# 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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### Cydney

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### Kim, Tunxis Community College in Farmington, CT

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 11, Problem 21

By kevin.rosario on Sat, 09/15/2018 - 23:12

I did part A on my calculator and I came up with 22.64942355. How did you come up with 2.29Hz?

Thank you.

### Giancoli 7th Edition, Chapter 4, Problem 63

By chaegyunkang on Sat, 09/15/2018 - 16:18

How do we know that the initial velocity is zero? If it was given a push I thought it would be a non-zero velocity.

By Mr. Dychko on Sat, 09/15/2018 - 20:52

Hi chaegyunkang, the presumption here is zero initial velocity, after which a force is applied in order to provide some acceleration. Bobsleds start from rest and the athletes run beside it, pushing it along, before jumping inside (which they do at the 75 meter mark in this case). I suppose some familiarity with bobsledding is called for here, but nevertheless, if an initial speed isn't given in a question, chances are it must be zero. The force applied gives extra acceleration (beyond that provided by gravity), not initial velocity.

Best wishes with your studies,
Mr. Dychko

### Giancoli 7th Edition, Chapter 4, Problem 61

By chaegyunkang on Sat, 09/15/2018 - 16:03

If the car was skidding what difference would it make to the problem? Does it change to kinetic friction when it starts to skid?

By Mr. Dychko on Sat, 09/15/2018 - 20:46

Hi chaegyunkang, it's always true that kinetic friction is less than static friction, all else being equal. In this question, however, it doesn't really matter what type of friction is mentioned, since we're not given the coefficient anyway. What matters is the deceleration the car, from which we calculate the coefficient. Be it a static or kinetic coefficient is not important. The important thing is that the type of friction on the incline is the same type of friction on the level surface so that we can be sure that whatever we calculated on the level surface will still apply on the incline.

All the best,
Mr. Dychko

### Giancoli 7th Edition, Chapter 4, Problem 55

By chaegyunkang on Sat, 09/15/2018 - 14:39

How do we know that it is a frictionless ramp? Is it because the question does not give us the coefficient of friction so we assume that it is frictionless?

By Mr. Dychko on Sat, 09/15/2018 - 20:41

Hi chaegyunkang, thanks for the question. Since the question has a truck driving on the ramp, we assume there is only rolling friction, which we assume is negligibly small. There is no sliding, so there's no kinetic friction. While there is static friction, since we assume the brakes are not being applied (since the ramp is used when the brakes have failed), this means the tire is not applying a horizontal force due to static friction, and so the pavement is not in turn applying a Newton's 3rd law reaction force on the truck. In other words, static friction is not slowing down the truck. There is only rolling friction, which we can ignore.

Hope this helps,
Mr. Dychko

### Giancoli 7th Edition, Chapter 2, Problem 42

By EddieG on Wed, 09/05/2018 - 18:50

Question #39 a ball is thrown straight up with a speed of 36 m/s. how long does it take to return to its starting point. Question is my teacher teaches out of 7th edition I have 6th edition but can not find this question in my book or on this website.It says on top of my page ch# 2 and page ref 2-7 but cant find it also how come you don't post question you just go to answers only . need to see question sometimes so I can find which edition it is in.

By Mr. Dychko on Wed, 09/05/2018 - 21:00

Hi EddieG, thanks for the question. I would love to post the questions, but I've avoided that since it would be a copyright issue with the publisher since I didn't create the questions. Perhaps you can ask your teacher to photocopy just the problems from their 7th Edition text for your reference?
All the best,
Mr. Dychko

### Giancoli 6th Edition, Chapter 19, Problem 32

By thesouthportschool on Mon, 08/27/2018 - 18:10

this is epic

### Giancoli 7th "Global" Edition, Chapter 3, Problem 16

By hugokatich on Mon, 08/20/2018 - 19:43

Also, what rule did you use for the triangle in part b? Which rule indicates that Vrx = V1x + V2x?

By Mr. Dychko on Fri, 08/24/2018 - 22:51

Thank you for the question. Adding vectors using components is one of the methods for adding vectors. The component method, which is probably the most popular method to use, involves finding the sum of the component of each vector along an axis, and then taking the total as the component of the resultant along that axis. Vrx = V1x + V2x is just an algebraic way of saying that.

All the best,
Mr. Dychko

### Giancoli 7th "Global" Edition, Chapter 3, Problem 16

By hugokatich on Mon, 08/20/2018 - 19:39

Where did the answer of 62.2 units come from? Given that the square root of 90^2 - (-65^2) is equal to 111.02?

By Mr. Dychko on Fri, 08/24/2018 - 21:40

Hi hugokatich, thank you for the question. It's important to notice the minus sign in the expression there, which doesn't go away after squaring -65. Yes, -65 squared becomes positive, but the expression is for finding a leg of a right triangle, not for finding the hypotenuse, so the expression has a minus between the squared terms. $\sqrt{90^2-(-65)^2} = 62.2$.

Best wishes,
Mr. Dychko