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Essay heading: Science
 
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Issue: Biographies
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Date added: June 24, 1999
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Record this measurement. Data Part A: Mass of weight (m-2) = 100 grams Position string balanced = 36.4 cm Distance from center of meter stick to balance point. (L-1) = 13.6 cm Distance from balance point to suspended weight. (L-2) = 26.4 cm Mass of meter stick. (at center gravity) m1 = m2 (L1/ L2) Therefore: m1 = 100 (26...
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(at center gravity) m1 = m2 (L1/ L2) Therefore: m1 = 100 (26.4/13.6) m1 = 100(1.94111) m1 = 194.1176 grams (mass of the meter stick) Data Part B: Found natural torque (off set support string) = t = fl 85 grams placed at 100 cm balanced the off set support string at 65 cm. Therefore: t = 85 * (100 ? 65) t = 2975 Total torque of right side of support string: t = 90cm ? 65cm (500 g) t = 12,500 Then we calculated the left side torque: t = 65cm ? 40cm (100g) t = 2500 Then we took the right torque and subtracted the left torque: 9525 ? 2500 = 7025 (this is the missing force on the left side) Missing torque 7025 = 50cm ( ? ) 7025/50 = 140...
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