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Wednesday 4 April 2018

Physics alt practical now posted

[4/4, 9:10 AM] Dito Dito: Phy - Pract - Answers
1 )
Tabulate
s /n = 1 , 2 , 3 , 4 , 5
L ( cm )= 130 , 110 , 90 , 70 , 50
T 1 ( sec)= 45 . 70 , 42 . 00 , 38 . 00 , 33 . 10 , 28 . 70
T = 2 . 29 , 2 . 10 , 1 . 90 , 1 . 66 , 1 , 44
T ^ 2 = 5 . 22 , 4 . 41 , 3 . 61 , 2 . 74 , 2 . 06
L = 100 , 80 , 60 , 40 , 20
1 )
- precautions -
Take Any Two
i ) I avoided conical oscillation
ii ) I avoided drought
iii ) I avoided parallax error in taking reading from the metre rule
iv ) I avoided zero error in taking reading from the stop watch
pls keep calm we will update
the rest soon
============================
3 )
Tabulate
3 )
- precautions -
take any two
i ) I removed key when reading is not taking
ii ) I ensured tight connections at the terminals
iii ) I avoided parallax error when taking reading from the ammeter.
============================
[4/4, 9:11 AM] ‪+234 813 649 3840‬: 1b i.
focal length of an object is 20means that the distancebetween the optical centre  or middle of the lens and its principal  focus is 20cm

1bii.
given that u=x   f=20   v=5

using the mirror formula
1/u   +  1/v  = 1/F

1/x + 1/5 = 1/20

multiply through by the LCM
1/x *5  + 1/5 * 5 = 1/20 * 5

5/x =5/20
5/x =1/4
taking the reciprocal
x/5 =4/1

cross multiply
x * 1 = 5 * 4
x =20
[4/4, 9:11 AM] Dito Dito: 1b i.
focal length of an object is 20means that the distancebetween the optical centre  or middle of the lens and its principal  focus is 20cm

1bii.
given that u=x   f=20   v=5

using the mirror formula
1/u   +  1/v  = 1/F

1/x + 1/5 = 1/20

multiply through by the LCM
1/x *5  + 1/5 * 5 = 1/20 * 5

5/x =5/20
5/x =1/4
taking the reciprocal
x/5 =4/1

cross multiply
x * 1 = 5 * 4
x =20
[4/4, 9:12 AM] Dito Dito: 2aix)
*PRECAUTIONS*
I)  experiment must be carried out in a dark room
II)  lens, screen and object must be in line
[4/4, 9:27 AM] Dito Dito: Phy - Pract - Answers

1axi)
- *Precautions* -
Take Any Two

i ) I avoided conical oscillation
ii) I avoided drought
iii ) I avoided parallax error in taking reading from the metre rule
iv ) I avoided zero error in taking reading from the stop watch

1b i.
focal length of an object is 20means that the distancebetween the optical centre  or middle of the lens and its principal  focus is 20cm

1bii.
given that u=x   f=20   v=5

using the mirror formula
1/u   +  1/v  = 1/F

1/x + 1/5 = 1/20

multiply through by the LCM
1/x *5  + 1/5 * 5 = 1/20 * 5

5/x =5/20
5/x =1/4
taking the reciprocal
x/5 =4/1

cross multiply
x * 1 = 5 * 4
x =20

3aix)
- *precautions*-
take any two
i ) I removed key when reading is not taking
ii) I ensured tight connections at the terminals
iii ) I avoided parallax error when taking reading from the ammeter .

3bii)  I) length
        II) Thickness

More loading...
E
TYPING.....
[4/4, 9:29 AM] Dito Dito: WAEC Physics Practical Answers.
(1)
TABULATE:
i:1,2,3,4,5,
L(cm):80.00,70.00,60.00,50.00,40.00
t1(s):17.87,16.22,16.09,14.08,13.28
t2(2):17.56,16.46,15.09,14.77,13.36
t(s)mean:17.715,16.340,15.590,14.425,13.320
T(t/10)s:1.7715,1.6340,1.5590,1.4425,1.3320
LogT(s):0.2483,0.2133,0.1928,0.1591,0.1212
Log L(cm):1.9031,1.8451,1.7782,1.6990,1.6021
Log T*10^-2(s):24.83,21.33,19.28,15.91,12.12
Log L*10^-1(cm):19.031,18.451,17.782,16.990,16.021
SLOPE(s)=(LogT2*10^-2 – LogT1*10^-2)(/Log L2 *10^-1 – LogL1*10^-1)
=(21.33-12.12(s))*10^-2/(18.50-16.02(cm)*10^-1)
=3.714*10^-1
=0.3714cm^-1
(1axi)
(i) I ensured supports of pendula were rigid
(ii) I avoided parallax error on the meter rule
(1bi)
Simple harmonic motion is the motion of a body whose acceleration is always direct towards a fixed point and is proportional to the displacement from the fixed points
(1bii)
T=1.2secs
Log 1.2=0.079
0079 shown on graph with corresponding Log L read L correctly determined
==========================================
(3a)
TABULATE:
x(cm):10,20,30,40,50,60
V(v):0.65,0.75,1.00,1.20,1.45,1.55
I(A):0.20,0.30,0.35,0.40,0.45,0.55
LogV(v):-0.187,-0.125,0.000,0.079,0.161,0.190
LogI(ohm):-0.699,-0.523,-0.456,-0.396,-0.391,-0.360
SLOPE(s):=( LogI2-LOgI1)/(LogV2-LogV1)
=-0.2-(-0.7)/0.3-(-0.2)
=0.5/0.5
=1AV^-1
(3axi)
(i) I ensured neat and tight terminals
(ii) I opened the key when reading was taken
(3bi)
(i)The brightness of the bulb increases
(ii) The voltage and current through the bulb increases
(3bii)
(i) diode
(ii) transistor

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