Giancoli Solutions on Video

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  • 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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  • 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.
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Sample solution

Giancoli 7th Edition, Chapter 5, Problem 12


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Recent questions and answers

Giancoli 6th Edition, Chapter 7, Problem 34

By rodri100172 on Thu, 05/17/2018 - 18:51

Thanks bro! you're a lifesaver! Say hi to Brian for me.

Giancoli 7th Edition, Chapter 3, Problem 17

By nguyen112212 on Tue, 05/15/2018 - 19:51

Hi Sr. I am wondering what happened to the other "Y" when you were solving for time. It's the part where t = square root of 2Y/AY.

By Mr. Dychko on Tue, 05/15/2018 - 21:23

Hello, thanks for the question. The "y" in the denominator is just a subscript for the "a" (acceleration), to label the acceleration as "the acceleration in the y direction". It is not a separate factor, but maybe I wrote it a bit too big. The equation is $t = \sqrt{\dfrac{2Y}{a_y}}$.

All the best,
Mr. Dychko

Giancoli 6th Edition, Chapter 5, Problem 43

By pathak_s on Sun, 05/06/2018 - 17:44

Hey why doesn't the equation for velocity (2*pi*r)/T work for this question as well?

Giancoli 7th Edition, Chapter 19, Problem 24

By aquaoasis14 on Wed, 04/18/2018 - 18:01

why did the R disappear finding the currents to the after problem?(the last problem)

Giancoli 7th Edition, Chapter 22, Problem 14

By william.g.parker on Wed, 04/04/2018 - 15:46

You wrote 0.314 as the answer for B instead of 0.341

By Mr. Dychko on Wed, 04/04/2018 - 15:52

Fixed! Thank you!

Giancoli 7th Edition, Chapter 1, Problem 28

By abgoooor201596 on Tue, 04/03/2018 - 02:10

Why your not estimating the Numbers 362 and 76 to one SIGFIG. ?

By Mr. Dychko on Tue, 04/03/2018 - 09:57

Hello, these are good questions. 365 days / year is a definition, so it's precise to three significant figures. 78 years / lifetime is an estimate, and personal opinion says that being precise to the "ones" place is appropriate, but it's just an opinion. Keep in mind that you never round numbers until the final answer. If you round before, even if a number such as 78 should be 80 since it has one significant figure, suppose, you would still nevertheless do calculations with 78, not 80. Only final answers get rounded, otherwise your calculation has an intermediate rounding error, which is to say that the calculation will be dramatically different only on account of rounding, not because any quantities have actually changed.

Hope that helps,
Mr. Dychko

By Mr. Dychko on Fri, 04/06/2018 - 19:47

Hmmm, that's a thought provoking example. The goal is always to make your calculation as precise as appropriate. Avoiding intermediate rounding error is normally part of that. However, example 1-6 illustrates that if the surrounding values have only one significant figure, as is the case with educated guesses, having only one significant figure in pi is acceptable since, in this context, greater precision isn't called for on account of the other imprecise quantities in the calculation. Think of these significant figure rules as guidelines which are open to interpretation, not as firm rules.

All the best,
Mr. Dychko

By abgoooor201596 on Fri, 04/06/2018 - 00:50

Ok That is good idea

but remember the example (Example 1–6) (How much water is in this lake?) in the book . he is rounded the value of pi to 3
is he rounded for to be 1 SIGFIG or other reason since pi is also in term of (definition) as you mentioned above ?


Giancoli 7th Edition, Chapter 1, Problem 25

By abgoooor201596 on Tue, 04/03/2018 - 01:28

hi I've a question
How we gona apply this idea with this Number 3.0*10^0
Thank You

By Mr. Dychko on Tue, 04/03/2018 - 09:52

Hi abgoooor201596, $3.0 \times 10^0$ has 2 significant figures.

Mr. Dychko

Giancoli 7th Edition, Chapter 8, Problem 41

By theovilous on Sat, 03/31/2018 - 10:31

The actual marry-go-round disk is the total weight - two persons. Which is 560-50 = 510 kg. So, Wouldn't the moment of inertia have to be (2.5)^2(510/2 + 2(25)? I don't understand why you used 560 when calculating the moment of inertia.

By Mr. Dychko on Sat, 03/31/2018 - 22:06

Hi theovilous, thanks for the question. What I'm seeing is that the text says

Assume the merry-go-round is a uniform disk of radius 2.5 m and has a mass of 560 kg, and two children (each with a mass of 25 kg) sit opposite each other on the edge.

which gives 560kg as the mass of the merry-go-round disk, not including the children. Perhaps you read that differently and interpreted that the children were included in the 560kg?

Best wishes,
Mr. Dychko

By theovilous on Sun, 04/01/2018 - 10:44

I read the question wrong! thanks for pointing that out! Thank you so much as always!!

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