# 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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The videos were extremely helpful. You can play them over and over and pause them to review. The narrator explained the steps beautifully and provided details on performing the algebraic manipulations. Different colored pens made it easy to differentiate steps. My Physics teacher often moved through the material very fast in class. The videos allowed me to review areas I found difficult as many times as I needed. Giancoli Answers was a wonderful learning tool for understanding Physics.

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## Recent questions and answers

### Giancoli 7th Edition, Chapter 4, Problem 51

By theovilous on Fri, 02/16/2018 - 07:47

In this problem, can I use the formula, Ffr = μk*Fn? I calculated μk to be .2. So the answer should be .2*60kg*9.8m/s2 = 117.6N. Is this just another way of solving the problem or did I get it wrong? Thank you!!

By Mr. Dychko on Fri, 02/16/2018 - 13:59

Hi theovilous, ah, now I understand your approach. Yes, you can do what you're suggesting. I suspect that if you keep lots of digits in your answer for $\mu_k$ (to avoid intermediate rounding error), that your answer will match the answer here exactly to three significant figures. The approach you're suggesting takes more steps, but it's perfectly valid.

I'm glad you're enjoying the solutions!

All the best,
Mr. Dychko

By Mr. Dychko on Fri, 02/16/2018 - 09:00

Hi theovilous, thanks for the question. While $F_{fr} = \mu_k F_n$ is a correct formula, I'm not sure how you calculated $\mu_k$? It doesn't look possible with the information given, so I think using acceleration is the only method available for solve this one.

All the best,
Mr. Dychko

By theovilous on Fri, 02/16/2018 - 11:18

I considered retarding force as friction. So I set the formula Ffr = ma --> μkmg=ma --> μk = a/g --> μk = 2/9.8 = .2.

So based on the formula, Ffr=μkFn, Ffr=μkmg = .2 * 60kg * 9.8m/s2 = 117.6N.

Does this approach work as well? my result is very close to your answer which is 120N.

Thank you for your quick response! I am learning a lot from you!

### Giancoli 7th Edition, Chapter 8, Problem 59

By choianchoi on Sun, 01/21/2018 - 17:02

Why is kinetic energy not accounted just before it hits the ground? Also, is the velocity of the center of mass the same with the end of the pole?

By Mr. Dychko on Wed, 01/31/2018 - 07:45

Hi choianchoi, thanks for the question. Let's keep in mind that one end of the pole stays in contact with the ground. It's a pole falling over, in other words. The end touching the ground has velocity zero in that case, at all times. The other end of the pole, on the other hand, has the maximum velocity of any point on the pole. The end initially in the air has a higher linear velocity than the center of mass. Kinetic energy is accounted for, and it's the strategy for calculating the velocity of the end of the pole, but the kinetic energy formula doesn't look like what you're used to from linear problems. We don't use $\dfrac{1}{2}mv^2$. Instead, we use rotational kinetic energy $\dfrac{1}{2}I\omega^2$ instead, and use that to figure out the rotational velocity, which is then used to find the linear velocity of the end of the pole.

All the best,
Mr. Dychko

### Giancoli 7th Edition, Chapter 3, Problem 24

By samkvertus on Thu, 01/18/2018 - 20:54

amazing

By Mr. Dychko on Fri, 01/19/2018 - 07:00

Thanks!

### Giancoli 7th Edition, Chapter 3, Problem 7

By samkvertus on Fri, 01/12/2018 - 11:20

what about the last part of the problem that asked to find the magnitude and direction ?

By Mr. Dychko on Fri, 01/12/2018 - 11:26

Hi samkvertus, thanks for the question. Both the magnitude and directly are actually given here. Direction in this case is "to the left" when the resultant is negative, whereas it's "to the right" when positive. The magnitude is the number when ignoring the negative sign. This is explained more in the video, so consider giving it a second view.

All the best,
Mr. Dychko

### Giancoli 7th Edition, Chapter 3, Problem 44

By chaegyunkang on Wed, 01/10/2018 - 08:58

Dear Mr. Dychoko,

For part (a), I understand how you used the cosine rule to find the unknown vector C. However, I tried to solve the problem similar to what you did in problem 42 and divided the wind velocity into X and Y components. I used to pythagorean theorem to find the resultant vector but was not able to get the right answer. Can you please tell me why it is wrong to use this method.

Thank You

By Mr. Dychko on Wed, 01/10/2018 - 09:20

Hi chaegyunkang, thanks for the question. The method you describe works fine, so your sleuthing should look at how you implemented it. Pay careful attention to whether you subtracted the y-component of the velocity of the air with respect to the ground from the velocity of the plane with respect to the air (instead of adding it). Knowing that $\cos(45) = \dfrac{1}{\sqrt{2}}$, here's what the work will look like in your calculator:
$\Big(688 - \dfrac{90}{\sqrt{2}} \Big)^2 + \Big(\dfrac{90}{\sqrt{2}}\Big)^2 = \textrm{ some number}$
$\sqrt{ \textrm{ some number}} = 627.5953...$ which is the correct magnitude, and I'll leave the angle for you.

Hope this helps,
Mr. Dychko

By chaegyunkang on Wed, 01/10/2018 - 09:20

My mistake was adding the vectors. Thank you so much.

### Giancoli 7th Edition, Chapter 3, Problem 4

By chaegyunkang on Sun, 12/31/2017 - 10:07

Hello, I got a different answer for this question. I divided all the vectors into x and y components and used pythagorean theorem at the end to find the resultant vector. My answer is 22.53 km ( 22.53 degrees North of East ).

By Mr. Dychko on Wed, 01/03/2018 - 21:08

All the best,
Mr. Dychko

### Giancoli 7th Edition, Chapter 6, Problem 3

By missedinger on Sat, 12/02/2017 - 21:05

wondering why you stop @ question 71 for solutions, when the text books has 94 questions?

By Mr. Dychko on Sun, 12/03/2017 - 15:51

Hi missedinger, that's a totally fair question, which I've addressed in the FAQ: It would be enormously time consuming to also answer all the "General Problems" so I'm limiting coverage only to the regular "Problems", of which there are still more than 1700. If you need help with a "General Problem", my suggestion would be to try and find a regular "Problem" that is similar to the "General Problem", and see if you can apply the same problem solving technique.

All the best,
Mr. Dychko

### Giancoli 7th Edition, Chapter 10, Problem 9

By jr59j on Fri, 12/01/2017 - 11:55

Why is it 1/2 of m*g

By Mr. Dychko on Sun, 12/03/2017 - 15:52

Hello, thanks for the question. The lady has two heels, so the weight supported by one heel is half the lady's weight, which works out to $\dfrac{1}{2}mg$.

Cheers,
Mr. Dychko