A cell has an emf of 1.5 V and an internal resistance of 0.65 Q. The cell is connected to a resistor R.
[0[8].[4] State what is meant by an emf of 1.5 V.
[2 marks]
I
[0] 8]-[2] The current in the circuit is 0.31 A.
Show that the total power output of the cell is approximately 0.47 W.
[1 mark]
[o[s).[3] Calcutate the energy dissipated per second in resistor R.
[2 marks]
energy dissipated per second = Ist
[o[si.[4] The cell stores 14 kJ of energy when it is fully charged. The cell's emf and internal resistance are constant as the cell f discharged.
Calculate the maximum time during which the fully-charged cell can deliver energy to
resistor R.
[2 marks)
maximum time = 5

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg

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