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Unformatted text preview: Part 2: We balanced a meter stick with 2 given weights hanging on the left side and a variable on the other. First we found what position the variable weight at 30g had to be in to balance the meter stick. Then we moved the meter stick left 10 cm and repeated the process with a 40g weight. Third we figured out the weight of an unknown by balancing the meter stick and using the position the weight was in to calculate its mass in grams. Data: Part 1 Experimental: 1. 220g @ 293 degrees 2. 500g of force required to stop acceleration Part 2 Experimental 1. 92.4 cm 2. 86 cm 3. Position of ball = 21.2cm Part 2 Mathematical 1. x(.04825kg)=0.25m(.03825kg)+0.40m(.02825kg) 0.87% error x=43.2m + 50m = 93.2m 2. x(.05825kg)=0.25m(.03825kg)+0.40m(.02825kg) 0.23% error x=35.8m + 50m = 85.8m 3. 0.212m(.14825kg)=0.13148kg+0.25x x=0.073kg (weight of ball)...
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This note was uploaded on 04/21/2011 for the course PHY 3240 taught by Professor Moore during the Spring '11 term at W. Florida.
 Spring '11
 Moore
 Force, Static Equilibrium

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