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ANSWERNO:1(1aviii) time taken for m to reach the ground, t=35.23S (1aix) t^2=(35.23)^2=1241.15(S)^2 (1x) a=2h/t^2 a=((2*1.4)/1241.15) a=0.0023m/s^2 T=(m/1000)(10-a) T=0.05*9.9977 T=0.50S alpha=a/R=(0.0023/(6.55*10^-2) alpha=0.233/6.55 alpha=0.035m^2/S^2 (1axi) Tabulate: SN: 1,2,3,4,5 m(g): 50|70|90|110|130 t(s): 35.23|30.10|26.15|21.70|18.50 t^2(s^2): 1241.15|906.01|683.8225|470.89| 342.25| a(m/s^2): 0.0023|0.0031|0.0041|0.0059 Tabulate: T(s): 0.50|0.70|0.90|1.01|1.30 alpha(m^2/s^2): 0.035|0.047|0.063|0.090|0.125 (1xiv) slope(s)= ((0.090-0.047)/(1.01-0.70) =0.043/0.31 =0.139m^2/s^3 (xv) I=R/s= (6.55*10^-2)/(0.139) I=0.47m/s^2 XVI) (i) I ensured the pulley is attached firmly to the pin and the retort stand to avoid loosing of the set up (ii) I ensured the height is maintained constant throughout for accurate results(1bi) Centripetal force is defined as the force which helps to keep an object moving in a circular path continuously (1bii) Given: height(h)=2.0m Speech(v)=? g=10m/s^2 Initial velocity(u)=0 V^2=U^2 2gh V^2=2*10*2 V^2=40 V=root40 =6.32m/s (3aiv) voltmeter reading E=2.0V Tabulate SN: 1,2,3,4,5,6 R(ohms): 1.00|2.00|3.00|4.00|5.00|6.00 R^-1(ohm ^-1): 1.00|0.50|0.33|0.25|0.20|0.17 V(v): 1.80|2.10|2.30|2.40|2.50|2.60 V^-1(V^-1): 0.56|0.48|0.43|0.42|0.40|0.38 PLS NOTE THAT: ^ MEANS RAISE TO POWER ohms looks like U written upside dwn ie resistance symbol. (3aix) slope= ((0.56-0.38)/(1.00-0.17)) =0.18/0.81=0.22 ohm V^-1 intercept on vertical axis 0.35V^-1 (3ax) value of c^-1= c^-1=1/c c^-1=1/0.35 c^-1=2.857 (3axi) (i)I ensured that the connecting wires are tightly connected (ii)i avoided error due to parallax when taking the voltmeter by ensuring i viewed the scale directly in front of it(3b) Electromotive force(E.M>F) is more than the terminal pd when current is supplied from external resistance which drop the e.m.f of the cell
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