# 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.
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## Sample solution

Giancoli 7th Edition, Chapter 4, Problem 62

(1:36)

### Giancoli 6th Edition, Chapter 8, Problem 46

nevermind
- kind regards sir
william "tssacy" schwazz

### Giancoli 6th Edition, Chapter 8, Problem 46

shouldn't you use I = 2/5Mr^2 because it's a sphere?

### Giancoli 6th Edition, Chapter 8, Problem 43

Isn't it 1.4x10^3?

### Giancoli 7th Edition, Chapter 13, Problem 1

Hi Greg, thank you for your question. I think the difference between what I've done here and what is common in chemistry is that here I've used "atomic mass", whereas in chemistry "molar mass" is more common. When looking at the periodic table of elements, the mass given for each element can actually be interpreted either way. For gold, you can say it has an atomic mass (technically, this is a "relative atomic mass" which takes an average of all naturally occurring isotopes weighted by how common each isotope is) of $196.966569 \textrm{ u/atom}$ or you can read it as $196.966569 \textrm{ g/mol}$, depending on your preference. Since I went with u/atom, I needed a conversion factor to turn the atomic mass unit (that's the "u") into kilograms, so that the "kg" in the numerator would cancel with "kg" in the denominator. The is the number you asked about: $1.66 \times 10^{-27} \textrm{ kg/u}$, which is just a constant you can look up in the front cover of the textbook under the section "Fundamental Constants" and it tells you how many kilograms are in one atomic mass unit.

I can't really say why your approach didn't work since I don't understand what you mean by "I took it through the mole". In any case, the chemistry approach definitely works. If the video took that approach, the atomic mass would be replaced with "molar mass", which is the same number but just different units, and then divide it by Avogadro's number.

Probably the confusing thing about this question is having fractions within fractions. Perhaps I should have written it on one line by multiplying by the reciprocal of the denominator. When checking your work make sure you're putting things in the right place within the fractions.

Hope this helps,
Mr. Dychko

### Giancoli 7th Edition, Chapter 13, Problem 1

Hi, The process on this is a little confusing to me. In chemistry we had similar problems and we found them through a similar factor label method. I took 27.5 g of gold I took it through the mole, and divided by the atomic mass (199.9665). Then I multiplied by Avogadro number 6.022 *10^23. I did the same thing for the other then divided, however I got a very different answer. How come this way does not work? Also where did you get the 1.66*10^-27 kg/u?

Thank you!
Greg.