If the force F needed to move the wire in Fig. 10–34 is 3.4 x 10^{-3} N , calculate the surface tension \gamma of the enclosed fluid. Assume l = 0.070 m .

This is Giancoli Answers with Mr. Dychko. With this apparatus surface tension is force divided by 2l and this 2 is here because there's both a top and a bottom surface in that apparatus in figure [10-34]. And the force is 3.4 times 10 to the minus 3 newtons and divide that by 2 times 0.070 meters and that's 0.024 newtons per meter is the surface tension.

Find us on:

Giancoli Answers, including solutions and videos, is copyright © 2009-2024 Shaun Dychko, Vancouver, BC, Canada. Giancoli Answers is not affiliated with the textbook publisher. Book covers, titles, and author names appear for reference purposes only and are the property of their respective owners. Giancoli Answers is your best source for the 7th and 6th edition Giancoli physics solutions.

If the force FFF needed to move the wire in Fig. 10–34 is 3.4×10−3 N3.4 \times 10^{-3} \textrm{ N}3.4×10−3 N, calculate the surface tension γ\gammaγ of the enclosed fluid. Assume l=0.070 ml = 0.070 \textrm{ m}l=0.070 m.

Find us on:

If the force $F$ needed to move the wire in Fig. 10–34 is $3.4 \times 10^{-3} \textrm{ N}$ , calculate the surface tension $\gamma$ of the enclosed fluid. Assume $l = 0.070 \textrm{ m}$ .