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  • 1,930 video solutions for all regular problems in Giancoli's 7th Edition and 1,681 solutions for most regular problems in the 6th Edition.
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Sample solution

Giancoli 7th Edition, Chapter 5, Problem 12


Recent questions and answers

Giancoli 6th Edition, Chapter 20, Problem 1

By thesouthportschool on Mon, 08/22/2016 - 21:52

6.43 for part b

By thesouthportschool on Mon, 08/22/2016 - 21:54

no its not my bad

By Mr. Dychko on Tue, 08/23/2016 - 00:26

No worries at all. Thanks for digging into the problem, and good job finding the solution.

Giancoli 6th Edition, Chapter 17, Problem 8

By thesouthportschool on Tue, 08/02/2016 - 05:52

For the answer for this question it is shown as -3.25x10^4 instead of 10^3!

By Mr. Dychko on Tue, 08/02/2016 - 09:09

Ah, thanks for spotting that Southport School! It has been fixed.

Best wishes,
Mr. Dychko

Giancoli 7th Edition, Chapter 20, Problem 10

By livanessa98 on Tue, 07/12/2016 - 13:26

Why 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 - 21:01

Can 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.


By Mr. Dychko on Sat, 07/09/2016 - 14:32

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 - 09:52

Two 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?

By Mr. Dychko on Wed, 07/06/2016 - 12:09

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 all the current going through one resistor must also 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 - 16:23

how do I do better and pass my last exam and not have to repeat physics?

thank you for your solutions, they are terrific.

By Mr. Dychko on Tue, 07/05/2016 - 00:24

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 - 07:06

how would you find the length of the short wire?

By Mr. Dychko on Wed, 06/15/2016 - 13:29

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 - 09:54

Should there be 3x's as many field lines going into the -3Q charge compared to the +1Q charge?

By Mr. Dychko on Tue, 06/14/2016 - 22:10

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