# 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 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

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

By cm2hn on Mon, 11/20/2017 - 20:20

Why the speed of astronaut is negative? Would I get the same result if I use the capsule as the negative value?

By Mr. Dychko on Tue, 11/21/2017 - 13:20

The only thing that matters is that you're consistent with your coordinate system. This is to say that everything pointing in a particular direction needs the same sign (be it positive or negative doesn't matter, just that they're all the same), and things pointing in the opposite direction have the opposite sign. So, yes, you could make the capsule velocity negative, keeping in mind that this now means the force exerted on the capsule by the astronaut would also be negative in that coordinate system, since it would be directed to the right just the same as the now negative capsule velocity.

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

By cm2hn on Mon, 11/20/2017 - 20:06

Why average force is based on ma*vfa and not ms*vfs?

By Mr. Dychko on Tue, 11/21/2017 - 13:17

Hi cm2hn, we could have used the space capsule final speed in part b), as you suggest, but that's taking a bit of a risk since we calculated the space capsule speed ourselves in part a), and we might have made an error in that calculation. Using the astronaut speed provided by the question is a safer approach, but outside of that, either speed is fine.