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September 10, 2015

GET FREE WAEC GCE 2015 PHYSICS PRACTICAL HERE

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3bi) diameter(d)=1.09mm ii) R=eL/A R=(5.0*10^-2*0.01)/(pie/4)*(0.001)^2 R=(5*10^-4)/(0.7855*1*10^-6) =6.37*10^2ohm P=I^2R =10^2*6.37*10^2 =6.37*10^4w 

3a)(i)d1=1.40cm, d2=1.55cm, d3=1.75cm, d4=2.1cm, d5=2.3cm Real values of di d1=1.4*0.5=0.7mm d2=1.55*0.5=0.77mm d3=1.75*0.5=0.875mm d4=2.1*0.5=1.05mm d5=2.3*0.5=1.15mm (ii) Ia1=2.4A, Ia2=3.8A, Ia3=5.0A, Ia4=7.4A, Ia5=11.6A Ib1=2.4A, Ib2=3.6A, Ib3=5.2A, Ib4=7.5A, Ib5=11.6A (iii) I=(Ia+Ib)/2 I1=(2.4+2.4)/2=2.4A I2=(3.8+3.6)/2=3.7A I3=(5.0+5.2)/2=5.1A I4=(7.4+7.5)/2=7.45A I5=(11.6+11.6)/2=11.6A 

(3aiv) logd1=log0.7=-0.15mm logd2=log0.775=0.11mm logd3=log0.875=0.06mm logd4=log1.05=0.02mm logd5=log1.15=0.06mm logI1=log2.4=0.38 logI2=log3.7=0.57 logI3=log5.1=0.71 logI4=log7.45=0.87 logI5=log11.6=1.06 (3av) TABULATE S/N; 1,2,,3,4,5 di(cm);1.40,1.55,1.75,2.10,2.30 di(mm);0.70,0.78,0.88,1.05,1.15 Ia(A);2.4,3.8,5.0,7.4,11.6 Ib(A);2.4,3.6,5.2,7.5,11.6 I(A);2.4,3.7,5.1,7.5,11.6 logd!(mm);-0.15,-0.11,-0.06,0.02,0.06 logI1;0.38,0.57,0.71,0.87,1.06 (vii)|Slope(s) =change in logI/change in logd =(0.87-0.57)/(0.02-(-0.11)) =0.3/0.13=2.3A (vii) -I will ensure the circuit is open when no readings are not taken -i will ensure tight connection bi) diameter(d)=1.09mm ii) R=eL/A R=(5.0*10^-2*0.01)/(pie/4)*(0.001)^2 R=(5*10^-4)/(0.7855*1*10^-6) =6.37*10^2ohm P=I^2R =10^2*6.37*10^2 =6.37*10^4W °°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°° 

(1a) Tabulate. Under s/n: 1,2,3,4,5 Under M(N): 140,120,110,84,66 Under tita(o): 24,32,38,51,62 Under Sin tita: 0.4067, 0.5299,0.6156, 0.7771,0.8829. Note that 1cm =20N

 (1aviii) - I would ensure that the meter rules balanced before the readings. - I would avoid error due to parallax 

(1bi) i. Total forces in one direction are equal to the forces in opposite direction. ii. The algebraic sum of the moment of all forces about any point should be zero.

 (1bii.) The moment of force at equilibrium point o is equal to, sum of clockwise moment = sum of anti- clockwise moment OC*M=CD*W. id=12660#sthash.OOwjf2WR.dpufjahub.com/answers ====================== 

Practical Answers 3bi) diameter(d)=1.09mm ii) R=eL/A R=(5.0*10^-2*0.01)/(pie/4)*(0.001)^2 R=(5*10^-4)/(0.7855*1*10^-6) =6.37*10^2ohm P=I^2R =10^2*6.37*10^2 =6.37*10^4w 3a) (i)d1=1.40cm, d2=1.55cm, d3=1.75cm, d4=2.1cm, d5=2.3cm Real values of di d1=1.4*0.5=0.7mm d2=1.55*0.5=0.77mm d3=1.75*0.5=0.875mm d4=2.1*0.5=1.05mm d5=2.3*0.5=1.15mm (ii) Ia1=2.4A, Ia2=3.8A, Ia3=5.0A, Ia4=7.4A, Ia5=11.6A Ib1=2.4A, Ib2=3.6A, Ib3=5.2A, Ib4=7.5A, Ib5=11.6A (iii) I=(Ia+Ib)/2 I1=(2.4+2.4)/2=2.4A I2=(3.8+3.6)/2=3.7A I3=(5.0+5.2)/2=5.1A I4=(7.4+7.5)/2=7.45A I5=(11.6+11.6)/2=11.6A (3aiv) logd1=log0.7=-0.15mm logd2=log0.775=0.11mm logd3=log0.875=0.06mm logd4=log1.05=0.02mm logd5=log1.15=0.06mm logI1=log2.4=0.38 logI2=log3.7=0.57 logI3=log5.1=0.71 logI4=log7.45=0.87 logI5=log11.6=1.06 (3av) TABULATE S/N; 1,2,,3,4,5 di(cm);1.40,1.55,1.75,2.10,2.30 di(mm);0.70,0.78,0.88,1.05,1.15 Ia(A);2.4,3.8,5.0,7.4,11.6 Ib(A);2.4,3.6,5.2,7.5,11.6 I(A);2.4,3.7,5.1,7.5,11.6 logd!(mm);-0.15,-0.11,-0.06,0.02,0.06 logI1;0.38,0.57,0.71,0.87,1.06 (vii)|Slope(s) =change in logI/change in logd =(0.87-0.57)/(0.02-(-0.11)) =0.3/0.13=2.3A (vii) -I will ensure the circuit is open when no readings are not taken -i will ensure tight connection bi) diameter(d)=1.09mm ii) R=eL/A R=(5.0*10^-2*0.01)/(pie/4)*(0.001)^2 R=(5*10^-4)/(0.7855*1*10^-6) =6.37*10^2ohm P=I^2R =10^2*6.37*10^2 =6.37*10^4W °°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°° (1a) Tabulate. Under s/n: 1,2,3,4,5 Under M(N): 140,120,110,84,66 Under tita(o): 24,32,38,51,62 Under Sin tita: 0.4067, 0.5299,0.6156, 0.7771,0.8829. Note that 1cm =20N (1aviii) - I would ensure that the meter rules balanced before the readings. - I would avoid error due to parallax (1bi) i. Total forces in one direction are equal to the forces in opposite direction. ii. The algebraic sum of the moment of all forces about any point should be zero. (1bii.) The moment of force at equilibrium point o is equal to, sum of clockwise moment = sum of anti- clockwise moment OC*M=CD*W 

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