goes.All the best,

Mr. Dychko

Just for clarification the way I thought it would be was KEi = PEf and both energies were referring to the energy of the ball.

## Giancoli 7th Edition, Chapter 17, Problem 54

By kniffin.1 on Wed, 06/08/2016 - 15:36this is not the answer my homework is giving me...the answer they want is x*10^-6 J

Hi kniffin.1, thanks for the comment. If you have a specific question, please let me know. I haven't noticed any error in the solution here.

All the best,

Mr. Dychko

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