# 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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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.

## Recent questions and answers

7th Edition Solutions 6th Edition Solutions Global Edition Solutions

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

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

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 - 11:46Your 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.

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 - 06:04angular position should be in radians, so it's 23*2*3.14 not 23 revolutions.

Hi flsktkd, thanks for the comment. Putting angular position in radians is often the better choice, and it's the

officialbest 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 inrpm, 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 - 21:22if the elevator is falling, wouldn't H be negative?

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 levelwhen 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

## Giancoli 7th Edition, Chapter 11, Problem 4

By dshadowalker on Thu, 04/20/2017 - 10:14My answer comes out to 316N/m can you show how to input into the calculator? I like when you show the calculator because this assists me in using the calculator and making sure I have the equation right.

Hi dshadowalker, thanks for the question. I get 316 N/m too! :) The difference is that I've rounded it to two significant figures, which explains why the answer is $320 \textrm{ N/m}$. Yes, showing calcs. with the on-screen calculator is nice, but it was getting very time consuming doing that so I included it only for a few chapters.

Best wishes with your studies,

Mr. Dychko

## Giancoli 7th Edition, Chapter 21, Problem 44

By julia.wolfe on Sun, 04/16/2017 - 09:00Why don't you subtract the potential due to the inductance of the coil? I thought the minus sign in the E = -L (dI/dt) formula was meant to indicate that the voltage would be in the opposite direction following Lenz's law (current increasing > flux increasing > induced current needs to decrease flux to compensate > current flows in other direction > voltage drop in other direction). What am I missing?

Hi julia.wolfe, thanks for the question. What's important to notice is that the sign of the potential difference will be the same across the resistor as it is across the inductor. In both cases, the circuit element causes a potential

drop. (I say "circuit element" since it's OK to model this circuit as containing a single resistor and a single inductor connected by zero-resistance wires). Recall that when using Kirchoff's Laws to analyze a circuit, you make the potential across a resistor negative (it's a potential drop, in other words) when traversing the resistor in the same direction as the current. Because the inductor opposes the change in current, as you correctly mentioned, this inductor also causes a potential drop to oppose the change (note that, contrary to what you mentioned, the current does not end up flowing in the other direction. Rather, it's increase is opposed, and the inductor would disappear once the current stops changing). It would be fine to call both potential drops negative if you wish, but the question isn't specific about direction, so calling both positive is fine when answering a question asking for a "potential difference".Hope that helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 13, Problem 47

By elkinsk on Sat, 04/15/2017 - 18:32i think the temperature is 22° not 20°?

Hi elkinsk, thank you for noticing that! You're quite correct that the temperature should be $22^\circ\textrm{C}$ instead of $20^\circ\textrm{C}$. It turns out that the final answer is the same in either case since the 22 or 20 get added to 273 in order to convert to Kelvin, and the percent difference between 295 and 293 is small. I'll put a note for other students about this, and thanks again.

Best wishes with your studies,

Mr. Dychko

## Giancoli 7th Edition, Chapter 14, Problem 3

By girondebordeaux92 on Sun, 04/09/2017 - 03:19I think specific heat capacity of 4.186 is missing in the calculation? Can you please clarify

Bonjour girondebordeaux92, thanks for the question. The 4.186 figure you're quoting is, with units, $4.186 \dfrac{\textrm{kJ}}{\textrm{kg} \cdot \textrm{C}^\circ}$. Understanding units answers your question, since if you consult table 14-1 you'll notice that an alternative way to mention the specific heat of water is to write $1.00 \dfrac{\textrm{kcal}}{\textrm{kg} \cdot \textrm{C}^\circ}$. It turns out that the "kcal" way of writing specific heat, as shown in the video, if just convenient in this case since the energy of the candy bar is giving in kcal.

All the best,

Mr. Dychko

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