# Giancoli Solutions on Video

Learn physics easily with guided practice.

7th Edition Solutions 6th Edition Solutions Global Edition Solutions

## 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

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## Recent questions and answers

7th Edition Solutions 6th Edition Solutions Global Edition Solutions

## Giancoli 7th Edition, Chapter 2, Problem 23

By umajacobshome on Thu, 08/02/2018 - 20:28hi why in the bottom equation can you just say that average velocity is the two terms summed and divided by two if average velocity was previously defined as displacement over time? This seems like the equation for speed

Hi umajacobshome, thanks for the question. Average velocity is indeed defined as displacement over time, as you say. There is a special case though when acceleration is constant, in that the average velocity in this case can also be calculated as the sum of the initial and final velocities divided by 2. Written as an equation, that is to say $v_{avg} = \dfrac{v_i + v_f}{2}$ in the special case when acceleration doesn't change.

Hope this helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 24, Problem 29

By justin.sunny.bains on Mon, 07/30/2018 - 16:06Why is m 1.5 and not just 1?

Hi justin.sunny.bains, thanks for the questions. We're looking at a diffraction pattern here, and I would suggest referring to Figure 24-21 from the textbook to see the pattern of minima in the diffraction pattern. They occur at integer multiples of the wavelength. The twist in this question is that it tells us the position of the

maxima, for which we need to deduce a formula from Figure 24.21. The maxima occur half way between the minima, so we need to add 1/2, as in $D \sin{\theta} = (m + \dfrac{1}{2})\lambda$.Best wishes with your studies,

Mr. Dychko

## Giancoli 7th Edition, Chapter 21, Problem 12

By justin.sunny.bains on Tue, 07/24/2018 - 17:23Should the answer be negative?

Hi justin.sunny.bains, thanks for your question. I suppose the better answer would have a negative sign, but I didn't feel it was that important in this case. The negative sign is meant to indicate that the induced EMF creates an induced magnetic flux that opposes the change in magnetic flux in the solenoid. This means the induced EMF will have a direction opposite to the EMF that increased to the 5.0 A current. Outside of that, we don't really have a circuit diagram with a specific direction specified, so it's not really clear otherwise what direction a negative sign would mean. Nevertheless, given a choice between the two answers, including a negative would be a marginally better answer.

Best wishes with your studies,

Mr. Dychko

## Giancoli 7th Edition, Chapter 4, Problem 22

By chaegyunkang on Sun, 07/01/2018 - 15:55where is the normal force for part a) ?

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

Yep, you got it! :)

## Giancoli 7th Edition, Chapter 19, Problem 20

By cm2hn on Sat, 06/30/2018 - 21:13Hello, I was trying yo watch the video, but it is unable to play. Thanks.

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 - 20:21I 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.

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 - 20:49If 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?

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 - 19:58I'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?

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.

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