Giancoli Solutions on Video

Learn physics easily with guided practice.

Features

• 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.
• Videos are delivered with a high performance content delivery network. No waiting for videos to load or buffer.
• Pause, rewind, repeat, and never miss what is being said.

Sample solution

Giancoli 7th Edition, Chapter 5, Problem 12

(4:49)

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Giancoli 7th Edition, Chapter 4, Problem 22

By chaegyunkang on Sun, 07/01/2018 - 15:55

where is the normal force for part a) ?

By chaegyunkang on Sun, 07/01/2018 - 18:57

Oh, I realized that free body diagram only shows the forces acting on the object, not on other objects....

By Mr. Dychko on Mon, 07/02/2018 - 15:21

Yep, you got it! :)

Giancoli 7th Edition, Chapter 19, Problem 20

By cm2hn on Sat, 06/30/2018 - 21:13

Hello, I was trying yo watch the video, but it is unable to play. Thanks.

By Mr. Dychko on Mon, 07/02/2018 - 15:20

Hello cm2hn, have you tried this video again? In case the Javascript enabling the PDF embedded display was interfering with viewing the video, I have made the PDF into a clickable link instead. I havn't heard reports from other students about difficult viewing videos, so I suspect it's an isolated issue. Please try the video again and let me know if there are still any problems.

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 3, Problem 16

By chaegyunkang on Wed, 06/27/2018 - 20:21

I think the answer to this question is wrong. The coordinate of the resultant vector is (-142.2 , 65) which means that it is located in quadrant 2. Since the value of tangent is negative in quadrant 2 it makes sense as well. The angle is 24.56 degrees above the negative x - axis, not positive x-axis.

By Mr. Dychko on Fri, 06/29/2018 - 18:19

Hi chaegyunkang, thank you for spotting that, and for the thoughtful analysis. You're quite right! I'll change the written answer, and made a note about the video.
All the best with your studies,
Mr. Dychko

Giancoli 7th Edition, Chapter 3, Problem 48

By brysongrondel on Sat, 06/02/2018 - 20:49

If I am not familiar on the sine law and I am not being instructed in the sine/cosine law, what formulas from the book can be used to arrive at the same solutions?

By Mr. Dychko on Sat, 06/02/2018 - 21:17

So... the alternative to using the sine law here (or cosine law in other questions) is to do a lot of work resolving vectors into components, adding those components, then using the Pythagorean Theorem to turn the resultant components into the final resultant vector. The sine and cosine laws are designed to make the solution much, much quicker, and I'm quite sure the time you put into learning them will pay off by avoiding the much larger amount of time dealing with components. Places to start would be https://en.wikipedia.org/wiki/Law_of_sines and https://en.wikipedia.org/wiki/Law_of_cosines. Short term pain for long term gain!
Good luck,
Mr. Dychko

Giancoli 7th Edition, Chapter 3, Problem 47

By brysongrondel on Sat, 06/02/2018 - 19:58

I'm confused on why the last problem did not utilize the Pythagorean theorem to find the actual speed traveled by the swimmer and the actual distance traveled, but for this problem, we did both... I solved the last problem (#46) using the a^2 + b^2 approach and found the angle using the inverse tangent, but my answer was wildly different. Any insight into how to reconcile this in my brain?

By Mr. Dychko on Sat, 06/02/2018 - 21:12

Hi brysongrondel, thanks for the question. I understand that something here is confusing, but it isn't clear to me which question/solution you're comparing with, since #46 isn't really comparable to this one... I notice you asked another question, so I'll follow up there.

Giancoli 7th Edition, Chapter 2, Problem 12

By brysongrondel on Mon, 05/28/2018 - 16:30

So, first off, I have to say that I love this site so far. I am kind of struggling with the issues that arise from rounding errors. I am familiar with significant figures, but I am wondering if there is a better rule of thumb. Normally I don't round until I get the final answer, but I got problem #7 wrong. On this problem, I did all of the steps correctly, but I rounded up the times t1 and t2 to 2 sig figs and ended up getting 62 km/h. Is there anything I should know that might be helpful as I prepare for tests and future problem sets?
Thanks!

By Mr. Dychko on Wed, 05/30/2018 - 10:21

Hi brysongrondel, thank you very much for the nice feedback, and your question. The general idea is to avoid rounding until the final answer. Rounding values before a final answer is called intermediate rounding error, and causes the final answer to differ as a result of the rounding. So, generally, avoid rounding until the end. It's a topic that calls for a bit of patience since most numbers need rounding eventually, and the answer you're comparing to may have chosen a different number of digits than you, resulting in a different answer than what you obtained despite your following the best practice of keeping "lots" of digits until the end. Different people could have different opinions on what "lots" means. I would say that two additional digits, beyond those that are significant , should be kept with intermediate numbers.
Hope that helps,
Mr. Dychko

Giancoli 7th Edition, Chapter 16, Problem 29

By cm2hn on Mon, 05/28/2018 - 12:12

Hello, in the part to replace cos45 if I write sq root 2 / 2 (that is what the calculator gives me) instead of 1 / sq root of 2, at the end I guess it will give me a different answer. Right?

By Mr. Dychko on Mon, 05/28/2018 - 13:23

Hello cm2hn, thanks for your question. Well, the ultimate proof is always in trying your variation and see the effect, but I can say that $\dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$. If you multiply $\dfrac{1}{\sqrt{2}}$ by "1", you won't change it's value, but if you make "1" look funny as, say, $\dfrac{\sqrt{2}}{\sqrt{2}}$, you'll find that it turns $\dfrac{1}{\sqrt{2}}$ into $\dfrac{\sqrt{2}}{2}$. We know that they're equivalent since you can turn one into the other by multiplying by 1.
All the best,
Mr. Dychko

Giancoli 6th Edition, Chapter 7, Problem 34

By rodri100172 on Thu, 05/17/2018 - 18:51

Thanks bro! you're a lifesaver! Say hi to Brian for me.