# Giancoli Solutions on Video

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## 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 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.
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• Pause, rewind, repeat, and never miss what is being said.

## Sample solution

Giancoli 7th Edition, Chapter 4, Problem 62

(1:36)

### Giancoli 7th Edition, Chapter 16, Problem 12

Hi dewald, thanks for spotting that, and you're quite right. I've updated the quick text answer to reflect the correct answer in the video.

All the best,
Mr. Dychko

### Giancoli 7th Edition, Chapter 16, Problem 12

I think the third charge should also be 2*10^2, typo probably?

### Giancoli 7th Edition, Chapter 9, Problem 12

Hello. There is a mistake in the snippet answer. F2 should be, indeed, 2900N.

Moreover, I wanted to know why you decided to take equation 2 and divide it by equation 1? If I am not mistaken, you already all have all of the information that you need to find F1 with equation 2, given that theta=33, mass m=190kg. All you would need to find F2 is to plug in F1 into equation 1. Just a thought...

### Giancoli 6th Edition, Chapter 6, Problem 63

Hi thesouthportschool,

Strictly speaking, yes, I suppose $v_f - v_i$ should be $65 - 85$. The resulting negative sign would correctly indicate that the force on the car (ie friction) while coasting is in the opposite direction to it's initial velocity, which would be taken to be positive (the car is slowing down in other words). The force applied by the car will be in the opposite direction when it's travelling at a constant velocity, and so the force applied by the car will be in the same direction as the velocity, and therefore positive. In the end, the car's applied force will be positive, which is what $85 - 65$ will give, but perhaps I should have arrived at a positive sign with this explanation.

Hope this helps,
Mr. Dychko

### Giancoli 7th Edition, Chapter 2, Problem 23

Hi Anne-Lelin,

Thanks for your comment, and you're quite right that $\dfrac{14 + 21}{2} \times 6.0 = 105$. $105$ would be the correct answer in a mathematics class. However, since this is physics, we need to account for significant figures since the values are not infinitely precise as they would be in math class. The sum of the two values in the numerator will be precise to the "ones" place, giving $14 + 21 = 35$, which has two significant figures. This is then divided by a "mathematical factor" $2$. I call this a "mathematical factor" since it isn't measured, so it does in fact have infinite precision. This results in $\dfrac{35}{2} = 17.5$, which has only two significant figures. We won't round it yet, however, since rounding happens only in the final answer, and we'll just keep the fact of 2 significant figures in our mind. Rounding now would be called an "intermediate rounding error" since this is an intermediate calculation done in the process of getting the final answer. The next calculation is $17.5 \times 6.0 = 105$, as you mentioned. And here's the real point: $105$ should have only 2 significant figures, which rounds to $110 \textrm{m/s}$.

I know that's a long answer, but hopefully it helps illustrate the process of considering significant figures in physics calculations.

All the best with your studies,
Mr. Dychko