# data[].attributes.body
ACBDEF

# data[].attributes.body
A, C, B, D, E, F

# data[].attributes.body
A,C,B,D,E,F

# data[].attributes.body
P=IV= 24X3.4=79.2w

E=PT=79.2X10=792j

# data[].attributes.body
E=Pt
=79.2x10
=792J

# data[].attributes.body
E=Pt
=79.2(10)
=792J

# data[].attributes.body
P=IV
=24x3.3
=79.2W

# data[].attributes.body
P=VI
=24(3.3)
=79.2W

# data[].attributes.body
A C B D E F

# data[].attributes.body
E = Pt
= (79.2)(10)
792J

# data[].attributes.body
v=f(landa)
f=3x10^8/1.2x10^-6
=2.5x10^14 Hz

# data[].attributes.body
P=VI
= (24)(3.3)
= 79.2W

# data[].attributes.body
infrared radiation

# data[].attributes.body
f=(3)(10^8)/91.2(10^-6)
=2.5(10^14) Hz

# data[].attributes.body
infra red

# data[].attributes.body
1.2x10^-6 m

# data[].attributes.body
v=f(landa)
(3.0)(10^8) = f(1.20)(10^-6)
f = (2.5)(10^14) Hz

# data[].attributes.body
infrared

# data[].attributes.body
1.2(10^-6) m

# data[].attributes.body
1.2(10^-6) m

# data[].attributes.body
P1V1=P2V2

# data[].attributes.body
infrared

# data[].attributes.body
As the switch is open, the effective resistance increases. The e.m.f does not change. So when V=IR, as the V is constant and R increases, I flowing through A2 decreases

# data[].attributes.body
V=FrequencyxWavelength
3x10⁸=Fx91.2x10⁻⁶
=2.5x10¹⁴Hz

# data[].attributes.body
I= 0.5x3
=1.5A

# data[].attributes.body
1.2x10-6m

# data[].attributes.body
As the switch is opened, if forms an open circuit an thus current cannot flow through ammeter 1. since the reading of ammeter 1 is now 0A, the current flowing through ammeter 2 decreases as the total current of an parallel circuit is divided equally among each circuit.

# data[].attributes.body
The effective resistance increases while the voltage stays constant. This causes the total cuurent , and thus the current flowing through A2 to decrease..

# data[].attributes.body
when the switch is opened, effective resistance increases and the emf remains constant. Thus using the formula V=RI, the current will decrease resulting in the amount of current flowing through A2 to decrease.

# data[].attributes.body
Q=It
t=10/0.5
=20s

# data[].attributes.body
10/1/2=20s

# data[].attributes.body
A,C,BD,E,F

# data[].attributes.body
E=IVt
=79.2x10
=792J

# data[].attributes.body
P=IV
=24x3.3
=79.2W

# data[].attributes.body
0.5(3)=1.5A

# data[].attributes.body
0.5(3)
1.5A

# data[].attributes.body
V=f入
f=(3x10^8)/91.2x10^-6)
=2.5x10^14Hz

# data[].attributes.body
Q=It
10=0.5t
t=20s

# data[].attributes.body
infra red

# data[].attributes.body
1.2x10^-6m

# data[].attributes.body
There is a vaccum in the tyre??

# data[].attributes.body
Q=It
t=10/0.5
=20s

# data[].attributes.body
there are space for the air to go into the tyre, thus, less effort is needed for the student to push the pump, thus, less force is needed

# data[].attributes.body
there is trapped air in the tyre which helps to make the piston move a larger distance

# data[].attributes.body
There was vacuum in the pump

# data[].attributes.body
the effective resistance increases. Hence since V=RI, with the same emf, the current would decrease hence current flowing A2 decreases

# data[].attributes.body
0.5x3=1.5A

# data[].attributes.body
P=F/A
380 000 = F/(6.1)(10^-4)
F = 232N (3sf)

# data[].attributes.body
Q=It
t=10/0.5
=20s

# data[].attributes.body
p=F/a
F=pa
=(380x10^3)(6.1x10^-4)
=232N (3sf)

# data[].attributes.body
P=F/A
(380x1000)=F/(6.1x10^-4)
F=232N

# data[].attributes.body
f=pa
=(380000)(6.1(10^-4))
=232N(3sf)

# data[].attributes.body
it was a slight vacuum in the pump

# data[].attributes.body
let force = s
W=sd
2720000 - (0.5)(85)(47^2) = s(3200)

s = 820.66N (5sf)
= 821N (3sf)

# data[].attributes.body
ke=1/2x85x47^2
=93882.5
F=W/d
=(2.72x10^6-93882.5)/3200
=820N (3sf)

# data[].attributes.body
(380x1000)x(6.1x106-40
=232N

# data[].attributes.body
F=W/d
=2720000-(0.5x85x47^2)/3200
=820N

3d)

# data[].attributes.body
mgh=85x10x3200
= 2.72x10^6J

# data[].attributes.body
85x10x3200=2720000

1/2x85x47^2=93882.5

F=W/d
=2720000-93882.5/3200
=820N

# data[].attributes.body
GPE = mgh
= (85)(10)(3200)
= 2720000J
thus, decrease in GPE = 2720000-0
= 2720000

# data[].attributes.body
Ep=mgh
=85(10)(3200)
=2720000J

# data[].attributes.body
47x(2.8x60)=7896
h=7896xsin24=3200

# data[].attributes.body
W=Fd
F=W/d
=2720000-(0.5x85x47^2)/3200
=820N

# data[].attributes.body
d= (speed)(time)
= (47)(2.8)(60)
=7896
7896(sin24) = 3211.6m (5sf)
= 3200m (3sf) (shown)

# data[].attributes.body
mgh=3200(10)(85)
=2720000J

# data[].attributes.body
constant velocity means there is no resultant force acting on the object, thus, the rate of change of displacement is constant

# data[].attributes.body
2.8min=168seconds
47(168)=7896
ab=7896
h=(sin 24)(7896)
=3200m

# data[].attributes.body
dist = area under graph
= (0.5)(2)(9)
= 9m

# data[].attributes.body
AB= 47x(2.8x60)
=7896
h= 7896xsin24
=3200m (shown)

# data[].attributes.body
constant velocity means that there is no acceleration and the rate of change in displacement is constant.

# data[].attributes.body
The speed and the direction of the man does not change

# data[].attributes.body
d=st
=47x168
=7896m
7896xsin24=3200m(shown)

# data[].attributes.body
the graph shows a constant increasing velocity as the gradient is increasing constantly. Thus, this shows that the acceleration is constant

# data[].attributes.body
constant velocity means that there is no acceleration

# data[].attributes.body
constant rate of change of displacement

# data[].attributes.body
0.5x2x9=9m

# data[].attributes.body
0.5(2)(9)=9m

# data[].attributes.body
Dist travelled=1/2x9.0x2.0
= 9.0m

# data[].attributes.body
The gradient of the speed time graph, which is equal to the acceleration, constantly increases

# data[].attributes.body
.5x9x2=9m

# data[].attributes.body
constant increasing gradient

# data[].attributes.body
The gradient of the speed time graph show a constant increase for the first 2s so the acceleration is uniform

# data[].attributes.body
The gradient of the velocity time graph is the acceleration of the athlete and the graph shows a constant gradient which means a constant acceleration.