HW #4-2-solutions

HW #4-2-solutions - pasha(sep635 – HW 4-2 – Antoniewicz...

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Unformatted text preview: pasha (sep635) – HW 4-2 – Antoniewicz – (56445) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 8.0 points If a chain of 30 identical short springs linked end to end has a sti ff ness of 260 N / m, what is the sti ff ness of one short spring? Correct answer: 7800 N / m. Explanation: Springs in series have the e ff ective sti ff ness K , where K is given in terms of the sti ff nesses of the individual springs as follows: 1 /K = 1 /k 1 + 1 /k 2 + · · · + 1 /k N . For identical springs, we have k = NK , where k is the sti ff ness of a single spring. Therefore k = (30)(260 N / m) = 7800 N / m . 002 8.0 points You place 50 identical springs side by side (in parallel) and connect them to a large, massive block. The sti ff ness of the 50-spring combi- nation is 20790 N / m. What is the sti ff ness of one of the individual springs? Correct answer: 415 . 8 N / m. Explanation: Springs in parallel have an e ff ective sti ff ness K , where K is given in terms of the sti ff nesses of the individual springs as follows: K = k 1 + k 2 + · · · + k N . For identical springs, we have K = Nk , or k = K/N , where k is the sti ff ness of a single spring. Therefore, k = 20790 N / m 50 = 415 . 8 N / m . 003 (part 1 of 4) 3.0 points One mole of tungsten (6 . 02 × 10 23 atoms) has a mass of 184 g, and its density is 19 . 3 g / cm 3 , so the center-to-center distance between atoms is 2 . 51 × 10- 10 m. You have a long, thin bar of tungsten, 2 . 5 m long, with a square cross section, 0 . 18 cm on a side. You hang the rod vertically and attach a 470 kg mass to the bottom, and you observe that the bar becomes 1 . 3 cm longer. From these measurements, it is possible to determine the sti ff ness of one interatomic bond in tungsten.ness of one interatomic bond in tungsten....
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This note was uploaded on 11/17/2011 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.

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HW #4-2-solutions - pasha(sep635 – HW 4-2 – Antoniewicz...

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