Giancoli 6th Edition, Chapter 10, Problem 19

Problem 19

1.41 \times 10^5 \textrm{N/m}^2

9.8 \times 10^4 \textrm{N/m}^2

Note: For part b), while the working in the video is correct, I forgot to include the

\times 10^3

after the

13.6 \textrm{ kg/m}^3

when plugging into my calculator for the final answer. The final part b) answer written in the quick answer above is correct, not the final part b) answer in the video.

Giancoli 6th Edition, Chapter 10, Problem 19 solution video poster

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COMMENTS

By monkboy11 on Tue, 6/14/2011 - 11:37 PM

Your answer for part b is incorrect. I think you inputted the rhoHg incorrectly.

(13.6X10^3)(9.8)(-0.042) = 98402.24 N/m^2

(13.6)(9.8)(-0.042) = 103994.40 N/m^2

Methodology is correct though! =) Thanks so much for doing this!!

By mireyaluis on Wed, 12/12/2012 - 6:42 AM

I agree! Part b was done incorrectly. Please fix whenever possible.

By rohanblazers on Tue, 9/13/2016 - 3:18 AM

The textbook denotes the answer as 9.8 x 10^4 Pa

By Mr. Dychko on Wed, 9/14/2016 - 7:27 AM

Hi rohanblazers, thanks for reminding me about this error. You're correct that the part b) answer is $9.8 \times 10^4 \textrm{ Pa}$ . I'm away from my recording equipment at the moment, but I'll make a note in the quick answer that the working in the video is correct, but I forgot to include the $\times 10^3$ after the $13.6 \textrm{ kg/m}^3$ when plugging into my calculator for the final answer.

All the best,
Mr. Dychko

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