# 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 28, Problem 8

By ZreLL on Wed, 11/28/2018 - 15:57We are missing a squared sign for C in the equation

## Giancoli 7th Edition, Chapter 8, Problem 3

By maizer01 on Sun, 11/11/2018 - 07:27wait but isn't the arclength not the diameter itself though? wouldn't you have to set it to 2pi (because you found the arclength of the crater it would be 180 degrees) and then solve for r, and follow up with multiplying it by 2?

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

By cmorales on Thu, 11/01/2018 - 14:22I have some arguments from trigonometry students and their teacher reasoning that for #24 (chapter 3) the answer would be 0. I know this text states on page 63 that Sin2theta is 1 but their reasoning being supported by the calculator is convincing. Would you be able to give more insight to this question?

Thank you.

Hi cmorales, thank you for your question. The textbook question asks "What is the maximum horizontal range of your gun?". The gun is able to shoot, since we're given data about how long a shot straight up takes to return to the gun. This data is useful to determine the initial velocity that's possible for the gun. With that initial velocity in mind, we can calculate the maximum possible range, and it will certainly be more than zero. The angle at which the range is maximum is $45^\circ$, and taking the sine of two times that gives "one", which is the greatest possible sine value of any angle. Keep in mind that we're maximizing horizontal range, not vertical height... I'm just guessing at where the misunderstanding might be...

## Giancoli 7th Edition, Chapter 26, Problem 26

By ZreLL on Thu, 11/01/2018 - 13:29My homework wanted to see the answer for part B in N*s. My homework is through Pearson Mastering Physics so, I'm not sure if this would apply to everyone.

## Giancoli 7th Edition, Chapter 16, Problem 13

By ragnarlaki on Fri, 10/19/2018 - 06:58Oh I’m sorry, I didn’t realize I had the global edition

No worries, and thanks for letting me know that solved the problem!

## Giancoli 7th Edition, Chapter 16, Problem 13

By ragnarlaki on Thu, 10/18/2018 - 09:18This is not problem 13, 7th edition

Hi ragnarlaki, thanks for the comment. This is #13 in my copy of the 7th Edition. Are you using the "Global Edition", perhaps? Solutions for that are here: https://www.giancolianswers.com/giancoli-physics-7th-global-edition-chap.... If not, I'd be curious as to what the ISBN is for your copy of the textbook.

Best wishes,

Mr. Dychko

## Giancoli 7th Edition, Chapter 5, Problem 12

By chaegyunkang on Wed, 10/17/2018 - 17:00For part c, how can you be certain to use 233.33 m/s as your velocity if this is the velocity give by the problem for the lowest point of the loop? I solved the problem by using 6.0g as my radial acceleration (i got the correct answer but I'm not sure why). Thank You.

Oops I misread the part where it said "same speed".

## Giancoli 7th Edition, Chapter 6, Problem 36

By maizer01 on Sun, 10/14/2018 - 16:23what about the 1/2's? don't they cancel?

Hi maizer01, thanks for the question. Are you referring to the 1/2 in front of each kinetic energy term in part b)? They don't cancel since 1/2 is not a common factor of

allterms. There is a potential energy term which is not multiplied by 1/2, which means 1/2 is not acommonfactor.All the best,

Mr. Dychko

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