Giancoli Solutions on Video

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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.
  • Final answer provided in text form for quick reference above each video, and formatted nicely as an equation, like $E=mc^2$. This is useful if you are in the library or have a slow internet connection.

  • Pen colors make the step-by-step solutions clear. Red is used to illustrate algebra steps, and to substitute numeric values in the final step of a solution. When a solution switches to a new train of thought a different pen color emphasizes the switch, so that solutions are very methodical and organized.
  • Solutions are classroom tested, and created by an experienced physics teacher.
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Sample solution

Giancoli 7th Edition, Chapter 5, Problem 12

(4:49)

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Recent questions and answers

Giancoli 7th "Global" Edition, Chapter 8, Problem 19

By lina09037788 on Tue, 04/25/2017 - 06:04

angular position should be in radians, so it's 23*2*3.14 not 23 revolutions.

Giancoli 7th Edition, Chapter 6, Problem 43

By suriyak786 on Sat, 04/22/2017 - 21:22

if the elevator is falling, wouldn't H be negative?

Giancoli 7th Edition, Chapter 11, Problem 4

By dshadowalker on Thu, 04/20/2017 - 10:14

My answer comes out to 316N/m can you show how to input into the calculator? I like when you show the calculator because this assists me in using the calculator and making sure I have the equation right.

Giancoli 7th Edition, Chapter 21, Problem 44

By julia.wolfe on Sun, 04/16/2017 - 09:00

Why don't you subtract the potential due to the inductance of the coil? I thought the minus sign in the E = -L (dI/dt) formula was meant to indicate that the voltage would be in the opposite direction following Lenz's law (current increasing > flux increasing > induced current needs to decrease flux to compensate > current flows in other direction > voltage drop in other direction). What am I missing?

By Mr. Dychko on Mon, 04/17/2017 - 06:10

Hi julia.wolfe, thanks for the question. What's important to notice is that the sign of the potential difference will be the same across the resistor as it is across the inductor. In both cases, the circuit element causes a potential drop. (I say "circuit element" since it's OK to model this circuit as containing a single resistor and a single inductor connected by zero-resistance wires). Recall that when using Kirchoff's Laws to analyze a circuit, you make the potential across a resistor negative (it's a potential drop, in other words) when traversing the resistor in the same direction as the current. Because the inductor opposes the change in current, as you correctly mentioned, this inductor also causes a potential drop to oppose the change (note that, contrary to what you mentioned, the current does not end up flowing in the other direction. Rather, it's increase is opposed, and the inductor would disappear once the current stops changing). It would be fine to call both potential drops negative if you wish, but the question isn't specific about direction, so calling both positive is fine when answering a question asking for a "potential difference".

Hope that helps,
Mr. Dychko

Giancoli 7th Edition, Chapter 13, Problem 47

By elkinsk on Sat, 04/15/2017 - 18:32

i think the temperature is 22° not 20°?

By Mr. Dychko on Mon, 04/17/2017 - 06:18

Hi elkinsk, thank you for noticing that! You're quite correct that the temperature should be $22^\circ\textrm{C}$ instead of $20^\circ\textrm{C}$. It turns out that the final answer is the same in either case since the 22 or 20 get added to 273 in order to convert to Kelvin, and the percent difference between 295 and 293 is small. I'll put a note for other students about this, and thanks again.

Best wishes with your studies,
Mr. Dychko

Giancoli 7th Edition, Chapter 14, Problem 3

By girondebordeaux92 on Sun, 04/09/2017 - 03:19

I think specific heat capacity of 4.186 is missing in the calculation? Can you please clarify

By Mr. Dychko on Mon, 04/17/2017 - 06:33

Bonjour girondebordeaux92, thanks for the question. The 4.186 figure you're quoting is, with units, $4.186 \dfrac{\textrm{kJ}}{\textrm{kg} \cdot \textrm{C}^\circ}$. Understanding units answers your question, since if you consult table 14-1 you'll notice that an alternative way to mention the specific heat of water is to write $1.00 \dfrac{\textrm{kcal}}{\textrm{kg} \cdot \textrm{C}^\circ}$. It turns out that the "kcal" way of writing specific heat, as shown in the video, if just convenient in this case since the energy of the candy bar is giving in kcal.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 18, Problem 35

By cj_vana on Mon, 04/03/2017 - 01:41

For part b, should the 120V be squared?

By Mr. Dychko on Mon, 04/17/2017 - 06:35

Hi cj_vana, thanks for pointing this out. It turn out that I did, in fact, square the $120 \textrm{ V}$ when doing the calculation, even though I forgot to write that the number is squared in the working. I'll make a note about this for other students.

All the best,
Mr. Dychko

Giancoli 6th Edition, Chapter 5, Problem 21

By thesouthportschool on Wed, 03/29/2017 - 22:17

FN = ... +mgcosO

By Mr. Dychko on Mon, 04/17/2017 - 06:45

Hi thesouthportschool, I'm not really sure what the comment is here. I could maybe help if the comment was more specific.

All the best,
Mr. Dychko


7th Edition Solutions 6th Edition Solutions Global Edition Solutions