# Giancoli Solutions on Video

Learn physics easily with guided practice.

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## 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)

### Giancoli 7th Edition, Chapter 8, Problem 18

By carolsilber on Wed, 05/04/2016 - 17:55

wouldnt it be 3.0x10^5 because the answer is 30,000

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

By daniel.weiss1 on Wed, 05/04/2016 - 07:22

For part b, how do we know were are looking for the radial acceleration and not the tangential acceleration?
Thanks

### Giancoli 7th Edition, Chapter 4, Problem 40

By mafiadx83 on Sat, 04/30/2016 - 06:49

why friction force equal mass * acceleration ?

By Mr. Dychko on Sat, 04/30/2016 - 21:20

Hi mafiadx83, the rule for all force questions (and this rule goes by the name Newton's Second Law) is that net force equals mass * acceleration. In this particular question there is only one force, namely friction, and for this reason the net force is made up only of the friction force. Since the friction force is the net force in this particular question, that's why friction force equals mass * acceleration for this particular question.

All the best,
Mr. Dychko

### Giancoli 7th Edition, Chapter 4, Problem 29

By mafiadx83 on Fri, 04/29/2016 - 10:45

why initial velocity equal zero ? is it suppose that final velocity equal zero because he is saying in the question what speed did the sprinter leave the starting block

By Mr. Dychko on Sat, 04/30/2016 - 21:16

Hi mafeadx83, yes, that's right. It's just implied that the sprinter is initially at rest on the starting block, and the question is asking for the speed with which the sprinter launches off the starting block after applying a force.

Best wishes,
Mr. Dychko

### Giancoli 7th Edition, Chapter 20, Problem 31

By merkinthedark on Mon, 04/25/2016 - 13:39

can you check your calculations because I get my answer in the order of 5.0437x10^-7

By Mr. Dychko on Sat, 04/30/2016 - 21:13

Hi merkinthedark, I ran the final numbers shown in the video through the calculator again and confirmed the answer, so please let me know if you find any specific error. It looks fine so far.

Best,
Mr. Dychko

### Giancoli 6th Edition, Chapter 6, Problem 10

By mikepeter on Sun, 04/24/2016 - 15:23

why does friction force opposite to the applied force?
In question 8 , applied force and friction force are in the same direction /
can you please elaborate this difference?

By Mr. Dychko on Sat, 04/30/2016 - 21:07

Hi mikepeter, thanks for the good question. A sliding friction force is always opposes motion. Sliding friction slows things down. This means the direction of a sliding friction force is in the opposite direction as the direction of motion. The direction of the applied force in problem 8, or here in problem 10, actually doesn't matter at all. It's only the direction of motion that matters in determining the direction of the sliding friction force.

All the best with your studies,
Mr. Dychko

### Giancoli 7th Edition, Chapter 19, Problem 34

By bmuniz8219 on Fri, 04/15/2016 - 14:16

Hi i have a quick question. If I1 points towards the left how does it in include 12OHMs

By Mr. Dychko on Fri, 04/22/2016 - 20:20

Hi bmuniz8219, that's a good question. Each current "starts" at a junction. While the diagram draws $I_1$ to the left along the very top, it's understood (even though not explicitly drawn) that the current originates at the junction below the $12\Omega$ resistor where the $I_2$ and $I_3$ currents also meet. So $I_1$ initially goes up through the $12\Omega$ resistor, and THEN, left along the top.

All the best with your studies,
Mr. Dychko

### Giancoli 6th Edition, Chapter 6, Problem 29

By hsumal on Wed, 04/13/2016 - 22:37

*Work-Energy Theorem