CURRENT EXAMINATION: WAEC GCE PHYSICS PRACTICAL; (1i.)Experimental Result h1=5.1cm, h2=4.25cm, h3=3.65cm, h4= 2.95cm, h5=1.7cm h1=0.051m, h2= 0.0425m, h3=0.0365m, h4=0.0295,h5=0.017m Real values of h h1=0.2295m, h2=0.19125m, h3=0.1643m, h4=0.13275m, h5=0.0765m (1ii.) first Set L1=2.7cm, L2=2.9cm, L3=3.1cm, L4=3.4cm, L5=3.55cm Second Set L1=2.8cm, L2=2.9cm, L3=3.0cm, L4= 3.15cm, L5=3.6cm (1iii) Mean Value L1 = 2.7+2.8/2 = 2.75cm L2= 2.9+ 2.9/ 2 = 2.9cm L3 = 3.1 +3.0/2 = 3.05cm L4 = 3.4+3.15/2 = 3.275cm L5= 3.5+3.6/2= 3.575cm (1iv) e= Lo - Lmean e1= 6 - 2.75= 3.25cm e2= 6 -2.9 = 3.1cm e3= 6 - 3.05= 2.95cm e4= 6 - 3.275 = 2.725cm e5= 6 - 3.575 = 2.425cm e1^2= 10.56cm^2, e2^2= 9.61cm^2, e3^2 = 8.7025cm^2, e4^2 = 7.426cm^2, e5^2= 5.881cm^2 Quote NOTE THAT ^ MEANS "RAISE TO POWER SIGN" (1v) Tabulate SN: 1, 2,3,4,5 h(m): 0.2295,0.19125,0.1643,0.1327,0.0765 h(cm): 22.95,11.13,16.43,13.28,7.65 L(cm): 2.70,2.90,3.10,3.40,3.55 L(cm): 2.80,2.90,3.00,3.15,3.60 e(cm): 3.25,3.10,2.95,2.73,2.43 e^2(cm): 10.56,9.61,8.70,7.43,5.88 (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. (1bi) 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 (1bix)*I would avoid error due to parallax when taking readings on the meter rule. *I would ensured that the mass is released from an accurate height. (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 Quote NOTE THAT U ARE TO DRAW A TABLE AND USE DIS GUIDELINE I GAVE YOU IN NUMBER 3. (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 (3) (3i.) lo = 0, l1= 0.5A, l2= 1.0A, l3= 1.5A, l4=2.0A, l5=2.5A, l6=3.0A (3ii.) Vo= 1.3v, V1= 1.28v, V2=1.25v, V3=1.23v, V4=1.2v,V5=1.17v, V6=1.15v (3iii.) Tabulate SN: 0,1,2,3,4,5,6 V(v): 1.30,1.28,1.25,1.23,1.20,1.17,1.15 l(A): 0.00,0.50,1.00,1.50,2.00,2.50,3.00 Quote =============== NOTE THAT comma means NEXT LINE, Draw a table, under I, put 0.50, 1.00, 1.50 etc.. ================ (3viii) (a) i will ensure that the key is opened when no reading is taken to avoid battery wastage (b) i will ensure that the circuit element are tightly connected to ensure accurate result (3bi) i. Temperature ii. Nature of material iii. cross-Sectional Area iv. Length of conductor (3bii) Electromotive force is the bpotential difference between the terminals of a cell when no current is flowing in the circuit. GOODLUCK!!!!!!!!... NEXT EXAMINATION: WAEC GCE English Language; Sat. 7th Sept 2013; English Lang. (Essay) = 9am - 11:30am English Language (Obj) = 2:30pm - 3:45pm English Lang. (Test Of Oral) = 4pm - 4:45pm =============== NOTE: Our Direct Mobile Subscribers Always Get All Solved WAEC GCE Answers 15mins Immediately Exam Starts. Call 08148898459
Posted on: Thu, 05 Sep 2013 15:07:16 +0000
Recently Viewed Topics
© 2015