ANSWER QUESTION 1 AND 3 ONLY
^ means raise to power. i : means
that under i , u will get 1 , 2 , 3 , 4 , 5 . .
=============================PHYSICS
PRACTICAL======
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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)ensured clean/tight terminals
ii)open key when reading was taken
3bi)
i)brightness of the bulb increases
ii)voltage/current through the bulb increases
3bii)
i)diode
ii)transistor
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(1)
Tabulate
i= L(cm)| |t1(s)| |t1(s)| |mean t(s)| |t(t/10)s| |logt
(s)| |log Lcm| |Log T*10^-2|(s)| |Log
L*10^-2(cm)
1= |80.00| |17.87| |17.56| |17.715|
|17.715| |0.2483| |1.9031| |24.83|
|19.031|
2= |7000 1622 1646 16340 16340 02133 18457
2133 18451
3= |6000 1609 1509 15590 15590 01928 17782
1928 17782
4= |5000 1408 1477 14425 14425 01591 16990
5= |4000 1328 1336 13320 13320 01212
1591 16990 1212 16021
slope (s) Log T2*10^-2/Log T2*10^-1 - Log
T2*10^-2/Log T2*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)ensured supports of pendula were rigged
ii)avoided parallax error on meter rule/stop
watch
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
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Lorem ipsumLoremANSWER QUESTION 1 AND 3 ONLY^ means raise to power. i : meansthat under i , u will get 1 , 2 , 3 , 4 , 5 . .=============================PHYSICSPRACTICAL======
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::3a)TABULATEx(cm):10,20,30,40,50,60V(v):0.65,0.75,1.00,1.20,1.45,1.55I(A):0.20,0.30,0.35,0.40,0.45,0.55LogV(v):-0.187,-0.125,0.000,0.079,0.161,0.190LogI(ohm):-0.699,-0.523,-0.456,-0.396,-0.391,-0.360SLOPE(s):= ( LogI2-LOgI1)/(LogV2-LogV1)=-0.2-(-0.7)/0.3-(-0.2)=0.5/0.5=1AV^-13axi)i)ensured clean/tight terminalsii)open key when reading was taken3bi)i)brightness of the bulb increasesii)voltage/current through the bulb increases3bii)i)diodeii)transistor:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::(1)Tabulatei= L(cm)| |t1(s)| |t1(s)| |mean t(s)| |t(t/10)s| |logt(s)| |log Lcm| |Log T*10^-2|(s)| |LogL*10^-2(cm)1= |80.00| |17.87| |17.56| |17.715||17.715| |0.2483| |1.9031| |24.83||19.031|2= |7000 1622 1646 16340 16340 02133 184572133 184513= |6000 1609 1509 15590 15590 01928 177821928 177824= |5000 1408 1477 14425 14425 01591 169905= |4000 1328 1336 13320 13320 012121591 16990 1212 16021slope (s) Log T2*10^-2/Log T2*10^-1 - LogT2*10^-2/Log T2*10^-1 =(21.33-12.12(s))*10^-2/(18.50-16.02(cm)*10^-1)=3.714*10^-1=0.3714cm^-11axi)i)ensured supports of pendula were riggedii)avoided parallax error on meter rule/stopwatch1bi)simple harmonic motion is the motion of a bodywhose acceleration is always direct towards afixed point and is proportional to thedisplacement from the fixed points1bii)T=1.2secsLog 1.2=0.0790079 shown on graph with corresponding Log Lread L correctly determined:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
