neco physics gce alt

QUESTION 1 Solutions Reading S/n:1,2,3,4,5,6 L (cm):150.0,120.0,100.0,80.0,60.0,40.0 t (sec):4.800,4.100,3.300,2.600,1.900,1.200 t^-1(Sec^-1):0.208,0.244,0.303,0.385,0.526,0. 833 t^2(sec^2):23.40,16.81,10.89,6.76,3.61,1.44 L/t^2(m/­ s^2):6.510,7.139,9.183,11.83,16.6,27.7 ==Precautions== - I ensured the experiment was not affected by air interference when determining the balance points - I ensured the mass roll down freely as being influenced under gravity - I ensured that the readings were taken without any error.... (3ai) Experimental Results m1=35.1cm, M1=35.15cm m2=35.17cm, M2=35.46cm m3=35.21cm, M3=35.54cm m4=35.28cm, M4=35.84cm m5=35.32cm, M5=35.97cm Real values of mi's m1=(35.1* 0.1)/2 =1.755gm m2=(35.17 * 0.1)/2=1.759gm m3=(35.21 * 0.1)/2= 1.761gm m4=(35.28 * 0.1)/2=1.764gm m5=(35.32 * 0.1)/2=1.766gm Real Values of Mi's M1=(35.15 *0.1)/2=1.758gm M2=(35.46 * 0.1)/2=1.773gm M3=(35.54 * 0.1)/2=1.777gm M4=(35.84 * 0.1)/2=1.792gm M5=(35.97 * 0.1)/2=1.799gm (3aii) Values of t t1=105 t2=125 t3=155 t4=17.55 t5=205 (3aiii) Values of I I1=0.75A I2=1.00A I3=1.25A I4=1.5A I5=1.75A (3aiv) Values of d d= M - m d1= M1-m1=1.758 - 1.755=0.003gm d2=M2-m1=1.773 - 1.758=0.014gm d3=M3-m3=1.777 - 1.761=0.016gm d4=M4-m4=1.792 - 1.764=0.028gm d5=M5-m5=1.799 - 1.766=0.033gm (3av) Z=It Z1= I1T1= 0.75 * 10= 7.5As Z2= I2T2=1.00 * 12=12As Z3=I3T3= 1.25*15=18.75As Z4=I4T4=1.5 * 17.5=26.25As Z5=I5T5=1.75 * 20=35As (3vi) S/N: 1,2,3,4,5 mi(gm): 1.755, 1.759, 1.761, 1.764, 1.766 Mi(gm): 1.758, 1.773, 1.777, 1.792, 1.799 t(s): 10.000, 12.000, 15.000, 17.500, 20.000 I(A): 0.750, 1.000, 1.250, 1.500, 1.750 d(gm): 0.003, 0.014, 0.016, 0.028, 0.033 Z(As): 7.500, 12.000, 18.750, 26.250, 35.000 (3axi) 1. i will ensure tight connection to get accurate result 2. i will avoid error due to pallarax (3B-) Electromotive force is the potential difference between the terminals of a cell when no current is flowing (3bii) 1. Lead-Acid Accumulator 2. Nife- iron Accumulator 1ai) d1=5.9cm d2=4.8cm d3=3.65cm d4=3.7cm d5=1.50cm Real values of di d1=73.75cm d2=60cm d3=45.625cm d4=46.25cm d5=1.50*12.5=18.75cm (1aii) t1=5s t2=4s t3=3.5s t4=3s t5=2.5s (1aiii) W=d/t w1=d1/t1 = 73.75/5=14.75cm/s w2=d2/t2 = 60/4= 15cm/s w3= d3/t3 = 45.625/3.5= 13.04cm/s w4=d4/t4= 46.25/3 = 15.42cm/s w5=d5/t4 = 18.75/2.5 =7.5cm/s (1aiv) v=2w v1= 2*14.75 =29.50cm/s v2= 2*15= 30cm/s v3=2*13.04=26.08cm/s v4=2*15.42=30.84cm/s v5=2*7.5=15.00cm/s (1av) Tabulate Reading S/N: 1,2,3,4,5 d(cm): 5.90, 4.80, 3.65, 3.70 ,1.50 di(cm): 73.75, 60.00, 45.63, 46.25, 18.75 t(s): 5.00, 4.00, 3.50, 3.00, 2.50 w(cm/s): 14.75, 15.00, 13.04, 15.42, 7.5 (1avi) slope(s)= v2 -v1/t2 - t1 = 30.84 -30.00/3-4 =-0.84cm/s (1aviii) Slopes shows the decebration of the spherical marble (1ix) 1. i will ensure the spherical marble does not slip from the inclined plane 2. i will ensure the pile of tissue stop the spherical marble to obtain accurate result