Exam_3_form D Solutions

23 109 n 10 a force of 250 n is applied to a

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Unformatted text preview: 1.6 × 108 N/m 2 A C) 4.80 × 1012 N G) 8.70 × 105 N F < (1.6 × 108 N/m 2 )(7.70 × 10−4 m 2 ) = 1.23 × 105 N D) 3.00 × 103 N H) 4.23 × 109 N 10. A force of 250 N is applied to a hydraulic jack piston that is 0.02 m in diameter. If the piston that supports the load has a diameter of 0.15 m, approximately how much mass can be lifted by the jack? Ignore any difference in height between the pistons. A) 1400 kg E) 5600 kg 2 B) 700 kg F) 3300 kg ⎛ A2 ⎞ ⎛ d2 ⎞ C) 250 kg G) 990 kg P = P2 , F1 / A1 = F2 / A2 , F2 = ⎜ ⎟ F1 = ⎜ ⎟ F1 1 ⎝ A1 ⎠ ⎝ d1 ⎠ D) 2800 kg H) 1700 kg 2 2 F2 ⎛ d2 ⎞ F1 ⎛ 0.15 ⎞ 250 N m= = = = 1400 kg g ⎜ d1 ⎟ g ⎜ 0.02 ⎟ 9.81 N/kg ⎝⎠ ⎝ ⎠ 2 Form D Exam 3...
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This document was uploaded on 03/03/2014 for the course PHY 231 at Michigan State University.

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