# 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 57

By idan on Thu, 06/22/2017 - 14:40

I found a good explanation online so I'm all set!

### Giancoli 7th Edition, Chapter 4, Problem 57

By idan on Thu, 06/22/2017 - 14:16

How do you know the angle theta used to resolve the components is the same as the angle the incline makes with the horizontal? Please refresh my geometry!

### Giancoli 7th Edition, Chapter 16, Problem 30

By aheumangutman on Mon, 06/05/2017 - 19:27

Hi Professor Dychko,
I'm confused on why you would not solve for the x and y components of both electric fields and then take the magnitude of the net electric field.
Thanks!

### Giancoli 7th Edition, Chapter 16, Problem 30

By aheumangutman on Mon, 06/05/2017 - 19:27

Hi Professor Dychko,
I'm confused on why you would not solve for the x and y components of both electric fields and then take the magnitude of the net electric field.
Thanks!

### Giancoli 7th Edition, Chapter 28, Problem 2

By kbick on Sun, 05/07/2017 - 11:33

Hi! Just wondering when we can use the small angle approximation - you say here that it is when theta is less than zero, but how can that be?

By Mr. Dychko on Sat, 05/13/2017 - 13:03

Hi kbick, this is a good question since the small angle approximation is often a very useful trick to make the algebra much easier. Yes, I know it seems unbelievable that you can just replace $sin(x)$ with simply $x$, so what you should do to convince yourself that it's acceptable when $x$ is small (I'm saying $x$, but $\theta$, or whatever variable represents the angle in your equation) is this: plot the graph of $y=x$ and $y=sin(x)$ on the same graph, but there's a catch. Zoom in on the graph so that you can only see $x$ between, say, -1 and 1. You'll notice the graphs look the same! This is the reason the approximation works, is that within this restricted domain of $-1 \leq x \leq 1$, both functions $y=x$ and $y=sin(x)$ give the same result. When you zoom out and look at larger values of $x$, then of course the graphs do not look the same, and you can no longer claim that $x \approx sin(x)$, like you can for small $x$.

All the best,
Mr. Dychko

### Giancoli 7th Edition, Chapter 31, Problem 14

By fridley.26 on Wed, 04/26/2017 - 11:46

Your video says released and the answer says required. Just wanted to point it out was confused at first but then after watching the video I understood what happened.

By Mr. Dychko on Fri, 04/28/2017 - 12:24

Ah, thanks fridley.26! Sorry for the confusion. I've updated the quick answer to say released.

Cheers,
Mr. Dychko

### Giancoli 7th "Global" Edition, Chapter 8, Problem 19

By lina09037788 on Tue, 04/25/2017 - 06:04

angular position should be in radians, so it's 23*2*3.14 not 23 revolutions.

By Mr. Dychko on Fri, 04/28/2017 - 12:22

Hi flsktkd, thanks for the comment. Putting angular position in radians is often the better choice, and it's the official best unit for expressing angular position, that's true. However, revolutions are still used frequently since it's a unit that people can better understand. When looking at the tachometer in a car, it shows the engine speed in revolutions per minute, for example. Since this question asks for the final angular speed in rpm, it makes sense to use revolutions throughout, although it doesn't specify units for part (a) so I suppose it would be fine to use radians per second squared there.

Best wishes,
Mr. Dychko

### Giancoli 7th Edition, Chapter 6, Problem 43

By suriyak786 on Sat, 04/22/2017 - 21:22

if the elevator is falling, wouldn't H be negative?

By Mr. Dychko on Fri, 04/28/2017 - 12:14

Hi suriyak786, thanks for the question. Yes, since we haven't specified the coordinate system, it's true that up is the positive direction, and down is negative. The elevator would have a negative velocity while it's falling. However, it's initial displacement is positive since we've done one unconventional thing here: the origin (in other words the location where the vertical position is zero) is not where the elevator starts. Instead we've chosen the origin, also called the reference level when talking about vertical position, to be at the height of the fully compressed spring, which is below where the elevator starts. This choice of reference level simplified our algebra a bit. $h$ is the displacement of the elevator compared to the reference level, and it is positive since it's above the reference level, and we have the conventional coordinate system with up being positive.

All the best,
Mr. Dychko