Physics practical answers

1(a)
V(cm)=10.0,15.0,20.0,25.0,30.0,35.0
U(cm)=47.0,48.5,50.0,51.5,53.0,54.5 respectively
Precautions
i I avoided drought to prevent swinging of the mass Q
ii I avoided parallax error in reading the metre rule
iii I avoided that the mass did not rest on the table

(1) b(i)  Moment  of a force about a point is the product of the force sF and the perpendicular distance from the point to the line of force.
Example opening  of door, turning of tap,riding of bicycle.


1 b(ii) (a) The principle of moment states that if a body is in equilibrium the the sum of the clockwise turning moments acting upon it about any point equals the sum of the anticlockwise turning moment about the same point.
(b) The term equilibrium meabs that when a body acted on by several forces and it does not accelerate or rotate, it is said to be at equilibrium.
2bi)Heat Capacity: is the amount of heat required to change its temperature by one degree, and has units of energy per
degre

4)
Ohm ’ s Law .
The current flow through a
metallic conductor
is directly proportional to
the potential
difference provided that
temperature and
other physical conditions
are constant.
(4b)
Factors that determine
resistance of
conduction .
PL
===
A
i – Nature Of the materials .
ii – Length of the conducting
wire.
iii – Cross setional area .
iv – Temperature .
How Temperature affects
Conduction.
=======================
Answer => How
Temperature affects
Conduction Is that The more
the temperature
will be the resistance . (If
neccessary) .
(4c)
Since the lost volt have
developed across the
cell when the supplied volt
across the
resistance onces will be
affected

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=0/2
I1=4.8/2=2.4A
I2=7.4/2=3.7A
I3=10.2/2=5.1A
I4=14.9/2=7.45A
I5=23.2/2=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
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
(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
=0.3/-0.09)
=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=0.48/0*(0.001)^2
R=46/1.855
=6.37*10^2ohm
P=I^2R
=10^2*6.37*10^2
=6.37*10^4W