(1ix) a. i will ensure the spring is not slanted to avoid error of reading length L b. i will avoid error of parallax due to reading the instrument. (1b) Given, Force constant(K)= 300N/ m Extension(e)= 3.0cm = 0.03m Energy stored (E)= 1/2Ke^2 E= 1/2 *300*0.03*0.03 E = 150*0.03*0.03 E =0.135J (1bii) F1/L1-Lo = F2/L2-Lo = F3/L3-Lo 5/12-8 = 10 / 30-12 = F/30-12 5/4 =F/18 F = 5*18/4 = 22.5N (2a) sn:1,2,3,4,5,6 M/g:20,30,40,50,60,70 tita(degree celcius) :39.2,41.6,43.2,44.2,45.4,45.8 xi(degree) :14.2,16.6,18.2,19.2,20.4,20.8 yi(degree):10.8,8.4.6.8,5.8,4.6,4.2 Ti:1.31,1.98,2.68,3.31,4.43,4.95 (2v) slope= deltay/delta x =2.68-1.31/40-20 =0.69 (2vi) -i would ensure that parallax errors were avoided -i would ensure that i continuously stir the iron fillings (2b) -it means that iron requires 470 joules of heat energy to raise 1kg of iron by 1 degree kelvin of temperature
Posted on: Thu, 05 Sep 2013 14:56:56 +0000
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