# 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 14, Problem 26

By chaegyunkang on Wed, 03/07/2018 - 18:41I am very confused about how the water can evaporate in room temperature. I know that the water cannot reach 100 celsius from our skin but how can water just evaporate in room temp?

Hi chaegyunkang, nice question! It turns out that temperature is an

averagekinetic energy of the particles of the substance. Water at $100^\circ \textrm{ C}$ has molecules with an average kinetic energy sufficient to turn into gas, so many of them do really quickly, and we observe this as boiling. Water at, say, $20^\circ \textrm{ C}$alsohas some molecules with enough kinetic energy to turn into gas, but not very many, so we don't see anything special such as bubbles or steam. Nevertheless those few molecules with enough energy to do so turn into gas. This is what's responsible for evaporation, and eventually the luke warm water will all turn to gas for this reason.Even ice evaporates! A full ice cube tray in your freezer will eventually have much smaller ice cubes after months since some of the molecules have enough kinetic energy to turn into a gas, even though the average kinetic energy is such that the molecules are in the ice phase.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 8, Problem 58

By aquaoasis14 on Sun, 03/04/2018 - 19:42why did you multiply it by 4?

Hi aquaoasis14, thanks for the question. At the 2:00 minute mark in the video, I multiply by 4 since 4 is the lowest common denominator of the fractions in the equation. Doing this gets rid of the fractions since, as a matter of personal opinion, they're annoying! It would be possible to solve the equation just fine while keeping the fractions, if you prefer. The last term in the equation has 1/2 multiplied by 1/2 (making 1/4), keep in mind, so that's why I needed to multiply by 4 instead of 2.

Hope that helps,

Mr. Dychko

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

By chaegyunkang on Thu, 03/01/2018 - 16:50If the question asked for amount of water, is it okay to end with kg of water? Thank You.

Hi chaegyunkang, it looks to me like the question is asking for mass per time, which means there needs to be units of mass divided by time, which could be kg / hr, but not just kg.

All the best,

Mr. Dychko

## Giancoli 6th Edition, Chapter 21, Problem 16

By mbnoroozi on Sun, 02/25/2018 - 14:51Hi Dr. Dychko - I might be missing something obvious but can you explain why the final answer is negative?

Thanks!

Hi mbnoroozi, thanks for the question. The final answer I have is positive, but I think you're referring to the negative in the formula for the induced EMF? That negative is there to remind us that the induced EMF causes a current directed such that it produces a magnetic flux in the coil that opposes the change in flux that caused the EMF in the first place. Negative in this case can be thought of as "opposite to", in the same way as a car going in the opposite direction to a car with velocity 30 m/s will have a velocity -30 m/s.

Hope this helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 4, Problem 51

By theovilous on Fri, 02/16/2018 - 07:47In this problem, can I use the formula, Ffr = μk*Fn? I calculated μk to be .2. So the answer should be .2*60kg*9.8m/s2 = 117.6N. Is this just another way of solving the problem or did I get it wrong? Thank you!!

Hi theovilous, ah, now I understand your approach. Yes, you can do what you're suggesting. I suspect that if you keep lots of digits in your answer for $\mu_k$ (to avoid intermediate rounding error), that your answer will match the answer here exactly to three significant figures. The approach you're suggesting takes more steps, but it's perfectly valid.

I'm glad you're enjoying the solutions!

All the best,

Mr. Dychko

Hi theovilous, thanks for the question. While $F_{fr} = \mu_k F_n$ is a correct formula, I'm not sure how you calculated $\mu_k$? It doesn't look possible with the information given, so I think using acceleration is the only method available for solve this one.

All the best,

Mr. Dychko

I considered retarding force as friction. So I set the formula Ffr = ma --> μkmg=ma --> μk = a/g --> μk = 2/9.8 = .2.

So based on the formula, Ffr=μkFn, Ffr=μkmg = .2 * 60kg * 9.8m/s2 = 117.6N.

Does this approach work as well? my result is very close to your answer which is 120N.

Thank you for your quick response! I am learning a lot from you!

## Giancoli 7th Edition, Chapter 8, Problem 59

By choianchoi on Sun, 01/21/2018 - 17:02Why is kinetic energy not accounted just before it hits the ground? Also, is the velocity of the center of mass the same with the end of the pole?

Hi choianchoi, thanks for the question. Let's keep in mind that one end of the pole stays in contact with the ground. It's a pole falling over, in other words. The end touching the ground has velocity zero in that case, at all times. The other end of the pole, on the other hand, has the maximum velocity of any point on the pole. The end initially in the air has a higher linear velocity than the center of mass. Kinetic energy is accounted for, and it's the strategy for calculating the velocity of the end of the pole, but the kinetic energy formula doesn't look like what you're used to from linear problems. We don't use $\dfrac{1}{2}mv^2$. Instead, we use rotational kinetic energy $\dfrac{1}{2}I\omega^2$ instead, and use that to figure out the rotational velocity, which is then used to find the linear velocity of the end of the pole.

All the best,

Mr. Dychko

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

By samkvertus on Thu, 01/18/2018 - 20:54amazing

Thanks!

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

By samkvertus on Fri, 01/12/2018 - 11:20what about the last part of the problem that asked to find the magnitude and direction ?

Hi samkvertus, thanks for the question. Both the magnitude and directly are actually given here. Direction in this case is "to the left" when the resultant is negative, whereas it's "to the right" when positive. The magnitude is the number when ignoring the negative sign. This is explained more in the video, so consider giving it a second view.

All the best,

Mr. Dychko

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