361_Physics ProblemsTechnical Physics

361_Physics ProblemsTechnical Physics - Chapter 12 P12.34...

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Chapter 12 363 P12.34 Let the 3.00 kg mass be mass #1, with the 5.00 kg mass, mass # 2. Applying Newton’s second law to each mass gives: ma T mg 11 =− (1) and ma mg T 22 (2) where T is the tension in the wire. Solving equation (1) for the acceleration gives: a T m g 1 , and substituting this into equation (2) yields: m m TmgmgT 2 1 −=− . Solving for the tension T gives T mmg mm = + == 2 2 3 00 5 00 9 80 800 36 8 12 21 ... . . kg kg m s kg N 2 bg ej . From the definition of Young’s modulus, Y FL AL i = af , the elongation of the wire is: L TL YA i ×× = 36 8 2 00 2 00 10 2 00 10 00293 11 3 2 .. . N m Nm m mm 2 a fa f e j π . P12.35 Consider recompressing the ice, which has a volume 1 09 0 . V . PB V V i F H G I K J = −× 200 10 0090 109 165 10 9 8 . . 2 2 *P12.36 B P PV V V V i i (a) V PV B i × × 113 10 1 021 10 00538 8 10 . . . Nm m m 23 2 3 (b) The quantity of water with mass 1 03 10
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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